Physics 251b Lecture Notes, spring 2011. Section 1 through 4 are primarily based on

the content of Gottfried & Yan, Quantum Mechanics: Fundamentals. The exposition of

section 5 is based on Faddeev’s lectures on Bethe ansatz and papers of Zamolodchikov’s.

The exposition of section 6 is based on the appendix of Polchinski’s string theory volume

I, and Coleman’s Aspects of Symmetry, chapter 7.

1 The atom

1.1 The hydrogen atom revisited

We begin by reviewing the spectrum of bound states of the hydrogen atom, and then

explain the role of a hidden symmetry of the system. Recall the Schrodinger equation

for the wave function ψ of an energy eigenstate[− ~2

2µ∇2 + V (r)

]ψ = Eψ (1.1)

of a charged particle in a Coulomb field with potential

V (r) = −Ze2

4πr(1.2)

Here −e is the electron charge, +Ze the charge of the nucleus, and µ the effective

reduced mass in the center of mass frame:

µ =mM

m+M(1.3)

where m is the electron mass and M the nucleus mass. The two body problem will be

treated in the center of mass frame as the system of a single particle of reduced mass

µ moving in a fixed potential V (r). In spherical coordinates (r, θ, φ) we can write

∇2 =1

r2

∂

∂rr2 ∂

∂r− L2

r2(1.4)

where the angular momentum square L2 = ~L2 is given by

L2 = − 1

sin θ

∂

∂θsin θ

∂

∂θ− 1

sin2 θ

∂2

∂φ2(1.5)

The spatial components of the angular momentum are

L3 = −i ∂∂φ,

L± ≡ L1 ± iL2 = ±e±iφ(∂

∂θ± i cot θ

∂

∂φ

).

(1.6)

1

Eigenfunctions of (~L2, L3) are the spherical harmonics Y`m(θ, φ), with ` = 0, 1, 2, · · ·and m = −`,−` + 1, · · · , `. Writing ψ = R(r)Y`m(θ, φ), the Schrodinger equation

reduces to the radial equation[~2

2µ

(− 1

r2

∂

∂rr2 ∂

∂r+`(`+ 1)

r2

)+ V (r)− E

]R(r) = 0, (1.7)

Substituting

E = −~2β2

2µ, V (r) =

~2

2µU(r), R(r) =

1

ru(r), (1.8)

the equation simplifies a bit to[d2

dr2− `(`+ 1)

r2− U(r)− β2

]u(r) = 0. (1.9)

where

U(r) = −2µ

~2

Ze2

4πr= − 2

ar(1.10)

Here a ≡ 4π~2/µZe2 is the characteristic length scale of the problem. After a further

rescaling r → ar, β → κ/a, we have[d2

dr2− `(`+ 1)

r2+

2

r− κ2

]u(r) = 0, (1.11)

whose solutions are expressed in terms of the confluent hypergeometric function

u(r) ∼ r`+1e−κr1F1

(`+ 1− 1

κ; 2`+ 2; 2κr

)(1.12)

Here 1F1(a; c; z) obeys the equation[z∂2

z + (c− z)∂z − a]

1F1(a; c; z) = 0. (1.13)

and has the power series expansion

1F1(a; c; z) =∞∑n=0

(a)nn!(c)n

zn (1.14)

where (a)n = a(a+ 1) · · · (a+ n− 1). The bound states are described by normalizable

wave functions, which occur when the series terminates, corresponding to values of

1/κ:1

κ≡ n = `+ 1, `+ 2, · · · (1.15)

The full radial wave function is then

Rn`(r) = Cn`r`e−r/n1F1(`+ 1− n; 2`+ 2; 2r/n) (1.16)

2

the energy being

E = − ~2

2µa2

1

n2= − 1

2n2

Ze2

4πa= − 1

n2

Z2µ

mRy (1.17)

where Ry = me4/32π2~2 = 13.606 eV is the Rydberg unit of energy.

At each energy level n of the Coulomb problem, there are

n−1∑`=0

(2`+ 1) = n2 (1.18)

degenerate states. Rotational invariance symmetry [H, ~L] = 0 implies that the energy

eigenstates labelled by the total angular momentum ` have degeneracy 2` + 1. The

extra degeneracy is due to a new hidden symmetry - the Lenz vector. Working in

atomic units, we can write the Hamiltonian as

H =1

2p2 − 1

r(1.19)

We write ~p, ~r as operators, and ~L = ~r× ~p. The Lenz vector

~M =1

2~p× ~L− 1

2~L× ~p− r (1.20)

is then conserved, namely

[H, ~M] = 0. (1.21)

Indeed,

[H, ~M] = −[

1

2p2, r

]−[

1

r,1

2~p× ~L− 1

2~L× ~p

]= −1

2[p2,

~r

r]− 1

2[1

r, ~p]× ~L +

1

2~L× [

1

r, ~p]

= − 1

2r[p2,~r]− 1

2[p2,

1

r]~r +

i

2

~r

r3× ~L− i

2~L×

~r

r3

=i

r~p− i

2(~p ·

~r

r3+~r

r3· ~p)~r +

i

2

~r

r3× ~L− i

2~L×

~r

r3= 0.

(1.22)

It is straightforward to check that ~M obeys the commutation relations

[Mi,Mj] = −2iεijkHLk, [Li,Mj] = iεijkMk (1.23)

and~L · ~M = 0, M2 = 1 + 2H(L2 + 1). (1.24)

The system is therefore integrable and the Hamiltonian H can be diagonalized simply

using the algebra of ( ~M, ~L). Working in the eigenspace of energy −E (E being the

binding energy), we have

[Mi,Mj] = 2iεijkELk (1.25)

3

Define the rescaled operator

Ni =Mi√2E

, (1.26)

then we have

[Ni, Nj] = iεijkLk, E =1

2(L2 +N2 + 1). (1.27)

where N2 ≡ ~N2. Now, observe that ~F+ and ~F− defined by

~F± ≡ 1

2(~L± ~N) (1.28)

obey two independent set of commutation relations for the angular momentum, and

commute with each other. The eigenvalues for ((F+)2, F+3 , (F

−)2, F−3 ) are

(f+(f+ + 1), µ+, f−(f− + 1), µ−) (1.29)

The constraint ~L · ~N implies

(~F+)2 − (~F−)2 = 0 (1.30)

and hence f+ = f− ≡ f = 0, 12, 1, · · · . We then have

L2 +N2 = 2(F+)2 + 2(F−)2 = 4f(f + 1) (1.31)

and so

E =1

2(2f + 1)2(1.32)

1.2 Fine structure

The degeneracy among energy eigenstates of different values angular momentum ` is

a consequence of the special symmetry of the Coulomb potential and does not follow

from rotational symmetry. A generic perturbation to the Hamiltonian that preserves

rotational symmetry will break this degeneracy. Such effect indeed occurs in the hy-

drogen atom due to relativistic corrections. The leading order corrections are due to

three effects: the relativistic correction to the energy-momentum dispersion relation,

the spin of the electron and spin-orbit interaction, and the so called Darwin term which

has no classical analog. All of these can be understood in terms of the Dirac equation

which describes the relativistic electron.

Note that with the relativistic corrections to the Hamiltonian, the Lenz vector is

no longer conserved. Classically, the Lenz vector points along the axis of the elliptical

orbit. A small non-conservation of the Lenz vector implies precession of the elliptical

orbit. The fine structure is a quantum version of this phenomenon.

4

A fully consistent treatment of the relativistic quantum theory requires second

quantization, which we shall defer to later discussion. For now we will work with the

ordinary (1 particle) wave function that obeys the Dirac equation (in ~ = c = 1 units)

γµPµΨD = mΨD. (1.33)

where ΨD is a four-component wave function and the Gamma matrices γµ are written

in a particular basis as

γ0 = γ0 =

(1 0

0 −1

), γi = −γi =

(0 σi−σi 0

)(1.34)

The defining property of the Gamma matrices is that they satisfy the Lorentzian Clif-

ford algebra

γµ, γν = 2ηµν (1.35)

with ηµν the Lorentzian metric of (+,−,−,−) signature. The Dirac equation is par-

tially motivated by trying to come up with a relativistic equation for the wave function

that is first order in time (just like Schrodinger equation), hence by Lorentz covariance

must be first order in spatial coordinates as well. In the absence of electromagnetic field,

the Dirac equation simply implies that each component of ΨD obeys P 2 = m2, which

is just the Klein-Gordon equation (for the two independent components of ΨD). It is

only when coupled to background electromagnetic field that the Dirac equation starts

to exhibit different dynamics for the spin-12

particle than the Klein-Gordon equation

which describes a spin-0 particle.

Note that in our convention ~P = (P 1, P 2, P 3) = (−P1,−P2,−P3). And so γµPµ =

γ0P 0 − ~γ · ~P . For the free electron, (P 0, ~P ) = (i∂t,−i~∇) when acting on the wave

function, as in the nonrelativistic convention.

To see that the Dirac equation is indeed Lorentz covariant (and independent of

choice of basis for the Gamma matrices), consider a Lorentz transformation

xµ → x′µ = Λµνx

ν (1.36)

where the matrix Λµν obeys

ΛTηΛ = η. (1.37)

In other words, Λ is an element of the Lorentz group SO(1, 3). If the wave function

transforms as

Ψ′D(x′) = S(Λ)ΨD(x) (1.38)

where the matrix S(Λ) is such that

S(Λ)γµS(Λ)−1 = Λµνγ

ν , (1.39)

5

then the Dirac equation transforms covariantly. Indeed such S(Λ) can be constructed -

this is the Dirac spinor representation of SO(1, 3). It suffices to exhibit the infinitesimal

form of S(Λ) (indices are raised and lowered using ηµν and ηµν):

Λµν = ηµν + ωµν +O(ω2), ωµν = −ωνµ,

S(Λ) = I − 1

8ωµν [γ

µ, γν ] +O(ω2).(1.40)

To recover the finite form of S(Λ) is left as an exercise.

Now writing

ΨD =

(Ψ

Φ

), (1.41)

where (Ψ, Φ) are two-component wave functions, the Dirac equation splits into com-

ponents(P0 −m)Ψ− ~σ · ~PΦ = 0,

(P0 +m)Φ− ~σ · ~PΨ = 0.(1.42)

In the non-relativistic limit, it is appropriate to work with energy relative to m. Define

Ψ = e−imtΨ, Φ = e−imtΦ, (1.43)

then Dirac equation becomes

P0Ψ− ~σ · ~PΦ = 0,

P0Φ− ~σ · ~PΨ = −2mΦ.(1.44)

In the non-relativistic limit, it is natural to solve Φ in terms of Ψ, and Ψ will behavior

analogously to the Shrodinger wave function. To go from Ψ to the Schrodinger wave

function ψ, one must note that |Ψ|2 is not the probably density - it cannot be because

it isn’t conserved. The conserved four-current under Dirac equation is

jµ = ΨDγµΨD, ΨD ≡ Ψ†Dγ

0, (1.45)

Check: using ㆵ = γ0γµγ0, we have

∂µjµ = ΨDγ

µ(←−∂ µ +

−→∂ µ)ΨD = (γµ∂µΨD)†γ0ΨD + ΨDγ

µ∂µΨD

= −(imΨD)†γ0ΨD −ΨDimΨD = 0.(1.46)

In particular, the charge density is

ρ = j0 = |Ψ|2 + |Φ|2 (1.47)

6

In the relativistic limit, |P0|, |~P | m, we have

ρ ' |Ψ|2 +1

4m2|~σ · ~PΨ|2 (1.48)

Define Λ ≡ ~σ · ~P. To obtain the Schrodinger wave function with conserved probability,

we need to identify

ψ '(

1 +Λ2

8m2

)Ψ (1.49)

where Φ is related by

Φ '(

1− P0

2m

)Λ

2mΨ (1.50)

The non-relastivistic approximation of Dirac equation is then obtained as[P0 − Λ2

2m+

ΛP0Λ

4m2

]Ψ = 0 (1.51)

or in terms of the Schrodinger wave function ψ up to order O(Λ4/m3)[(P0 − Λ2

2m

)(1− Λ2

8m2

)+

ΛP0Λ

4m2

]ψ = 0 (1.52)

or (P0 − Λ2

2m

)ψ = δHψ (1.53)

where

δH =P0Λ2

8m2− Λ4

16m3− ΛP0Λ

4m2(1.54)

When coupled to electric-magnetic field, the momenta of the electron are related to

canonical momenta via (working in ~ = 1, c = 1 units)

P0 = i∂t − eV, ~P = −i~∇− e ~A (1.55)

And so the static Schrodinger equation takes the form(Λ2

2m+ eV + δH

)ψ = Eψ (1.56)

The non-relativistic part of the Hamiltonian

H0 =Λ2

2m+ eV (1.57)

contains the coupling of the electron magnetic moment to the magnetic field. This is

seen fromΛ2 = (~σ · ~P)2 = P 2 + iεijkσkPiPj

= P 2 − e~σ · ~B(1.58)

7

From this we read off the magnetic moment of the electron

~µ = − e

2m~σ. (1.59)

The relativistic correction to the Hamiltonian, under the assumption of static field, is

δH = −Λ2P0

8m2+

[P0,Λ2]

8m2− Λ4

16m3− Λ[P0,Λ]

4m2

= − Λ4

8m3+

e

4m2

(Λ[V,Λ]− 1

2[V,Λ2]

) (1.60)

The commutators are evaluated as

[V,Λ] = [V, ~σ · ~P] = i~σ · ∇V = −i~σ · ~E,

Λ[V,Λ]− 1

2[V,Λ2] =

1

2[Λ, [V,Λ]]

= − i2

[~σ · ~P, ~σ · ~E]

= − i2

([σiPi, σj]Ej − σj[σiPi, Ej])

= − i2

(2iεijkσkPiEj − (δij − iεijkσk)[Pi, Ej])

= ~σ · (~P× ~E)− 1

2~∇ · ~E +

i

2~σ · (~∇× ~E)

= −1

2~∇ · ~E− ~σ · (~E× ~P)

(1.61)

In a static Coulomb potential, we have

~∇ · ~E = ρsource, ~E× ~P = −1

r

dV

dr~r× ~p = −1

r

dV

dr~L. (1.62)

Hence the leading correction to the non-relativistic Hamiltonian is

δH = − p4

8m3− Ze2

8m2δ3(~r) +

e

4m2

1

r

dV

dr~σ · ~L

= − p4

8m3− Ze2

8m2δ3(~r) +

Ze2

16πm2r3~σ · ~L

(1.63)

The first term is due to the correction to the kinematic relation, the second term being

the Darwin term (with the nucleus treated as a point charge), and the third term the

spin-orbit coupling. The spin of the electron is ~s = 12~σ. Restoring ~ and c, we have

δH = − p4

8m3c2− Ze2~2

8m2c2δ3(~r) +

Ze2~2

8πm2c2r3~s · ~L (1.64)

whereas the nonrelativistic Hamiltonian is

H0 =p2

2m− Ze2

4πr(1.65)

8

The energy is of order(Ze

2

4π)2m

~2= (Zα)2mc2 (1.66)

where

α =e2

4π~c' 1

137.04(1.67)

is the fine structure constant. The leading relativistic correction is suppressed by a

factor (assuming that Z is not too big)

E

mc2∼ (Zα)2 (1.68)

At this order in perturbation theory, the mixing between states of different n quantum

numbers does not contribute to the energy eigenvalues (they would contribute at order

(Zα)4). At order (Zα)2, we only need to diagonalize δH on the subspace of n2 degen-

erate states with orbital angular momentum ` = 0, 1, · · · , n − 1. In terms of rescaled

variables

r =a0

Zr, p =

~Za0

p, E = (Zα)2mc2E (1.69)

where a0 = 4π~2/(me2) is the Bohr radius, we can write

δH = (Zα)2

[−1

8p4 − π

2δ3(r) +

1

2r3~s · ~L

](1.70)

While ~L no longer commutes with the Hamiltonian, the total angular momentum

~J = ~L +~s (1.71)

does. The new Hamiltonian is diagonalized in the basis |`jm〉 with L2 = `(` + 1),

J2 = j(j + 1) and J3 = m. It is conventional to label the states |n`jm〉 by n`j, and

use the letters s, p, d, f, g, h, · · · for ` = 0, 1, 2, · · · . Note that j = ` ± 12

is half integer

valued here.

On the state |`jm〉,

~s · ~L =J2 − L2 − s2

2=j(j + 1)− `(`+ 1)− 3

4

2=

`2, j = `+ 1

2

− `+12, j = `− 1

2

(1.72)

It remains to calculate the expectation value of p4, δ3(r), and r−3 in the state labeled

by |n`〉 (independently of j,m). We have described their wave function Rn`(r) earlier.

The results are

〈r−1〉n` =1

n2, 〈r−2〉 =

1

n3(`+ 12), 〈r−3〉 =

1

n3`(`+ 12)(`+ 1)

(1.73)

9

and

〈δ3(~r)〉 = |ψn`(0)|2 =δ`0πn3

(1.74)

The expectation value of p4 can be computed by replacing p2 → 2(E+ 1r) and then ap-

plying the above formulae. Collecting all these, the result for the order (Zα)2 corrected

energy eigenlevel is

Efs(n`j) = En

[1 +

(Zα)2

n2

(n

j + 12

− 3

4

)](1.75)

While the degeneracy in j is lifted, the resulting spectrum is still degenerate in `,

remarkably. For instance, the n = 2 levels in the hydrogen atom consist of 2s 12, 2p 1

2,

and 2p 32. The former two are split from the latter by

∆E =E2α

2

4' 0.453× 10−4eV = 1.095× 104MHz (1.76)

Relativistic quantum mechanics as described by Dirac equation would predict that 2s 12

and 2p 12

are degenerate. However, a small splitting of 1057MHz is observed between the

two levels. This is known as the Lamb shift. The effect is due to the vacuum fluctuation

of the electromagnetic field and virtual electron-position pairs. We will return to

this when we systematically quantize the electromagnetic and electron/position fields

(quantum electrodynamics).

There is also the hyperfine structure due to the interaction between the nucleus

and the electron magnetic dipole and electric quadrupole moments, which depend on

the nuclear state. We will not discuss it here.

1.3 Dirac equation in Coulomb field

While the Dirac equation correctly captures O(α2) corrections to the energy spec-

trum of the hydrogen atom, at higher order in α one must take into account vacuum

fluctuation and virtual pair creation/annihilation, which requires second quantization.

However, such effects are suppressed by powers of α rather than Zα. In the limit of

very small α but with Zα finite (say for a nucleus of large Z), the spectrum of the

atom would be accurately captured by the full solution to the Dirac equation. It turns

out that the Dirac equation in Coulomb field can be solved exactly. In particular, we

will see that the degeneracy among states with different ` quantum number but the

same j is not lifted.

The Dirac equation describing an electron in a Coulomb field is

[γµ(−i∂µ − eAµ) +m] Ψ = 0. (1.77)

10

We will take A0 = −V , ~A = 0, Ψ = e−iEtΨ(~r), and so the equation can be written as[eV (r) +

(m −i~σ · ∇

−i~σ · ∇ −m

)]Ψ(~r) = EΨ(~r) (1.78)

Write

Ψ =

(χ

η

), (1.79)

and then(E −m− eV )χ = −i~σ · ∇η,(E +m− eV )η = −i~σ · ∇χ.

(1.80)

Now define χ± = χ± η,

(E − eV + i~σ · ∇)χ+ = mχ−,

(E − eV − i~σ · ∇)χ− = mχ+.(1.81)

Substituting the second equation into the first one, we have

(E − eV − i~σ · ∇)(E − eV + i~σ · ∇)χ+ = m2χ+ (1.82)

or [∇2 + (E − eV )2 + ie~σ · ∇V −m2

]χ+ = 0. (1.83)

For the Coulomb potential eV = −α/r, the equation is[∇2 + (E +

α

r)2 + iα

~σ · rr2−m2

]χ+

=

[1

r2∂rr

2∂r −~L2 − α2 − iα~σ · r

r2+

2Eα

r+ E2 −m2

]χ+ = 0.

(1.84)

This is in fact identical to the Schrodinger equation for the non-relativistic Coulomb

problem provided that we replace ~L2 by ~L2−α2−iα~σ · r, replace the Coulomb potential

by 2Eα/r, and replace the non-relativistic energy by E2 −m2. Note that ~L does not

commute with ~σ · r, but the total angular momentum ~J = ~L+ 12~σ does. By rewriting

~L2 = ~J2 +3

4− ~σ · ~J, (1.85)

we simply need to determine the eigenvalues of

~σ · ~J + iα~σ · r (1.86)

11

Observe that(~σ · ~J + iα~σ · r − 1

2

)2

= (~σ · ~J)2 − α2 +1

4− ~σ · ~J − iα~σ · r + iα~σ · ~J, ~σ · r

= (~σ · ~J)2 − α2 +1

4− ~σ · ~J

= (~σ · ~L)2 + 2~σ · ~L+ 1− α2

= ~L2 + ~σ · ~L+ 1− α2

= ~J2 +1

4− α2 = (j +

1

2)2 − α2 ≡ v2.

(1.87)

So we have

~L2 − α2 + iα~σ · r = ~J2 +1

4− α2 − (~σ · ~J − iα~σ · r − 1

2)

= v(v + 1)(1.88)

The spectrum can be obtained from that of the non-relativistic Coulomb problem with

n replaced by

n→ k + v (1.89)

where k = 1, 2, · · · . The energy spectrum is given by

E2 −m2 = − (Eα)2

(k + v)2(1.90)

or

E = m

1 +α2(

k +√

(j + 12)2 − α2

)2

− 1

2

(1.91)

In the weak potential limit α→ 0, we see that k should be identified as k = n−(j+ 12).

Expanding to order O(α2) reproduces our earlier result on the fine structure. Note

that the degeneracy among states with the same j is not lifted even with the full Dirac

equation taken into account.

1.4 The helium atom

The Hamiltonian for the helium atom in atomic units is

H =p2

1 + p22

2− 2

(1

r1

+1

r2

)+

1

r12

(1.92)

The wave function for two electrons, Ψ(~x1, s1; ~x2, s2), where si denotes the spin, obeys

Fermi-Dirac statistics

Ψ(~x1, s1; ~x2, s2) = −Ψ(~x2, s2; ~x1, s1). (1.93)

12

We shall ignore spin-orbit coupling etc. for now. By rotational invariance, the total

spin ~S and total orbital angular momentum ~L are conserved. We may write the energy

eigenstates as

Ψ = U(~r1, ~r2)χSMS(1.94)

where χSMSis the spin wave function. For two electrons, S = 0 or 1, in which case

χSMSis anti-symmetric or symmetric under the exchange of the two electrons. Cor-

respondingly, U(~r1, ~r2) is symmetric or anti-symmetric under the exchange of ~r1 with

~r2.

The energy eigenstates may be approximated by each electron occupying an energy

level of a hydrogen-like atom with screened charge. Let us begin with the 1s level. The

trial wave function for an electron in the Coulomb field of a screened charge is

u(r) =α

32

√πe−αr (1.95)

The trial wave function for the ground state of the helium atom is then

Ψ0 = u(r1)u(r2)χ0 (1.96)

The ground state energy and wave function are approximated determined by mini-

mizing the energy expectation value with respect to the screening parameter α. We

have ⟨p2

2− 2

r

⟩u

=

∫4πr2u(r)

[− 1

2r2∂rr

2∂r −2

r

]u(r)dr =

α2

2− 2α (1.97)

and

〈Ψ0|1

r12

|Ψ0〉 =α6

π2

∫d3~r1d

3~r2e−2α(r1+r2)

r12

= 8α

∫dr1r

21dθ sin θdr2r

22

e−2(r1+r2)√r2

1 + r22 − 2r1r2 cos θ

= 32α

∫ ∞0

dr1

∫ r1

0

dr2r1r22e−2(r1+r2) =

5

8α.

(1.98)

So

〈Ψ0|H|Ψ0〉 = 2

(α2

2− 2α

)+

5α

8= α2 − 27

8α (1.99)

It is minimized at

α =27

16(1.100)

This can be compared to the unscreened value α = 2. The resulting approximate

ground state energy is

E0 = −729

256' −2.8477 (1.101)

13

To restore the units, this result is multiplied by 2Ry. This result is 2% away from the

observed binding energy of the helium atom.

The first excited energy levels can be approximated by introducing the 2p trial wave

function with screened charge,

vm(~r) =β

52

2√

6re−βr/2Y1m(r) (1.102)

It is conventional to label the energy levels in terms of the total spin S, orbital angular

momentum L, and total momentum J as n2S+1LJ . We are now interested in 21P and

23P , with trial wave functions

Ψ(21P ;m) =1√2

[u(r1)vm(~r2) + u(r2)vm(~r1)]χ0,

Ψ(21P ;mMS) =1√2

[u(r1)vm(~r2)− u(r2)vm(~r1)]χ1Ms .(1.103)

The expectation value of the Hamiltonian is

〈H〉 = 〈H1〉u + 〈H2〉vm + I ± J (1.104)

where 〈H1〉u is the expectation value of a single electron in the Coulomb field describe

by the wave function u(r), 〈H2〉vm that of the wave function vm(~r), and I, J are the

contribution due to the electrostatic repulsion between the two electrons,

I =

∫d3~r1d

3~r2|u(r1)|2|vm(~r2)|2

r12

,

J =

∫d3~r1d

3~r2u(r1)∗vm(~r2)∗u(r2)vm(~r1)

r12

.

(1.105)

J is called the “exchange integral”, and purely due to the spin statistics of electrons.

The integration can be performed using the representation

1

r12

=∑`,m

4π

2`+ 1

r`<r`+1>

Y ∗`m(r1)Y`m(r2), (1.106)

where r>, r< are the larger and smaller ones of r1, r2 respectively. So we proceed and

calculate

I = 4π

∫dr1r

21

∫r2<r1

dr2r22

|u(r1)|2|v(r2)|2 + |u(r2)|2|v(r1)|2

r1

=αβ(8α4 + 20α3β + 20α2β2 + 10αβ3 + β4)

(2α + β)6,

J =8π

3

∫dr1r

21

∫r2<r1

dr2r22

r2

r21

u(r1)∗v(r2)∗u(r2)v(r1) =112

3

α3β5

(2α + β)7,

(1.107)

14

where v(r) is the radial part of vm(r), namely vm(r) = v(r)Y1m(r). The total energy

expectation value

〈H〉 =α2

2− 2α +

β2

8− β

2+ I ± J (1.108)

is minimized at

21P : α = 2.00302, β = 0.96473, 〈H〉 = −2.12239,

23P : α = 1.99119, β = 1.08915, 〈H〉 = −2.13069.(1.109)

The fact α ' 2, β ' 1 indicates that the electron in 1s has screened the nucleus and

so the electron in 2p feels only the Coulomb field due to charge e as opposed to 2e.

Note that the splitting between 21P and 23P , though small compared to the Rydberg

energy, is still much bigger than the fine structure splittings.

2 Scattering theory

2.1 Partial waves and cross section

To begin let us consider the scattering problem of a particle by an infinitely massive

spherically symmetric object (potential). In experiments, the incident particle is typ-

ically described by a wave packet moving at some velocity v. Such a wave packet

does not describe an energy eigenstate, as the wave function of the latter would have

the phase factor e−iEt and time-independent modulus. Nevertheless, it is useful to

decompose the wave packet in terms of energy eigenstates, of the asymptotic form

ψ ∼∫dkf(k)eikz−iEt + · · · , z 0 (2.1)

where z is the incident direction, with the source at z = 0. f(k) is the wave packet

profile in momentum space, taken to peak around k0, with v = (∂E/∂k)|k0 . In position

space, such a wave packet peaks around

z − vt = z0, (2.2)

where z0 is determined by the k-dependence in the phase of f(k). We would like to

determine the late time behavior of such a wave function. Asymptotically at large

distance r away from the source, the component of the wave function with some fixed

angular momentum ` behaves as

ψ(~r, t) ∼ P`(cos θ)

∫dkf(k)

[ϕ+(k)

eikr−iEt

r+ ϕ−(k)

e−ikr−iEt

r

](2.3)

15

(This is because the radial Schrodinger equation becomes (r−1∂2rr + · · · + k2)ψ = 0,

where · · · are small at large r.) The amplitudes of the incoming and outgoing waves,

ϕ− and ϕ+, depend on k in some way which is constrained by the details of the

interaction. If the momentum spread of the incident wave packet is small, then the

k-independence of ϕ± is negligible when multiplied by f(k), and consequently the

incoming and outgoing waves peak around r + vt = r0 and r − vt = r0. So at fixed

large r, at early times one sees the incoming wave which is the angular momentum `

component of the incident wave packet, and at late times one finds the outgoing wave

after the scattering process. It is convenient to study the scattering process by looking

at an energy eigenstate wave function ψ(~r, t) = e−iEtψ(~r), that is asymptotically plane

wave eikz−iEt at z 0, with ϕ− determined by the incident plane wave, while ϕ+

will be solved through the Schrodinger equation and describes the superposition of the

incident plane wave and the scattered waves at late times.

To begin, it is useful to decompose a plane wave, describing the incident particle,

into angular momentum eigenstates. Let (x, y, z) be spatial coordinates, and a radial

coordinate system related by z = r cos θ. Then we have

eikz =∞∑`=0

(2`+ 1)i`j`(kr)P`(cos θ) (2.4)

where P`(cos θ) =√

4π2`+1

Y`0(θ) is the Legendre polynomial in cos θ, and j` is the spher-

ical Bessel function, related to ordinary Bessel function by

j`(ρ) =

√π

2ρJ`+ 1

2(ρ). (2.5)

In the limits ρ→∞ and ρ→ 0, j` has the asymptotic behavior

j`(ρ) ∼ 1

ρsin(ρ− `π

2), ρ→∞,

j`(ρ) ∼ ρ`

(2`+ 1)!!, ρ→ 0.

(2.6)

If the interaction is rotationally invariant, then different partial waves do not mixes.

The asymptotic form of the scattering wave function is

ψ(~r) =∞∑`=0

(2`+ 1)i`R`(k; r)P`(cos θ) (2.7)

where R`(k; r) is an unknown function with asymptotic behavior

R`(k; r) ∼ 1

2ikr

[η`(k)ei(kr−

`π2

) − e−i(kr−`π2

)], ρ→∞ (2.8)

16

The term e−i(kr−`π2

) represents the incoming wave and η`(k)ei(kr−`π2

) the outgoing wave.

Probability conservation requires |η`(k)| ≤ 1 for general inelastic scattering. If the

scattering is elastic, then |η`(k)| = 1. We can express it in terms of the partial wave

phase shift

η`(k) = e2iδ`(k) (2.9)

The asymptotic radial function is then

R`(k, r) ∼ eiδ`sin(kr − `π

2+ δ`)

kr(2.10)

The elastically scattering wave ψsc is defined by

ψ(~r) = eikz + ψsc(~r) (2.11)

It then follows that asymptotic behavior of ψsc as r →∞ has the partial wave decom-

position

ψsc ∼eikr

2ikr

∞∑`=0

(2`+ 1)(η` − 1)P`(cos θ) (2.12)

Now the asymptotic behavior of the total wave function is

ψ(~r) ∼ eikz +eikr

rf(k, θ) (2.13)

where

f(k, θ) =1

2ik

∞∑`=0

(2`+ 1)(η` − 1)P`(cos θ) ≡∞∑`=0

(2`+ 1)f`(k)P`(cos θ) (2.14)

is the elastic scattering amplitude. In experiments, what one typically measures is the

number of particles scattered per unit time, given some incoming particle flux. Let

Finc be the incoming particle flux, i.e. the number of incoming particles per unit area

per unit time. Let Nsc be the number of scattering particles per unit time. Then

σ =Nsc

Finc(2.15)

is the total cross section. In other words, the number of scattering particles is equal to

the flux through an effective area σ.

Better, we can measure the number of particles scattering into a solid angle dΩ =

sin θdθdφ per unit time, dNsc. The differential cross section is defined as

dσ

dΩ=

1

Finc

dNsc

dΩ(2.16)

17

Let v be the incident velocity of the particles. With the wave function normalized so

that |ψ(~r)|2 = 1 for the incoming plane wave, v is also the same as the flux Finc. The

number of outgoing particles in the solid angle Ω is

dNsc = v limr→∞|ψsc(~r)|2r2dΩ = v|f(k, θ)|2dΩ (2.17)

So we see that the differential cross section is given by the modulus square of the

amplitude,dσ

dΩ= |f(k, θ)|2 (2.18)

Using ∫dΩP`(cos θ)P`′(cos θ) =

4π

2`+ 1δ``′ , (2.19)

in terms of partial wave amplitudes f`(k), the total elastic cross section is

σel =

∫dΩ|f(k, θ)|2

=π

k2

∞∑`=0

(2`+ 1)|η` − 1|2(2.20)

In case of elastic scattering, η` = e2iδ` , the `-th partial wave cross section is given by

σ` =4π

k2(2`+ 1) sin2 δ` (2.21)

In particular, we see that

σ =4π

kImf(k, θ = 0) (2.22)

Namely the total cross section is determined in terms of the imaginary part of the

forward (θ = 0) scattering amplitude. This is called the optical theorem. So far we

consider elastic scattering but in fact the optical theorem generalizes to the inelastic

case as well.

If the scattering is inelastic, |η`| < 1, the flux of the outgoing `-th partial wave

is down by the factor |η`|2. The difference between the ingoing and outgoing flux is

accounted for by the absorption of particles by the scattering source. The inelastic

part of the cross section is therefore

σinel =π

k2

∞∑`=0

(2`+ 1)(1− |η`|2) (2.23)

and so that total cross section

σtot = σel + σinel =π

k2

∞∑`=0

(2`+ 1)(1− |η`|2 + |1− η`|2)

=4π

kImfel(k, θ = 0)

(2.24)

18

So we see that generally, the optical theorem relates the total (elastic plus inelastic)

cross section to the elastic forward scattering amplitude.

A little bit of kinematics: typically one scatters a particle of mass m1 off a rest

particle of mass m2. The scattering can be treated as one particle scattering off a

fixed potential, in the center of mass frame, with reduced mass µ = m1m2/(m1 +m2)

and relative coordinates/momenta. But then, we must translate the differential cross

section in the center of mass frame back to the rest/laboratory frame. Let θ be the

deflection angle of particle 1 in the center of mass frame, and θ1, θ2 be the angles of

the final momenta of the two particles in the laboratory frame with respect to the

incidental momentum. It is the easiest to derive the relations in a relativistic set up.

Let the initial momentum four-vectors of the two particles in the center of mass frame

be

(E1 =√k2 +m2

1, k, 0, 0), (E2 =√k2 +m2

2,−k, 0, 0) (2.25)

and the final momenta in the center of mass frame be

(E1, k cos θ, k sin θ, 0), (E2,−k cos θ,−k sin θ, 0) (2.26)

Going to the laboratory frame, we perform the Lorentz boost along x direction,

(E, px, py, pz)→ (E − βpx√

1− β2,px − βE√

1− β2, py, pz) (2.27)

with β = −k/E2, 1/√

1− β2 = E2/m2. This results in the final momenta in the

laboratory frame

(E1E2 + k2 cos θ

m2

,E1 + E2 cos θ

m2

k, k sin θ, 0), (E2

2 − k2 cos θ

m2

,1− cos θ

m2

E2k,−k sin θ, 0)

(2.28)

So we have

tan θ1 =m2 sin θ

E1 + E2 cos θ, tan θ2 =

m2

E2

sin θ

1− cos θ=m2

E2

cotθ

2. (2.29)

Taking the non-relativistic limit Ei → mi, we find

tan θ1 =m2 sin θ

m1 +m2 cos θ, θ2 =

π − θ2

. (2.30)

So the solid angles in the two frames are related by

dΩlab

dΩcom

=sin θ1dθ1

sin θdθ=

m22(m2 +m1 cos θ)

(m21 +m2

2 + 2m1m2 cos θ)32

(2.31)

19

and hence the differential cross sections are related by

dσlabdΩlab

=dσcomdΩcom

(m21 +m2

2 + 2m1m2 cos θ)32

m22(m2 +m1 cos θ)

(2.32)

In particular, when m1 = m2, we have θ1 = θ/2,

dσlabdΩlab

(θ1) = 4 cos θ1dσcomdΩcom

(θ = 2θ1) (2.33)

2.2 Example: hard sphere scattering

To illustrate let us consider the example of elastic scattering by a hard sphere of radius

a. The wave function is that of a free particle for r > a and vanishes for r ≤ a. For

r ≥ a, it takes the form

ψ(~r) =∞∑`=0

(2`+ 1)i` [A`j`(kr) +B`n`(kr)]P`(cos θ) (2.34)

Here j`(ρ) and n`(ρ) the spherical Bessel and Neumann functions, with asymptotic

behavior at large ρ

j`(ρ)→sin(ρ− `π

2)

ρ, n`(ρ)→

cos(ρ− `π2

)

ρ, (ρ→∞) (2.35)

Note that while j`(ρ) are regular at ρ = 0, n`(ρ) are not.

j`(ρ)→ ρ`

(2`+ 1)!!, n`(ρ)→ −(2`− 1)!!ρ−`−1, (ρ→ 0) (2.36)

This is not relevant in the hard sphere scattering problem, since the wave function

takes the form (2.34) only outside the sphere of radius a. ψ(~r) has the asymptotic

behavior

ψ(~r)→ 1

2ikr

∞∑`=0

(2`+ 1)i`[α`e

i(kr− `π2

) − β`e−i(kr−`π2

)]P`(cos θ) (2.37)

with

α` = A` + iB`, β` = A` − iB`. (2.38)

We demand the incoming part of the wave function to be that of a plane wave eikz,

namely

β` = 1 for all `. (2.39)

20

α` are then determined by the continuity of the wave function, ψ(r = a) = 0,

α` = η` = e2iδ` =n`(ka)− ij`(ka)

n`(ka) + ij`(ka). (2.40)

In the two limits, large ka and small ka, we have

δ` ' −ka+`π

2, ka `,

δ` ' −(ka)2`+1

(2`+ 1)!!(2`− 1)!!, ka `.

(2.41)

In the special case of s-wave, we have

j0(ρ) =sin ρ

ρ, n`(ρ) = −cos ρ

ρ, (2.42)

and the simple answer for δ0,

δ0 = −ka. (2.43)

The total cross section is

σ =4π

k2

∑`

(2`+ 1) sin2 δ` (2.44)

In the low energy limit, ka 1, the scattering is dominated by s-wave. The total cross

section

σ ' 4π

k2δ2

0 = 4πa2 (2.45)

is equal to the entire surface area of the sphere. This is in the extreme quantum regime,

where we don’t expect to see the classical result. On the other hand, in the high energy

limit ka 1, all partial waves up to `max ∼ ka contribute. For ` > `max, the partial

wave cross sections are suppressed exponentially. The total cross section is then

σ ' 4π

k2

ka∑`=0

(2`+ 1)1

2= 2πa2. (2.46)

While we are expecting the high energy scattering to reduce to the classical result, it

seems puzzling that we are discovering twice the classical result for the scattering cross

section. The resolution to this puzzled is revealed by examining the differential cross

section in this limit. The classical reflection differential cross section is

dσcldΩ

=2πa2 cos(π−θ

2)d( θ

2) sin π−θ

2

2πdθ sin θ=a2

4. (2.47)

The quantum scattering differential cross section

dσ

dΩ=

1

4k2

∣∣∣∣∣∑`

(2`+ 1)(η` − 1)P`(cos θ)

∣∣∣∣∣2

=1

k2

∣∣∣∣∣∞∑`=0

(2`+ 1)eiδ` sin δ`P`(cos θ)

∣∣∣∣∣2 (2.48)

21

in the high energy limit ka 1 approaches a2/4 everywhere except near θ = 0, where

it approaches a δ-function with cross section πa2 concentrated at near θ = 0. In

particular, since the forward scattering amplitude obeys the optical theorem

Imf(k, θ = 0) =k

4πσtot, (2.49)

we can write

dσ

dΩ

∣∣∣∣θ=0

= |f(k, 0)|2 =k2σ2

tot

16π2+ (Ref(k, 0))2 ≥ k2σtot

16π2σtot (2.50)

which is always much greater than the classical θ = 0 differential cross section when

ka 1. Since σtot =∫

dσdΩdΩ, we know that the large differential cross section near

θ = 0 must be concentrated in a small solid angle ∆Ω ∼ 8π2

k2a2 .

This forward scattering interferes with the plane wave to form the shadow behind

the hard sphere (with a bright spot at the center of the shadow).

2.3 General elastic scattering amplitude

Let us now investigate the general problem of elastic scattering of a particle of energy

E off a potential V . The Schrodinger equation is written as

(E −H0)|k,+〉 = V |k,+〉 (2.51)

where |k,+〉 represents a state of incoming plane wave at momentum k and outgoing

scattering waves. Let |k〉 be the plane wave state in the absence of the potential,

namely

(E −H0)|k〉 = 0. (2.52)

Treating the potential as a perturbation, we want to find the solution by, naively

|k,+〉 =? |k〉+1

E −H0

V |k,+〉 (2.53)

The inverse of E −H0 is not defined, however, since H0 has a continuous spectrum in

which E lies. We can fix this problem by taking

|k,+〉 = |k〉+1

E −H0 + iεV |k,+〉 (2.54)

where ε is a small positive number that will be taken to approach zero. This is called

the Lippmann-Schwinger equation. The second term corresponds to the scattered

wave function. We will now see that it indeed describes outgoing waves as opposed to

incoming waves, due to the positive sign of ε.

22

The propagator (E−H0 +iε)−1 has momentum space representation (with E = k2

2m)

〈~q| 1

E −H0 + iε|~q′〉 = δ(~q − ~q′) 2m

k2 − q2 + iε(2.55)

and position space representation

〈~r| 1

E −H0 + iε|~r′〉 = 2m

∫d3~q

(2π)3

ei~q·(~r−~r′)

k2 − q2 + iε

=2m

(2π)2

∫ ∞0

dqq2

∫ π

0

dθ sin θeiqR cos θ

k2 − q2 + iε

=2m

(2π)2

∫ ∞0

dqq2 1

k2 − q2 + iε

eiqR − e−iqR

iqR

= − 2im

(2π)2

∫ ∞−∞

dqq

R

eiqR

k2 − q2 + iε

(2.56)

where R ≡ |~r − ~r′|. Now deforming the integration contour in q to R + i∞ picks up

the residue at q = k + iε,

〈~r| 1

E −H0 + iε|~r′〉 = −2m

eikR

4πR. (2.57)

In terms of wave functions, then, we have

ψ+k (~r) = ei

~k·~r − 1

4π

∫d3~r′

eik|~r′−~r|

|~r′ − ~r|2mV (~r′)ψ+

k (~r′) (2.58)

which can be solved recursively.

Assuming sufficiently fast fall off of the potential V (~r) at large r, the solution has

asymptotic form

ψ+~k

(~r) ∼ ei~k·~r +

eikr

rf(~k, r) (2.59)

where

f(~k, r) = −2m

4π

∫d3~r′e−ikr·~r

′V (~r′)ψ+

~k(~r′) = −4π2m〈kr|V |~k,+〉 (2.60)

is what we have previously defined as the scattering amplitude. What we see now is that

the scattering amplitude is given by the matrix element of the interaction Hamiltonian

between the incoming state and outgoing plane wave states.

We could also use an alternative regularization of the propagator, namely

|~k,−〉 = |~k〉+1

E −H0 − iεV |~k,−〉 (2.61)

23

This will result in a state in which some incoming waves are prepared so that they

scattering into the momentum eigenstate |~k〉. Using

(E −H0 ± iε)|~k,±〉 = (E −H0 ± iε)|~k〉+ V |~k,±〉 (2.62)

and H = H0 + V , we can write

|~k,±〉 =

(1 +

1

E −H ± iεV

)|~k〉

=

(1 +

1

E −H0 ± iεV +

1

E −H0 ± iεV

1

E −H0 ± iεV + · · ·

)|~k〉

(2.63)

If the system has time reversal symmetry, then we expect time reversal to relate |~k,+〉to | − ~k,−〉. This is indeed the case. The time reversal operator It has the defining

property

It(eiHt/~|ψ〉) = e−iHt/~It|ψ〉 (2.64)

and that it reverses the momentum. Such an operator must be anti-unitary, namely

given two states |a〉, |b〉, the time reversed states |aR〉 = It|a〉 and |bR〉 obey

〈a|b〉 = 〈bR|aR〉 (2.65)

Assuming V is time reversal invariant, we then have

It|~k,+〉 = | − ~k,−〉. (2.66)

If we define the operator T as

T = V + V1

E −H + iεV, (2.67)

then the scattering amplitude can be written as

〈~p|V |~k,+〉 = 〈~p|T |~k〉 = 〈~p,−|V |~k〉 (2.68)

which takes a form that is symmetric in the incoming and outgoing states. Note that

the incoming and outgoing are constraint to have the same energy E.

When V is time reversal invariant,

ItT†I−1t = T. (2.69)

It then follows that

〈−~p|T | − ~k〉 = 〈~k|ItT †I−1t |~p〉 = 〈~k|T |~p〉. (2.70)

24

Namely, if we exchanging the incoming and outgoing states and flip the signs of the

momenta at the same time, the amplitude remains invariant.

There is an alternative way to formulate scattering theory by starting with a plane

wave in the absence of interactions, and turn on the interaction potential adiabatically.

Consider a time dependent Hamiltonian

H(t) = H0 + f(t)V (2.71)

where f(t) is a slowly varying function with f(−∞) = 0 and f(∞) = 1. Writing the

wave function as e−iEtΨ(t), the Schrodinger equation reads

i∂tΨ(t) = (H0 − E + f(t)V )Ψ(t). (2.72)

The solution that asymptotes to plane wave ψ0 (with H0ψ0 = Eψ0) as t→ −∞ is

Ψ(t) = P exp

[i

∫ t

−∞dt(E −H0 − f(t)V )

]ψ0, (2.73)

where P stands for time/path ordering of the exponential. Expanding the exponential

in V , and keeping track of the time ordering of the operators, we have

Ψ(t) = ψ0 + (−i)∫ t

−∞dt1e

i(E−H0)(t−t1)f(t1)V ψ0

+ (−i)2

∫ t

−∞dt1

∫ t1

−∞dt2e

i(E−H0)(t−t1)f(t1)V ei(E−H0)(t1−t2)f(t2)V ψ0 + · · ·

(2.74)

It suffices to consider the case t = 0 and choose the adiabatic function f(t) to be 1 at

t = 0. For example, we can take f(t) = eεt for a small positive ε. Then we have

Ψ(0) = ψ0 + (−i)∫ 0

−∞dt1e

−i(E−H0+iε)t1V ψ0

+ (−i)2

∫ 0

−∞dt1

∫ t1

−∞dt2e

−i(E−H0+2iε)t1V ei(E−H0+iε)(t1−t2)V ψ0 + · · ·

= ψ0 +1

E −H0 + iεV ψ0 +

1

E −H0 + 2iεV

1

E −H0 + iεV ψ0 + · · ·

(2.75)

In the adiabatic limit, ε approaches zero, and we can replace (E − H0 + 2iε)−1 by

(E − H0 + iε)−1 in the second term, and so on and so forth. In this way we recover

the Lippmann-Schwinger equation for the scattering wave function. We see that the

iε prescription has the physical interpretation of adiabatic parameter.

25

2.4 Bound state poles

We have seen that the scattering amplitude can be represented as

〈~k′|T |~k〉 = 〈~k′|V + V1

E −H + iεV |~k〉 (2.76)

Suppose there is a bound state |ψ∗〉 of energy E∗ = k2∗

2m< 0, k∗ = iα∗ purely imaginary.

Then

〈~k′|T |~k〉 ∼ 〈~k′|V |ψ∗〉〈ψ∗|V |~k〉

E − E∗+ (regular at k = iα∗). (2.77)

It is a general feature of the scattering amplitude is that its poles at purely imaginary

values of momentum, i.e. negative energy, correspond to bound states at this energy.

We will see this in explicit examples later.

2.5 Born approximation

We have written the scattering amplitude as

f(~k′|~k) = −4π2m〈~k′|V |~k,+〉 (2.78)

with

|~k,+〉 = |~k〉+1

E −H0 + iεV |~k,+〉 (2.79)

If the potential is weak, treating V as a perturbation to the Hamiltonian, the leading

order result is the Born approximation:

fB(~k′|~k) = −4π2m〈~k′|V |~k〉 = −m2π

∫d3~rei(

~k−~k′)·~rV (~r) ≡ −m2πV (~q), ~q = ~k − ~k′.

(2.80)

Here V is the Fourier transformed potential in momentum space, ~q is the momentum

transfer. For spherical symmetric potential, V (~q) = V (q), q = 2k sin θ2. As an example,

consider the Yukawa potential

V (r) = V0e−αr

αr. (2.81)

Its Fourier transform is

V (q) = 4π

∫ ∞0

dr r2 sin qr

qrV (r) =

4πV0

α

1

q2 + α2(2.82)

The differential cross section in Born approximation is

dσBdΩ

= |fB(k, θ)|2 =

(2mV0

α

)21

(4k2 sin2 θ2

+ α2)2(2.83)

26

and the total cross section

σB(k) =

(2mV0

α2

)24π

4k2 + α2(2.84)

Note that since fB is real here, the optical theorem is violated.

The Coulomb potential

V (r) =Ze2

4πr(2.85)

can be obtained from the Yukawa potential by taking the limit α→ 0, V0 → αZe2/4π.

This results in the scattering amplitude in Born approximation,

fB(k, θ) = − mZe2

8πk2 sin2 θ2

(2.86)

The square of this expression gives the Rutherford cross section. In fact, the exact

scattering amplitude in the Coulomb potential turns out to be this expression mul-

tiplied by a phase, and the resulting differential cross section is the same as in Born

approximation, as we shall see later. Note that the total cross section is infinite, due

to the infinite range of the Coulomb field.

2.6 Eikonal approximation

The eikonal approxmation can be applied when the incidental wave length is short

compared to the size of the object it scatters off. It is based on the wave function

ansatz

ψ~k(~r) = eikzΦ(~r). (2.87)

with Φ(~r) varying slowly over distances of order 1/k. The Schrodinger equation[−∇

2

2m+ V (~r)

]ψ~k(~r) =

k2

2mψ~k(~r) (2.88)

is written in terms of Φ(~r) as[−ikm∂z + V (~r)

]Φ(~r) = −∇

2

2mΦ(~r). (2.89)

The slowing varying assumption on Φ(~r) would allow us to ignore the RHS, and work

with the equation [−ikm∂z + V (~r)

]Φ(~r) = 0. (2.90)

27

This is easily integrated to give

Φ(~b, z) = exp

[−imk

∫ z

−∞dz′V (~b, z′)

](2.91)

where the impact parameter ~b is a vector perpendicular to the incidental direction z. If

the potential V has a finite range, then Φ(~b, z) is simply 1 when ~b is outside the range,

and one would not see any scattering at all. This is too drastic an approximation for

our purpose. What we can do instead is to take this form of ψ~k(~r) and calculate the

scattering amplitude from the matrix element 〈~k′|V |ψ~k〉. The result is

f(~k|~k) = −m2π

∫d~bdzeikz(1−cos θ)−i~kT ·~bV (~b, z)Φ(~b, z) (2.92)

where ~kT is the transverse component of ~k′ to the incident direction. When the wave

length is short compared to the range of V , the differential cross section is sharply

peaked near the forward direction. We are thus interested in the θ → 0 limit. In

this limit, 1 − cos θ ∼ θ2/2, whreas |~kT | = k sin θ ∼ kθ. We shall ignore the term

ikz(1− cos θ) in the exponent and keep only −i~kT ·~b, giving the approximate result

f(~k′|~k) = −m2π

∫d~be−i

~kT ·~b∫dzV (~b, z)Φ(~b, z)

= − ik2π

∫d~be−i

~kT ·~b∫ ∞−∞

dz∂z exp

[−imk

∫ z

−∞dz′V (~b, z′)

]=

k

2πi

∫d~be−i

~kT ·~b(e2iδ(~b) − 1

) (2.93)

where the phase shift δ(~b) is

δ(~b) = −m2k

∫ ∞−∞

dzV (~b, z). (2.94)

Now consider a spherically symmetric potential, in which case δ(~b) = δ(b). We have

f(k, θ) =k

2πi

∫ ∞0

db b(e2iδ(b) − 1)

∫ 2π

0

dφe−ikT b cosφ

= −ik∫ ∞

0

db b(e2iδ(b) − 1)J0(kT b).

(2.95)

with kT = kθ in the eikonal approximation. Now the total cross section is

σ ' 2π

∫ ∞0

dθ θ|f(k, θ)|2 (2.96)

28

Using the closure equation∫ ∞0

dθ θJ0(kθb)J0(kθb′) =δ(b− b′)k2b

, (2.97)

we arrive at the result

σeik = 8π

∫ ∞0

db b sin2 δ(b). (2.98)

The result of eikonal approximation, in particular, satisfies the optical theorem.

2.7 Coulomb scattering

Now we will solve the problem of scattering in a Coulomb potential exactly. The

Schrodinger equation for two particles of charges Z1e, Z2e in the center of mass frame

is [−~2∇2

2µ+Z1Z2e

2

4πr

]ψ(~r) =

~2k2

2µψ(~r) (2.99)

or (∇2 + k2 +

2γk

r

)ψ(~r) = 0, (2.100)

where

γ = −Z1Z2e2µ

4π~2k= −Z1Z2α

c

v(2.101)

Here α ' 1/137 is the fine structure constant.

Due to the long range nature of the Coulomb potential, the incidental wave function

can not strictly take the plane wave form ei~k·~r. To understand this, consider a classical

particle of incident momentum ~k along z direction, which travels along a hyperbolic

orbit

r =r0tan2 θ0

1 + cos(θ−θ0)cos θ0

(2.102)

This is the orbit of a particle coming in at angle θ = π, and impact parameter b =

r0 tan θ0. Here the energy of the particle is proportional to 1/r0, and does not depend

on θ0. In fact, r0 = γ/k. The wave fronts are then a family of surfaces that are

orthogonal to the hyperbolic orbits. The orbits can also be expressed as

r + z + tan θ0x = r0 tan2 θ0. (2.103)

The tangent vector (dx, dz) to the orbit at (x, z) is given by

xdx+ zdz

r+ dz + tan θ0dx = 0 ⇒ (dx, dz) ∼ (

r + z

xr,−tan θ0

x− 1

r). (2.104)

29

When the incident particle was far away from the source, z → −∞, x ∼ b = r0 tan θ0.

And so the tangent vector to the orbit is approximately along

(r + z

xr,− 1

r0

− 1

r) = (

x

r

1

r − z,− 1

r0

− 1

r) (2.105)

The wavefront orthogonal to this vector is given by

z − r0 ln(r − z) = const. (2.106)

This suggests that the incident wave should take the form

∼ eikz−iγ ln(k(r−z)) (2.107)

We will see later that this is indeed the correct asymptotic behavior of the quantum

wave function.

In fact, the hidden symmetries of the Coulomb problem allows the scattering wave

function to factorize. This is achieved through the ansatz

ψ~k(~r) = ei~k·~rχ(u), u ≡ i(kr − ~k · ~r) = ik(r − z). (2.108)

The Schrodinger equation becomes[∇2 + 2i~k · ∇+

2γk

r

]χ(u)

=

[(∇2u)∂u + (∇u)2∂2

u + 2i~k · (∇u)∂u +2γk

r

]χ(u)

=

[2ik

r∂u − 2k(k − ~k · r)∂2

u − 2~k · (kr − ~k)∂u +2γk

r

]χ(u)

=

[2ik

r∂u +

2ik

ru∂2

u − 2iuk

r∂u +

2γk

r

]χ(u)

= 0.

(2.109)

Factoring out 2iku/r, we see that the ansatz indeed solves the Schodinger, provided(∂2u +

1− uu

∂u −iγ

u

)χ(u) = 0. (2.110)

We will think of this equation as for a function χ(u) defined on the complex plane,

holomorphic everywhere except at the singularities u = 0 and u = ∞. To see the

behavior near u = ∞, let us go to the coordinate w = 1/u, and write χ(w) = χ(u),

then the equation for χ becomes(∂2w +

1 + w

w2∂w −

iγ

w3

)χ(w) = 0. (2.111)

30

Consider a Laplace transform

χ(u) =

∫C

dt eutf(t) (2.112)

where C is a to-be-determined contour in the complex t-plane. The equation is now

written as ∫C

dt

(t2 +

1− uu

t− iγ

u

)eutf(t) = 0. (2.113)

Multiplying by u, we have∫C

dt[ut2 + (1− u)t− iγ

]eutf(t)

=

∫C

dtf(t)[t2∂t + t(1− ∂t)− iγ

]eut

= 0.

(2.114)

Integrating by part, f(t) must satisfy

∂t [t(t− 1)f(t)] + (iγ − t)f(t) = 0. (2.115)

In addition, we need eutt(t− 1)f(t) to vanish at the end points of the contour C. The

equation for f(t) is first order, and easily solved by

f(t) = Atiγ−1(1− t)−iγ, (2.116)

where A is a constant. And so χ is solved as

χ(u) = A

∫C

euttiγ−1(1− t)−iγdt. (2.117)

In the physical kinematic regime, u ∈ iR+. As long as the contour C ends at t→ +i∞,

the boundary terms of the integration by part vanish. The boundary condition for the

scattering wave function is encoded in the behavior of χ(u) at large imaginary u.

Change the integration variable t to s = ut, and write

χ(u) = A

∫C′ds essiγ−1(u− s)−iγ (2.118)

C ′ should end at s = −∞. The integrand has two branch points, s = 0 and s = u. To

get a nontrivial solution χ(u), we need to choose the branch cuts to go from these two

points to s = −∞, and let the contour C ′ to go around the branch cuts. The large u

asymptotics is dominated by the value of the integrand near the branch points, namely

∼ u−iγ and ∼ euuiγ−1. (2.119)

31

Recall that γ is the strength of the potential, and is small when the potential is weak.

These two terms are precisely the incident plane wave and the scattering wave.

Which contour do we pick for C ′? Note that u → i0+ corresponds to θ → 0, and

we want χ(u) to behave regularly in this limit. The two branch cuts as we have chosen

so far would collide as u → i0+, and so we need C ′ to go around both branch cuts in

this limit. For instance, we can choose C ′ to come in from −∞ along R− − iε, around

0, u, and go back to infinity along R− + u+ iε. The large u asymptotics is given by

χ(u) ∼ A

[u−iγ

∫C1

ds essiγ−1 + euuiγ−1

∫C1

ds es(−s)−iγ]

(2.120)

where C1 is the contour going around the branch cut R−. We have∫C1

ds essiγ−1 =(eπγ − e−πγ

)Γ(iγ) = − 2πi

Γ(1− iγ)(2.121)

and ∫C1

ds es(−s)−iγ =(−1 + e2πγ

)Γ(1− iγ) = −2πieπγ

Γ(iγ)(2.122)

In the second integral, (−s)−iγ is continuously deformed from (u−s)−iγ which uniquely

determine its branch. We have chosen to work in the branch arg(u) = π/2. The

asymptotic form of χ(u) is then determined to be

χ(u) ∼ −2πiA

[u−iγ

Γ(1− iγ)+ eu+πγ u

iγ−1

Γ(iγ)

]∼ −2πiA

u−iγ

Γ(1− iγ)

[1 +

eu

u|u|2iγΓ(1− iγ)

Γ(iγ)

] (2.123)

Recall that |u| = k(r − z). So the full wave function has the asymptotic form

ψ~k(~r) ∼ −2πiAeπγ/2

Γ(1− iγ)

[eikz−iγ ln(k(r−z)) +

eikr

r − zeiγ ln(k(r−z))γ

k

Γ(1− iγ)

Γ(1 + iγ)

](2.124)

The logarithmic correction to the incident wave function and scattered wave function

is due to the long range nature of the Coulomb potential. It is natural to take

eikz−iγ ln(k(r−z)) (2.125)

as the incident wave, andeikr

reiγ ln(2kr) (2.126)

to be the outgoing wave. We hence define the Coulomb scatterig amplitude fC(k, θ)

through the r →∞ asymptotic behavior

ψ~k(~r) ∼ eikz−iγ ln(k(r−z)) +eikr+iγ ln(2kr)

rfC(k, θ) (2.127)

32

with

fC(k, θ) =1

1− cos θ

γ

k

Γ(1− iγ)

Γ(1 + iγ)eiγ ln sin2 θ

2 (2.128)

Apart from the phase factor, |fC(k, θ)| coincides with the amplitude obtained through

Born approximation. In particular, the differential cross section is identical to the

Rutherford cross section obtained in Born approximation. Note however that the

phase factor is important in more complicated situations, such as Coulomb scattering

of identical particles.

Of particular interest is the probably of finding the particle at the location of the

source for the Coulomb potential, namely ~r = 0 (and so u = 0). We have

χ(0) = A

∫C′ds essiγ−1(−s)−iγ = Aeπγ

∫C′dses

s= 2πiAeπγ (2.129)

where in the last step we simply picked up the residue at s = 0. This should be

compared to

χinc(u) = −2πiAu−iγ

Γ(1− iγ)= −2πiAeπγ/2

e−iγ ln |u|

Γ(1− iγ)(2.130)

and so the probability density of finding the particle at the origin is

|χ(0)|2

|χinc|2= eπγ|Γ(1− iγ)|2 =

2πγ

1− e−2πγ(2.131)

When the force is repulsive, γ < 0, and we have

|χ(0)|2

|χinc|2=

2π|γ|e2π|γ| − 1

(2.132)

Recall that |γ| = Z1Z2αc/v. So at low energies, the probability of finding the particle

at the origin is exponentially suppressed. This is particularly important for nuclear

reactions, where two nuclei scatter and the nuclear reaction occurs only when one

nucleus tunnels through the Coulomb potential and collide with the other nucleus.

Let us examine the poles of the Coulomb scattering amplitude at imaginary values

of γ, corresponding to (negative energy) bound state resonances. They occur at iγ =

n = 1, 2, 3, · · · . Recall that γ = −Z1Z2αc/v, and so the poles are located at

E =1

2mv2 =

m

2

(Z1Z2αc

γ

)2

= − 1

2n2(Z1Z2α)2mc2.

(2.133)

This is indeed the energy of the bound states of a single electron atom that we were

familiar with.

33

2.8 Scattering of particles with spin

Now let us consider the generalization to particle with internal degrees of freedom,

the simplest example being particles with nonzero spins. Consider the scattering of

two particles of spins s1 and s2. The spin states are labeled by |ν1〉 and |ν2〉. There

can be some nontrivial interaction involving the spins, which is taken as given in the

framework of nonrelativistic quatum mechanics. We write the full Hamiltonian as

H = H0 +V , where V has nontrivial matrix elements between states of different spins.

The scattering amplitude is still given by the matrix elements of the T operator,

f(~kf , νf |~ki, νi) = −4π2m〈~kf , νf |T |~ki, νi〉 (2.134)

with

T = V + V1

E −H + iεV (2.135)

The scattering wave function is then

Ψ = ei~ki·~r|νi〉+

eikr

r

∑νf

f(~kf , νf |~ki, νi)|νf〉, (2.136)

and the differential cross section for finding the final state in |νf〉 in a solid angle is

dσi→fdΩ

= |f(~kf , νf |~ki, νi)|2 (2.137)

If the interaction is parity or time reversal symmetric, the amplitude is then constrained

by the symmetry. Under parity (spatial reflection), the initial and final momenta flip

signs whereas the spins stay the same. Under time reversal, both momenta and spins

flip signs, and the initial and final states are exchanged.

A lot can be learned about scattering of particles with spins based on symmetries,

without knowing the details of the spin couplings. Let us consider the simplest non-

trivial example of the scattering of a spin 1/2 particle off a spin 0 target. Write the

scattering amplitude as a scattering matrix/operator acting on spin states,

〈νf |M |νi〉 = f(~kf , νf |~ki, νi) (2.138)

Rotational invariance implies that M must be a linear combination of the identity

matrix and Pauli matrices contracted with vectors constructed out of initial and final

momenta of the scattering particle. The most general such matrix is

M = g1 + g2~σ · (~ki × ~kf ) + g3~σ · (~ki + ~kf ) + g4~σ · (~ki − ~kf ) (2.139)

34

If we further impose parity symmetry, under which ~ki, ~kf flips sign and ~σ stays invariant,

then g3 and g4 must vanish. (Time reversal symmetry on the other hand would only

require g4 to vanish.) In this case, the scattering matrix reduces to the form

M = g(k, θ) + h(k, θ)~σ · n (2.140)

where n = ~ki × ~kf/|~ki × ~kf | is the unit normal vector to the scattering plane. Such

a scattering matrix can be generated by spin-orbit coupling. Consider an interaction

Hamiltonian of the form

V = V0(r) + ~σ · ~LV1(r). (2.141)

Under Born approximation, we have the scattering matrix

MB = −4π2m[〈kf |V0(r)|ki〉+ ~σ · 〈kf |V1(r)~L|ki〉

](2.142)

Let us compute the matrix element

〈kf |V1(r)~L|ki〉 = 〈kf |V1(r)~r|ki〉 × ~ki

=1

(2π)3

∫d3~rV1(r)ei(

~ki−~kf )·~r~r × ~ki

= − i

(2π)3∇~qV (q)× ~ki =

i

(2π)3

1

q∂qV (q)~kf × ~ki,

(2.143)

and so indeed the spin-orbit coupling gives rise to the term in the scattering matrix

proportional to ~σ · n. If we do not measure the polarization of the scattered particles

and sum over all final spins, the scattering cross section for polarized incident particle

isdσ

dΩ= 〈νi|M †M |νi〉

= 〈νi|(g∗ + h∗~σ · n)(g + h~σ · n)|νi〉= |g|2 + |h|2 + 2Re(g∗h)n · 〈~σ〉i

(2.144)

If the incident beam is unpolarized, the cross section is then evaluated by averaging

over the incident spins (described by a density matrix),

dσ

dΩ=

1

2Tr(M †M) = |g|2 + |h|2 (2.145)

In particular, this differential cross section is independent of the angle of the scattering

plane. Conversely, if the incident beam is polarized, we can measure its spin polar-

ization by measuring the dependence of the differential cross section on the azimuthal

angle.

35

Finally, consider the dependence of the polarization of the scattering particle on

the azimuthal angle. Assuming the incident beam is unpolarized, the polarization of

the scattered beam is given by

〈~σ〉sc =Tr(M †~σM)

Tr(M †M)= ~n

2Re(g∗h)

|g|2 + |h|2. (2.146)

2.9 Scattering of identical bosons

Let us begin by considering the scattering of two identical spin 0 bosonic particles.

The two-particle wave function ψ(~r1, ~r2, t) = ψ(~r2, ~r1, t) is written in center-of-mass

coordinates ~r = ~r1 − ~r2 and ~R = 12(~r1 + ~r2) as

Ψ(~r, ~R, t) = Ψ(−~r, ~R, t) (2.147)

In the center of mass frame, the scattering process can be formulated by considering

two incident particles with opposite momentum along z-axis,

ψ ∼ eikz1−ikz2−2iEkt + eikz2−ikz1−2iEkt

√2

or Ψ ∼ eikz−2iEkt + e−ikz−2iEkt

√2

(2.148)

By the linearity of Schrodinger equation, the full scattering wave function is a su-

perposition of the one-body scattering wave function and the one related by spatial

reflection,

Ψ =ψ(~r, t) + ψ(−~r, t)√

2= Ψinc + Ψsc (2.149)

with

Ψsc =f(k, θ) + f(k, π − θ)√

2

eikr−2iEkt

r(2.150)

While the scattered particle flux is given by

dNsc

dΩ=v

2|f(k, θ) + f(k, π − θ)|2, (2.151)

the incoming particle flux is halved compared to the one-body case, i.e. Finc = v/2.

The resulting differential cross section is then

dσ

dΩ= |f(k, θ) + f(k, π − θ)|2

= |f(k, θ)|2 + |f(k, π − θ)|2 + 2Ref ∗(k, θ)f(k, π − θ)(2.152)

The last term is due to the interference between the two particles, which has no classical

analog. Note that the total cross section is half of the integration of the differential

cross section over the 4π solid angle, because there are two scattering particles.

36

As an example, recall that in Coulomb scattering we found that the one-body

scattering amplitude is the same as the result of Born approximation (Rutherford

cross section) up to a phase that depends on the scattering angle. This phase does not

affect the cross section if the two particles are not identical, but does affect the result

if the two particles are identical. The differential cross section for Coulomb scattering

between two identical bosons is

dσ

dΩ=

(Z2e2

4π

)21

16E2

∣∣∣∣∣eiγ ln sin2 θ2

sin2 θ2

+eiγ ln cos2 θ

2

cos2 θ2

∣∣∣∣∣2

=

(Z2e2

4π

)21

16E2

[1

sin4 θ2

+1

cos4 θ2

+8 cos(γ ln tan2 θ

2)

sin2 θ

] (2.153)

Recall that

γ =Z2e2

4π~v(2.154)

with ~ dependence restored. The interference term is quantum and oscillates very fast

in the small ~ limit.

2.10 Scattering of identical fermions

Now let us turn to the scattering of two identical spin 1/2 fermionic particles. The

incident particles can be in either the singlet (S = 0) or triplet (S = 1) spin states.

In these two cases, the spatial wave function is then symmetric and anti-symmetric,

respectively. The former involves only the even partial waves, whereas the latter in-

volves only the odd partial waves. If there is no spin-orbit coupling and the interaction

is reflection symmetric, which we assume for now, then the singlet states will remain

in singlet and triplet states will remain in the triplet after the scattering, i.e. there is

no mixing between singlet and triplet states. The one-body scattering matrix would

take the form

M(kf , ki) = Ms(kf , ki) +Mt(kf , ki) (2.155)

For two identical fermions, we have

M(kf , ki) = Ms(kf , ki) +Ms(−kf , ki) +Mt(kf , ki)−Mt(−kf , ki) (2.156)

This allows us to compute the differential cross section with various polarizations for the

incident and scattering particles. Note that even if the interaction is spin independent,

the scattered state can be polarized.

37

We will illustrate with the simplest case, where the incident beam is unpolarized

and summing over all scattering polarization. Then

dσ

dΩ=

1

4TrM †M

=1

4|fs(k, θ) + fs(k, π − θ)|2 +

3

4|ft(k, θ)− ft(k, π − θ)|2

(2.157)

If the interaction is spin independent, fs = ft = f , we have

dσ

dΩ= |f(k, θ)|2 + |f(k, π − θ)|2 − Ref ∗(k, θ)f(k, π − θ). (2.158)

Note in particular that at θ = π/2, the interference term gives a negative contribution

to the differential cross section, in contrast to the bosonic case where the interference

contribution is always positive at θ = π/2.

2.11 The S-matrix and optical theorem revisited

We now give a general treatment of scattering theory that applies to inelastic scattering

of multiple particles. The key is to describe the evolution of asymptotic states of

multiple particles that are free when far away from each other and scatter through

some finite range interaction. This is most conveniently achieved by working with

plane wave states (energy eigenstates of the free Hamiltonian) and turning on and off

the interaction adiabatically. Let |Φi(t)〉 be the state describing a wave packet of the

projectile and some state of the target, far away from each other at early time t 0

and not interacting. It obeys the free Schrodinger equation

i∂t|Φi(t)〉 = H0|Φi(t)〉. (2.159)

The scattering state at some time t is obtained by acting on |Φi(t)〉 with the evolution

operator

|Ψ+i (t)〉 = U(t,−∞)|Φi(−∞)〉, U(t, t′) = P exp

[i

∫ t

t′H(t′′)dt′′

], (2.160)

where f(t′′) is a function ranging from 0 to 1 that adiabatically turns on the interaction

at an early time. Similarly, at late time, consider an asymptotic state |Φf (t)〉 which

also obeys the free Schrodinger equation

i∂t|Φf (t)〉 = H0|Φf (t)〉, (2.161)

and the associated scattering state

|Ψ−f (t)〉 = U(+∞, t)−1|Φf (+∞)〉. (2.162)

38

Here U involves a similar function f(t) that turns off the interaction adiabatically at

late times. The S-matrix is defined as

Sfi = 〈Ψ−f (0)|Ψ+i (0)〉. (2.163)

The choice of time t = 0 in computing the overlap is of course arbitrary, and the

result does not depend on this choice sice Ψ+i and Ψ−f evolve by the same Schrodinger

equation at finite time t.

For H = H0 + V with V time-independent, we have seen before that the adiabatic

scattering state (at time t = 0) has the more explicit expression

|Ψ+i 〉 =

(1 +

1

Ei −H + iεV

)|Φi〉,

|Ψ−f 〉 =

(1 +

1

Ef −H − iεV

)|Φf〉.

(2.164)

The S-matrix can be written as

Sfi = 〈Ψ+f |Ψ

+i 〉+ (〈Ψ−f | − 〈Ψ

+f |)|Ψ

+i 〉

= δfi + 〈Φf |V(

1

Ef −H + iε− 1

Ef −H − iε

)|Ψ+

i 〉

= δfi + 〈Φf |V(

1

Ef − Ei + iε− 1

Ef − Ei − iε

)|Ψ+

i 〉

= δfi − 2πi δ(Ef − Ei)Tfi

(2.165)

where the T -matrix is defined as before

Tfi = 〈Φf |V |Ψ+i 〉 = 〈Ψ−f |V |Φi〉 (2.166)

The S-matrix is unitary by construction. Now using the expression in terms of the

T -matrix, SS† = I is written

δfi =∑a

(δfa − 2πiδ(Ef − Ea)Tfa)(δai + 2πiδ(Ei − Ea)T †ai) (2.167)

which immediately implies

Tfi − T ∗if + 2πi∑a

δ(Ea − Ei)TfaT ∗ia = 0 when Ef = Ei. (2.168)

For forward scattering, f = i, and we have

ImTii = −π∑a

δ(Ea − Ei)|Tia|2. (2.169)

39

Let us now relate the T -matrix elements to the transition rate and scattering cross

section. Splitting H = H0 + V , let us try to relate U(t, t′) to U0(t, t′), the evolution

operator for the free Hamiltonian H0. We have treated this before in perturbation

theory. Here we present a slightly different and more general derivation. Once again

introduce the adiabatic function f(t). Begin with

∂t′′ [U0(t, t′′)U(t′′, t′)] = iU0(t, t′′)H0U(t′′, t′)− iU0(t, t′′)(H0 + f(t′′)V )U(t′′, t′)

= −if(t′′)U0(t, t′′)V U(t′′, t′)(2.170)

Integrate this expression in t′′ from t′ to t gives

U(t, t′)− U0(t, t′) = −i∫ t

t′dt′′f(t′′)U0(t, t′′)V U(t′′, t′) (2.171)

Acting this on |Φi(t′)〉 and taking t′ → −∞, we obtain

|Ψ+i (t)〉 = |Φi(t)〉 − i

∫ t

−∞dt′′f(t′′)U0(t, t′′)V |Ψ+

i (t′′)〉 (2.172)

This is just a slightly different form of Lippman-Schwinger equation. For a final state

|Φf (t)〉 orthogonal to |Φi(t)〉, the transition amplitude is

Ai→f (t) = 〈Φf (t)|Ψ+i (t)〉 = −i

∫ t

−∞dt′f(t′)〈Φf (t)|U0(t, t′)V |Ψ+

i (t′)〉

= −i∫ t

−∞dt′ei(Ef−Ei)t

′f(t′)〈Φf |V |Ψ+

i 〉

= −i∫ t

−∞dt′ei(Ef−Ei)t

′f(t′)Tfi.

(2.173)

In particular, the transition amplitude at late time when the interaction is turned off

by f(t), is given by

Ai→f = −i∫ ∞−∞

dtei(Ef−Ei)tf(t)Tfi. (2.174)

The transition probability is

Pi→f = |Ai→f |2 =

∣∣∣∣∫ ∞−∞

dtei(Ef−Ei)tf(t)

∣∣∣∣2 |Tfi|2→ T · 2πδ(Ef − Ei)|Tfi|2,

(2.175)

in the limit where f(t) takes the value 1 in a long time period T and goes to zero

adiabatically (but over a time period very small compared to T ) before and after. This

results in the transition rate

Γi→f = 2πδ(Ef − Ei)|Tfi|2. (2.176)

40

This is the transition rate of the initial state |Φi〉 into one particular final asymptotic

state |Φf〉. In practice, we are interested in the transition rate into a range of final

states, say some volume in the space of the momenta of escaped particles. Let ρf (E) be

the density of asymptotic states in energy E that obey certain criteria (e.g. momentum

range, scattering angle) so as to be detected. Namely, the number of states in the energy

interval (E,E + dE) that obey such criteria is ρf (E)dE. Then the transition rate of

the initial state into such final states is

dΓi→f = 2π|Tfi|2ρf (E), (2.177)

where E = Ei = Ef . If the initial state is an incident particle beam with flux Finc,

then the differential cross section is

dσ =dΓi→fFinc

= 2π|Tfi|2ρf (E)

Finc(2.178)

Suppose the final state consists of N + 1 escaping particles; N of them have momenta

~ps, s = 1, · · · , N , and the remaining one particle has its momentum constrained by

momentum conservation. Let us regularize the problem by putting the space in a large

box of size L. The number of states in the momentum range d3~ps is then

N∏s=1

L3

(2π)3d3~ps. (2.179)

In such situations we typically define the final states |f〉 to be delta function nor-

malizable, and so the factor L3/(2π)3 for each escaping particle is absorbed into the

normalization of |f〉. The density of final states is then

ρf =

∏Ns=1 d

3~psdEf

(2.180)

Similarly, if normalize the incident particle state this way, the flux is given by

Finc =vi

(2π)3. (2.181)

Thus we obtain the formula for the cross section of a general scattering process,

dσ = (2π)4|Tfi|2∏N

s=1 d3~ps

vidEf(2.182)

For example, if the final state is a two-particle state, of momenta ~kf and ~ki − ~kf , then

ρf =d3~kfdEf

=k2f

vfdΩf (2.183)

41

and the angular differential cross section is

dσ

dΩf

= (2π)4|Tfi|2k2f

vivf= (2π)4m2|Tfi|2

kfki

(2.184)

For elastic scattering kf = ki, this reduces to our familiar result derived from the

scattering wave function.

Now we see that the total cross section is

σtot =

∑f Γi→f

Finc=

(2π)4

vi

∑f

δ(Ef − Ei)|Tfi|2, (2.185)

and the optical theorem immediately follows from the unitarity of the S-matrix

ImTii = − vi16π3

σtot. (2.186)

2.12 A one-dimensional example

Let us consider the scattering of two particles of identical mass m moving in one

dimension, interacting through a delta function potential V (x1, x2) = V0δ(x1 − x2).

First suppose the two particles are distinct. The wave function Ψ(x1, x2) obeys[−∂2x1

2m−∂2x2

2m+ V0δ(x12)

]Ψ(x1, x2) = EΨ(x1, x2) (2.187)

Away from x1 = x2, Ψ(x1, x2) obeys the free Schrodinger equation. Going from x1 =

x2 − ε to x1 = x2 + ε, integrating the equation gives the matching condition

∂x1Ψ(x1, x2)|x1=x2+ε − ∂x1Ψ(x1, x2)|x1=x2−ε = 2mV0Ψ(x2, x2). (2.188)

If we take the incident particles to have momenta p1 and p2 respectively, then by

momentum and energy conservation, after the scattering the two particles either have

the same momenta or have their momenta exchanged. We can take the stationary

scattering wave function to be

Ψ+(x1, x2) = eip1x1+ip2x2 +Reip2x1+ip1x2 , x1 < x2,

Ψ+(x1, x2) = Teip1x1+ip2x2 , x1 > x2.(2.189)

The matching condition gives

1 +R = T,

ip1T − ip1 − ip2R = 2mV0T,(2.190)

42

and thus

R = − iγ/p12

1 + iγ/p12

, T =1

1 + iγ/p12

, (2.191)

where we defined γ ≡ 2mV0. This gives the transmission and reflection amplitudes.

Similarly, the reversed scattering wave function is

Ψ−(x1, x2) = T ∗eip1x1+ip2x2 , x1 < x2,

Ψ−(x1, x2) = eip1x1+ip2x2 +R∗eip2x1+ip1x2 , x1 > x2.(2.192)

The S-matrix acting on the basis of asymptotic states (|p1, p2〉, |p2, p1〉) is given by

S =

(T R

R T

)(2.193)

Indeed this is a unitary matrix.

Now let us consider the case of two identical bosons. Since Ψ(x1, x2) is symmetric

under exchange of x1 with x2, the scattering wave function takes the form

Ψ(x1, x2) = eip1x1+ip2x2 + Seip2x1+ip1x2 , x1 < x2,

Ψ(x1, x2) = eip1x2+ip2x1 + Seip2x2+ip1x1 , x1 > x2.(2.194)

The matching condition gives

ip2 + ip1S − ip1 − ip2S = γ(1 + S), (2.195)

and we can solve

S =1− iγ/p12

1 + iγ/p12

. (2.196)

In this case, there is only one asymptotic state of given energy and total momentum,

|p1, p2〉, and the S-matrix element on this state is a phase, as expected from unitarity.

This scattering wave function can be easily generalized to the case of n identical

bosons. Begin with

Ψ(x1, x2, · · · , xn) =∑σ∈Sn

S(σ) exp

(i

n∑i=1

pσ(i)xi

), x1 < x2 < · · · < xn, (2.197)

where the sum is over all n! permutations on x1, x2, · · · , xn; each σ can be obtained

by composing successive flipping of nearby neighbors (i, i + 1), and the phase S(σ) is

defined as the product of all phases

S(pi, pi+1) =1− iγ/pi,i+1

1 + iγ/pi,i+1

= S(pi+1, pi)−1. (2.198)

43

associated with such flips. We then completely symmetrize the wave function with

respect to x1, x2, · · · , xn.

This result has important applications. For instance, suppose we put n such iden-

tical bosons in a one dimensional box of periodic boundary condition (i.e. on a circle),

and ask what are is the energy spectrum of the system, say with zero total momentum.

The wave function is periodic in each of the xi’s. Let L be the size of the box, and so

we identify xi ∼ xi + L. This periodicity condition amounts to demanding

Ψ(x1, x2, · · · , xn) = Ψ(x2, · · · , xn, x1 + L), x1 < x2 < · · · < xn. (2.199)

Successively moving x1 through x2, · · · , xn results in a phase which must be equal

to e−ip1L and cancel the phase in the plane wave, so that the wave function is periodic

in x1. Namely,

eip1L

n∏i=2

S(p1, pi) = 1. (2.200)

This must also hold if take xi instead of x1 around the circle, which gives the equations

that constrain the momenta pi,

eipiLn∏j 6=i

S(pi, pj) = 1, i = 1, · · · , n (2.201)

These give n−1 independent equations (the product of all of the n equations is trivial).

They are known as Bethe equations or Bethe ansatz. Together with the condition on

total momentum∑n

i=1 pi = 0, this allows us to solve for the set pi, and obtain the

energy spectrum of the system

E =n∑i=1

p2i

2m. (2.202)

We will leave it to the reader to explore this further.

2.13 S-matrix and resonance in elastic scattering

Let us briefly revisit the partial wave expansion in the case of elastic scattering, where

the asymptotic scattering wave function for the `-th partial wave takes the form (with

delta function normalization for the plane wave)

ψ+k,`(~r) ∼

1

(2π)32

2`+ 1

2ik

[e2iδ`(k) e

ikr

r− (−)`

e−ikr

r

]P`(cos θ). (2.203)

The reversed scattering wave function, on the other hand, is

ψ−k,`(~r) ∼1

(2π)32

2`+ 1

2ik

[eikr

r− (−)`e−2iδ`(k) e

−ikr

r

]P`(cos θ). (2.204)

44

The S-matrix element on the partial wave basis, which is given by the overlap

between |ψ+k,`〉 and |ψ−k′,`′〉, is a delta function in the radial momenta multiplied by the

integration in the asymptotic region (the overlap between the incoming and outgoing

waves do not contribute to the delta function)

Sk′`′,k` = 〈ψ−k′,`′|ψ+k,`〉 =

4π(2`+ 1)δ``′

(2π)34k2e2iδ`(k)

∫ ∞0

r2drei(k−k

′)r + e−i(k−k′)r

r2

= (2`+ 1)δ``′δ(k − k′)

4πk2e2iδ`(k)

= (2`+ 1)δ``′δ(k − k′)

4πk2− 2πiδ(E − E ′)T`′,`(k)

(2.205)

where the T -matrix elements are

T`′,`(k) =iv

8π2k2(2`+ 1)δ``′

[e2iδ`(k) − 1

](2.206)

By optical theorem, the imaginary part of T`,`(k) is indeed equal to −v/16π3 times the

`-th partial wave cross section

σ` =4π

k2(2`+ 1) sin2 δ`. (2.207)

The `-th partial wave S-matrix element is given by

S` = e2iδ`(k). (2.208)

The unitarity of the S-matrix implies in this case that S` is a phase.

In elastic scattering, the potential may allow the existence of metastable bound

states. These will lead to poles in the S-matrix elements at complex energy E = E∗−iΓ,

where E∗ is the energy of the bound state and Γ its decay width. In terms of the

momentum k, the pole is off the positive real k-axis with a negative imaginary part.

This is in contrast to the positive imaginary momentum pole due to bound states.

Let us explore this phenomenon in the example of a potential shell V (r):

V (r) = V0 δ(r − a), (2.209)

with a > 0. We will restrict our attention to the s partial wave. The s-wave scattering

wave function takes the form

ψk(r) = A0j0(kr) +B0n0(kr) =S0(k)eikr − e−ikr

2ikr, r > a,

ψk(r) = ξ(k)j0(kr) = ξ(k)sin(kr)

kr, r < a.

(2.210)

45

where ξ(k) is a complex coefficient that depends on k. It follows from the continuity

of the wave function that

2iξ(k)sin(ka) = S0(k)eika − e−ika. (2.211)

Integrating the radial equation for the s-wave[− 1

2m

1

r∂2rr + V0δ(r − a)− k2

2m

]ψk(r) = 0 (2.212)

from r = a− ε to r = a+ ε gives the matching condition at the shell:

1

2m[ψ′k(a+ ε)− ψ′k(a− ε)] = V0ψk(a). (2.213)

This gives a second relation involving ξ and S0,

S0(k)eika + e−ika

2a− ξ(k)

cos(ka)

a= 2mV0ξ(k)

sin(ka)

ka(2.214)

From these we solve

ξ(k) =

[1− imV0

k(e2ika − 1)

]−1

,

S0(k) =1 + imV0

k(e−2ika − 1)

1− imV0

k(e2ika − 1)

.

(2.215)

While S0(k) is indeed a phase in the physical region k > 0, it has poles on the complex

k-plane, given by the solutions to

ik

mV0

= 1− e2ika. (2.216)

First let us examine purely imaginary poles k = ik∗, with

− k∗mV0

= 1− e−2k∗a. (2.217)

We see that there are poles on the positive imaginary k-axis only when V0 < 0, with

|V0| > (2ma)−1. This is when there are bound states localized near the shell.

For positive V0, there are no bound states. Let us write k = q− iw, with q > 0 and

w > 0. Such poles obey

w

mV0

= 1− e2aw cos(2aq),q

mV0

= −e2aw sin(2aq) (2.218)

Consider the limit V0 (ma)−1. We have then

w ' −2a ln cos(2aq), tan(2aq) ' − q

mV0

. (2.219)

46

The first such solution is approximately at

q ' π

a

(1− 1

2amV0

),

w ' −2a ln cosπ

amV0

' π2

am2V 20

.

(2.220)

If the potential barrier were infinitely high, there would be bound states confined to

r < a, with the first bound state at k = π/a. If the potential shell had finite width,

the tunneling probability would be exponentially small, and correspondingly the decay

width ∼ qw/m would be exponentially small. Here in the example of a delta function

potential, the decay width is only suppressed by a negative power of the potential

strength.

2.14 A toy model for inelastic resonance

Let us illustrate the S-matrix in a simple toy model. Consider the process in which

a particle A is absorbed by a target object, which then emits either A or a different

particle B. Denote by |A(k)〉 and |B(k)〉 the state of A and B particle at momentum~k, and by |C〉 the state in which the particle is absorbed by the target. The free

Hamiltonian takes the form

H0 = E0|C〉〈C|+∫d3kEA(k)|A(k)〉〈A(k)|+

∫d3kEB(k)|B(k)〉〈B(k)| (2.221)

Here EA(k) and EB(k) are the energy of A,B particle at momentum k. The interaction

Hamiltonian is taken to be

V =

∫d3k [f(k)|A(k)〉〈C|+ g(k)|B(k)〉〈C|+ f ∗(k)|C〉〈A(k)|+ g∗(k)|C〉〈B(k)|]

(2.222)

In particular, we have

1

E −H0 + iεV |C〉 =

∫d3k

[f(k)

E − EA + iε|A(k)〉+

g(k)

E − EB + iε|B(k)〉

]≡ |C〉,

1

E −H0 + iεV |C〉 =

1

E −H0 + iε

∫d3k

[|f(k)|2

E − EA + iε+

|g(k)|2

E − EB + iε

]|C〉 =

Σ(E)

E − E0 + iε|C〉,

(2.223)

where

Σ(E) =

∫d3k

[|f(k)|2

E − EA(k) + iε+

|g(k)|2

E − EB(k) + iε

]. (2.224)

47

We can now compute the T -matrix elements for asymptotic states |A(k)〉, |B(k)〉,

TAA(k′, k) = 〈A(k′)|V(

1 +1

E −H0 + V + iεV

)|A(k)〉

= f(k′)∞∑n=0

〈C|(

1

E −H0 + iεV

)2f ∗(k)

E − E0 + iε|C〉

=f(k′)f ∗(k)

E − E0 + iε

1

1− Σ(E)E−E0+iε

=f(k′)f ∗(k)

E − E0 − Σ(E) + iε,

(2.225)

and similarly

TBA(k′, k) =g(k′)f ∗(k)

E − E0 − Σ(E) + iε, (2.226)

etc. We will assume rotational invariance of the interaction, i.e. f(k), g(k) are functions

of k = |~k| only. The total cross section for the (elastic) A→ A process is given by

σA→A = (2π)4M2A

∫dΩ′|TAA(~k′, ~k)|2

= 64π5M2A

|f(k)|4

(E − E0 − ReΣ(E))2 + (ImΣ(E))2

(2.227)

where

ImΣ(E) = −π∫d3k

[δ(E − EA(k))|f(k)|2 + δ(E − EB(k))|g(k)|2

](2.228)

Suppose B particle has higher rest energy than A particle, with EB(k) = k2

2MB+µ, and

EA(k) = k2

2MA. Then

ImΣ(E) = −4π2pAMA|f(pA)|2 − 4π2pBMB|g(pB)|2θ(E − µ). (2.229)

where pA and pB are the momenta at which the energy of the particle is E. Define

ΓA(E) = 8π2pAMA|f(pA)|2,ΓB(E) = 8π2pBMB|g(pB)|2.

(2.230)

We can then write

ImΣ(E) = −1

2ΓA(E)− 1

2ΓB(E)θ(E − µ), (2.231)

and

σA→A =π

p2A

Γ2A

(E − E0 − ReΣ(E))2 + 14(ΓA + ΓBθ(E − µ))2

. (2.232)

48

Similarly, we can derive the inelastic A→ B cross section,

σA→B =π

p2A

ΓAΓBθ(E − µ)

(E − E0 − ReΣ(E))2 + 14(ΓA + ΓBθ(E − µ))2

. (2.233)

These are know as Breit-Wigner formulae. We see that ΓA and ΓB are the resonance

widths associated with A and B channels, respectively.

We see clearly that the scattering amplitude has poles in E off the real axis at

the resonance, with the imaginary part given by minus the resonance width. Let us

examine the behavior at energy E slightly above the threshold energy µ. We have

Σ(E) = 4π

∫k2dk

[|f(k)|2

E − k2

2MA+ iε

+|g(k)|2

E − µ− k2

2MB+ iε

]' Σ(µ)− i4π2M

32B |g(0)|2

√2(E − µ)

(2.234)

Due to this singular behavior as E → µ, the elastic cross section σA→A(E) exhibits a

cusp at E = µ.

2.15 Scattering with an atom

Now let us consider the scattering of a charged particle off an atom. This is an inelastic

process, in which the atom can be excited to a different state. We will label the energy

eigenstates of the atom by |χµ〉, µ being some arbitrarily defined quantum number;

µ = 0 will be the ground state. Let ∆µ be the excitation energy of the state |χµ〉. The

incident momentum ki and final momentum kf obey the energy conservation relation

k2i

2M=

k2f

2M+ ∆µ, (2.235)

where M is the mass of the incident particle (we assume that the incident particle itself

does not have any internal degrees of freedom that would be excited in the scattering

process). We will also assume that the target atom is heavy and ignore the recoil. Let

~q = ~ki−~kf be the momentum transfer. The scattering amplitude from the initial state

|ki, χ0〉 to a final state |kf , χµ〉 is given by the matrix element of the T operator as

before,

Tfi = 〈kf , χµ|(V + V

1

E −H + iεV

)|ki, χ0〉 (2.236)

where the interaction Hamiltonian V has nontrivial matrix elements between different

states of the atom. For instance, for an atom with nucleus charge Z (and Z electrons),

V takes the form (ignoring the size of the nucleus)

V (~r) =e2

4π

[Z

r−

Z∑s=1

1

|~r − ~rs|

](2.237)

49

where ~rs are the positions of the electrons. In the Born approximation, we have

TBfi =1

(2π)3

∫d3~r〈χµ|ei~q·~rV (~r)|χ0〉 (2.238)

Using the Fourier transform of the Coulomb potential∫d3~r

ei~q·~r

4π|~r − ~x|=ei~q·~x

q2, (2.239)

assuming that the nucleus state is not excited, we can write

TBfi =Ze2

(2π)3q2δµ0 −

e2

(2π)3q2〈χµ|

Z∑s=1

ei~q·~rs|χ0〉 (2.240)

where |χµ〉 describes a state of the Z electrons. Note that |χµ〉 need not be a bound

state! It has a wave function of the form

φµ(~r1, · · · , ~rZ) (2.241)

or in momentum space,

φµ(~p1, · · · , ~pZ) = 〈~p1, · · · , ~pZ |χµ〉. (2.242)

The last term in the Born amplitude involves a sum of terms

〈χµ|ei~q·~rs|χ0〉 =

∫ Z∏i=1

d3~piφ∗µ(~p1, · · · , ~ps + ~q, · · · , pZ)φ0(~p1, · · · , ~pZ). (2.243)

In other words, in the Born approximation, the amplitude is a superposition of the

amplitude in which the momentum transfer is carried by one electron. If we go to

higher order in the perturbative expansion in V , then the momentum transfer will be

distributed among more electrons. An improvement upon Born approximation, called

the impulse approximation, is obtained by superposing the exact two-body amplitudes

of the incident particle scattering with each electron.

It is conventional to define the elastic (µ = 0) and inelastic (µ 6= 0) form factors

Fµ(q) =1

Z〈χµ|

Z∑s=1

ei~q·~rs|χ0〉, (2.244)

and write the scattering amplitude as

TBµ (q) = − Ze2

(2π)3q2[δµ0 − Fµ(q)] . (2.245)

50

The elastic part of the amplitude,

TB0 (q) = − Ze2

(2π)3q2[1− F0(q)] (2.246)

gives rise to the elastic differential cross section

dσeldΩ

=dσRdΩ|1− F0(q)|2 (2.247)

where dσR is the Rutherford cross section for the nucleus with charge Z. Keep in mind

that we are still working in Born approximation, which is valid when |γ| = Zαc/v 1.

The inelastic cross section, whose relation to the amplitude was derived earlier, is

dσµdΩ

=dσRdΩ

kfki|Fµ(q)|2. (2.248)

First, consider the elastic scattering amplitude. In the limit where the momentum

transfer is small compared to the characteristic momentum of the atom (or inverse of

the size of the atom), F0(q) can be expanded as

F0(q) =1

Z〈χ0|

∑s

ei~q·~rs|χ0〉

= 1− qiqj2

1

Z

∑s

〈χ0|risrjs|χ0〉+O(q4)

= 1− q2

6

1

Z

∑s

〈χ0|r2s |χ0〉+O(q4)

(2.249)

In the last step we have made use of the spherical symmetry of the ground state wave

function. So the elastic amplitude takes the form

TB0 (q) = − Ze2

6(2π)3〈r2s〉0 +O(q2). (2.250)

Unlike the Rutherford cross section, which has infinite forward scattering amplitude,

now the forward amplitude is finite because the interaction range due to a neutral atom

is finite.

If the momentum transfer is large, on the other hand, the elastic form factor F0(q)

is suppressed due to the fast oscillatory phase of ei~q·~r, and the elastic cross section will

be dominated by the Rutherford cross section.

Now turn to the inelastic form factor. We will work with the simplest example of

the hydrogen atom. First, suppose that |χµ〉 is a bound state (which is valid when

51

the momentum transfer is small), labeled by |n`m〉, with wave function un`(r)Y`m(r).

Recall the partial wave expansion of ei~q·~r,

ei~q·~r =∞∑`=0

(2`+ 1)i`j`(qr)P`(r · q)

= 4π∞∑`=0

i`j`(qr)∑m=−`

Y`m(r)Y ∗`m(q)

(2.251)

Consequently,

〈χµ|ei~q·~r|χ0〉 = 4πi`Y ∗`m(q)

∫ ∞0

un`(r)j`(qr)u10(r)dr ≡√

4πi`Y ∗`m(q)In`(q). (2.252)

If we do not distinguish the angular momentum projection m of the final state and

sum over m, then the cross section for the atom going into the excited state |n`〉 is

dσn`dΩ

= (2`+ 1)dσRdΩ

kfki|In`(q)|2 (2.253)

Next, let us consider large momentum transfer under which an electron is ejected

through the collision. When the projectile is not an an electron, the final state of the

ejected electron |χµ〉 is described by a (reversed) scattering state |pf ,−〉. In the high

energy limit, we may ignore the influence of the Coulomb field on the ejected electron,

and approximate |pf ,−〉 by the plane wave state |pf〉. In the Born approximation, the

scattering amplitude is determined by the matrix element

〈~pf |ei~q·~r|χ0〉 = φ0(~pf − ~q) (2.254)

The differential cross section for the projectile scattered into solid angle dΩ and the

ejected electron in the momentum range dp3f is

dσBdΩ

=dσR(~q)

dΩ

kfki|φ0(~pf + ~kf − ~ki)|2d3pf (2.255)

Here the Rutherford cross section dσR is evaluated at the momentum transfer ~q =~ki − ~kf .

When the projectile is an electron, it is indistinguishable from the ejected electron.

The initial state is given by

1√2

(|ki〉 ⊗ |χ0〉+ |χ0〉 ⊗ |ki〉) (2.256)

and the final state1√2

(|kf〉 ⊗ |χµ〉+ |χµ〉 ⊗ |kf〉) (2.257)

52

The scattering amplitude therefore is the superposition of the “direct” amplitude (as in

the case when the projectile is distinguishable from the electron) and the “exchange”

amplitude (in Born approximation):

TBdir =e2

(2π)3q2

∫d3p φ∗µ(p+ q)φ0(p),

TBex =e2

(2π)3

∫d3p′〈χµ|e−ip

′·~r1|ki〉〈kf |eip′·~r2|χ0〉

p′2

=e2

(2π)3

∫d3p

φ∗µ(p+ q)φ0(p)

|kf − p|2.

(2.258)

Note that if the final momentum kf is large compared to the characterizing momentum

of the atom, the exchange amplitude will be suppressed by 1/k2f , as opposed to 1/q2

in the direct amplitude. In this limit, the scattering is well approximated by the Born

amplitude with the projectile treated as distinguishable from the electron.

It is often useful to consider the scattering cross section when we do not distinguish

the final states of the atom and the momentum of the scattered projectile, but only keep

track of the scattering angle. This total inelastic scattering differential cross section,

in Born approximation, is given by

dσBineldΩ

=M2e4

4π2ki

∑µ 6=0

kfq4|〈χµ|n~q|χ0〉|2 (2.259)

where q and kf are determined by ki, µ and the scattering angle θ, while n~q is defined

as

n~q ≡Z∑s=1

ei~q·~rs . (2.260)

In the high energy limit, kf ' ki; at fixed scattering angle θ away from the forward

direction, the dependence of q on the excitation energy of |χµ〉 can be ignored. We

then havedσBineldΩ

' dσRdΩ

∑µ 6=0

|〈χµ|n~q|χ0〉|2

=dσRdΩ

[〈n†~qn~q〉0 − |〈n~q〉0|

2] (2.261)

where 〈· · · 〉0 stands for the expectation value in the ground state |χ0〉. Since the Z

electrons in the atom are indistinguishable, we have

〈n†~qn~q〉0 − |〈n~q〉0|2 = Z + Z(Z − 1)〈ei~q·(~r1−~r2)〉0 − Z2|〈ei~q·~r1〉0|2

= Z(1− |F0(q)|2

)+ Z(Z − 1)

[〈ei~q·(~r1−~r2)〉0 − |〈ei~q·~r1〉0|2

] (2.262)

53

The last term is due to the correlation between different electrons in the ground state

wave function.

When high energy particles run through a medium, as the scattering cross section

is sharply peaked near the forward direction, most of the particles escape within a

small scattering angle. In experiments, a quantity of interest is the energy loss per

unit path length, due to inelastic scattering in which the particles are scattering into a

small angle Θ. Let N be the number of atoms per unit volume. The ratio of particles

that undergo inelastic scattering per unit length dx is then given by Ndσdx. Here dσ

stands for the inelastic cross section for which the atom is excited from the ground

state to some |χµ〉, and the energy loss will be Eµ − E0. This leads to the energy loss

of the projectile particle

dE

dx= N

∑µ

(Eµ − E0)

∫∆Ω

dΩdσµdΩ

. (2.263)

where ∆Ω is the solid angle within θ < Θ. Suppose the projectile has charge ze. Using

our expression for the inelastic cross section in the Born approximation, we have

dE

dx= N

z2e4

2π

∑µ

(Eµ − E0)M2kfki

∫ Θ

0

sin θdθ|〈χµ|n~q|χ0〉|2

q4 (2.264)

At given energy loss, i.e. fixed k2f , the momentum transfer is related to the scattering

angle θ via

q2 = (~ki − ~kf )2 = k2i + k2

f − 2kikf cos θ (2.265)

We have

qdq = kikf sin θdθ. (2.266)

Using this, the θ integral can be turned into a q-integral

dE

dx= N

z2e4

2πv2

∑µ

(Eµ − E0)

∫ qmax

qmin

|〈χµ|n~q|χ0〉|2dq

q3 (2.267)

The q-integral is evaluated over the range (qmin, qmax), with qmax given by the mo-

mentum transfer at angle θ = Θ, whereas qmin is given by the momentum transfer for

forward scattering, obeying

qmin = ki − kf =2M(Eµ − E0)

ki + kf' Eµ − E0

v(2.268)

in the limit where the momentum transfer is small compared to the incident momentum.

For small Θ, in this limit we also have qmax 'MvΘ.

54

Next, let us analyze the matrix elements of n~q. Observe that∑µ

(Eµ − E0)|〈χµ|n~q|χ0〉|2

=∑µ

〈χ0|n†~q(Eµ − E0)|χµ〉〈χµ|n~q|χ0〉

= 〈χ0|[n†~q, H]n~q|χ0〉

=1

2〈χ0|[[n†~q, H], n~q]|χ0〉

(2.269)

The commutator of n†~q with H is very simple:

[n†~q, H] =∑s

[e−i~q·~rs ,p2s

2m]

=1

2m

∑s

e−i~q·~rs , ~q · ~ps

(2.270)

and so

[[n†~q, H], n~q] =1

2m

∑s

[e−i~q·~rs , ~q · ~ps

, ei~q·~rs

]=q2Z

m(2.271)

Be aware that m here is the mass of the electron in the atom, not the mass of the

projectile. This gives the exact result∑µ

(Eµ − E0)|〈χµ|n~q|χ0〉|2 =q2Z

2m. (2.272)

This result cannot however be applied straightforwardly to evaluate the q-integral,

because qmin depends on Eµ. Note that qmax on the other hand is insensitive to Eµ.

Note however in the high energy limit, qmin is very small. In the small q regime, we

can make use of the approximation

〈χµ|ei~q·~rs|χ0〉 ' i〈χµ|~q · ~rs|χ0〉 (2.273)

and therefore

〈χµ|n~q|χ0〉 ' i~q · 〈χµ| ~D|χ0〉 (2.274)

where ~D =∑~rs is the dipole moment operator. While the matrix element of ~D

depends on the state |χµ〉, after summing over all excited states of energy Eµ, by

rotational symmetry the result for |〈χµ|~q · ~D|χ0〉|2 cannot depend on the orientation of

~q. Therefore we can replace |〈χµ|~q · ~D|χ0〉|2 by q2|〈χµ|Dz|χ0〉|2 in the integration over

small q regime. By splitting the q-integral over a “large” momentum regime q > q0 and

55

a small momentum regime q < q0, and approximating the small momentum integral as

above, we have

dE

dx= N

z2e4

2πv2

[Z

2mln

(MvΘ

q0

)+∑µ

(Eµ − E0)|〈χµ|Dz|χ0〉|2 ln

(q0v

Eµ − E0

)](2.275)

This would be a good approximation if q0 is small compared to the characterizing

momentum of the atom. The RHS is independent of q0 because of (2.272). We may

then choose q0 to be such that the last term vanishes, namely

ln(q0v) =

∑µ(Eµ − E0)|〈χµ|Dz|χ0〉|2 ln (Eµ − E0)∑

ν(Eν − E0)|〈χν |Dz|χ0〉|2

=2m

Z

∑µ

(Eµ − E0)|〈χµ|Dz|χ0〉|2 ln (Eµ − E0) .(2.276)

Define I ≡ q0v at this value of q0. This is a parameter that depends only on the

property of the atom. We then arrive at Bethe’s formula for the energy loss

dE

dx= N

z2e4

2πv2

Z

2mln

(Mv2Θ

I

). (2.277)

3 Quantization of radiation field

3.1 Quantization of free field

The quantum theory of photons begins with the quantization of the free electromagnetic

field. We will work in c = 1 units, and (−+ ++) signature for the Lorentzian metric.

The electromagnetic field is described by a Lorentzian four-vector potential

Aµ = (−V, ~A) (3.1)

The field strength

Fµν = ∂µAν − ∂νAµ (3.2)

contains the electric and magnetic fields

Fi0 = −F0i = Ei, Fij =∑k

εijkBk. (3.3)

Maxwell’s equations are

∂µFµν = 0. (3.4)

56

Aµ is subject to the gauge symmetry/redundancy

δAµ = ∂µΛ. (3.5)

By choosing Λ, we can go to the temporal gauge

A0 = 0, (3.6)

in which ~E = − ~A, ~B = ~∇× ~A. We can further choose the time independent part of Λ

to set ~∇ · ~A|t=t0 = 0 (at time t = t0). When there are sources of charge density ρ, ~E

obeys the equation ~∇ · ~E = −~∇ · ~A = ρ, and so ~∇ · ~A = 0 cannot be maintained at all

time. In the absence of sources, on the other hand, we can choose Λ to put Aµ in the

so called radiation gauge

A0 = 0, ~∇ · ~A = 0. (3.7)

In this gauge, Maxwell’s equations reduce to

~A ≡ (∂2t −∇2) ~A = 0. (3.8)

The free Maxwell’s equations can be derived from the Lagrangian density

L = −1

4FµνF

µν , (3.9)

or the Hamiltonian

H =

∫d3~r(AiΠi − L

)=

∫d3~x

(1

2Π2i +

1

4F 2ij

)=

1

2

∫d3~r(~E2 + ~B2

) (3.10)

where the canonical momentum density Πi is

Πi =∂L∂Ai

= Fi0. (3.11)

We will work in Heisenberg picture. In order to canonically quantize the theory, we can

either work with canonical fields Ai and their conjugate canonical momentum density

Πi (not to be confused with spacetime momentum), by imposing

[Ai(~r, t),Πj(~r′, t)] = i~δ3(~r − ~r′), (3.12)

or we can put the system in a box and canonically quantize the discrete Fourier modes.

For pedagogy we will illustrate the latter approach.

We shall put the space in a box of size L and impose periodic boundary condition

on the fields. In the end we will take L→∞ and the effect of the boundary condition

57

will disappear. We will write ωk = |~k| for the frequency of the mode (in c = 1 units).

With periodic boundary condition, ~k takes value on a lattice

~k =2π

L(n1, n2, n3), ni ∈ Z. (3.13)

We will reserve the notation k · x = ~k ·~r−ωkt for the Lorentzian inner product of four

vectors. To go back to continuous variables in the L→∞ limit, we have

L3

(2π)3δkk′ → δ3(~k − ~k′),

1

L3

∑k

→∫

d3~k

(2π)3.

(3.14)

Now we can expand ~A(x) on spatial Fourier modes as

~A(x) = L−32

∑k

~A~k(t)ei~k·~r. (3.15)

In radiation gauge, the Fourier modes of ~A are transverse to their momentum, namely~k · ~A~k(t) = 0. Since ~A(x) is real, we need to impose the condition ~A∗~k(t) = ~A−~k(t).

Correspondingly, the electric and magnetic field strengths are

~E(x) = −L−32

∑k

~A~k(t)ei~k·~r,

~B(x) = iL−32

∑k

~k × ~A~k(t)ei~k·~r.

(3.16)

In terms of ~A~k, the Lagrangian is written∫d3~rL =

1

2

∫d3~r(~E2 − ~B2

)=

1

2

∑k

[~A∗~k(t) · ~A~k(t)− ~k

2 ~A∗~k(t) · ~A~k(t)] (3.17)

The canonical momentum conjugate to ~A~k(t) (which is restricted to be transverse to ~k

by our gauge condition) is~Π~k(t) = ~A∗~k(t).

(3.18)

Canonical quantization then demands

[Ai~k(t), A∗j~k′

(t)] = i~δijδkk′ , (3.19)

where i, j label the two polarization directions transverse to ~k.

58

We now make use of the equations of motion and write (keep in mind the reality

condition)

~A~k(t) =

√~

2ωk

(~ake

−iωkt + ~a†−keiωkt), (3.20)

The reason for our choice of normalization in defining ak will be evident momentarily.

We can further expand ~a~k on an orthonormal basis of two polarizations vectors ε~k,1 and

ε~k,2 that are transverse to ~k. We will choose them by demanding

ε1 × ε2 = k. (3.21)

This condition fixes ε1,2 up to rotation in the plane transverse to ~k. Alternatively, we

may work with circular polarization vectors

e~k,± =~ε1 ± i~ε2√

2(3.22)

They are defined so as to have the property

ik × e~k,± = ±e~k,±. (3.23)

For instance, suppose we take the four-vector kµ to be kµ = (k, k, 0, 0), then a set of

polarization vectors are

ε1 = (0, 0, 1, 0), ε2 = (0, 0, 0, 1), or e+ =1√2

(0, 0, 1, i), e− =1√2

(0, 0, 1,−i).

(3.24)

Now decomposing the Fourier modes of ~A(t) on the two circular polarization vectors,

~a~k =∑α

a~k,λ(t)e~k,λ, (3.25)

we can write the mode expansion for the free vector potential as

~A(x) = L−32

∑k,λ

√~

2ωk

(a~k,λe~k,λe

ik·x + c.c.). (3.26)

The canonical commutation relation is now written as

[a~k,λ, a†~k′,λ

] = δkk′δλλ′ , [a~k,λ, a~k′,λ′ ] = [a†~k,λ, a†~k′,λ′

] = 0. (3.27)

Now it is clear that a~k,λ has the interpretation of the annihilation operator of a photon

of momentum/wave vector ~k and polarization e~k,λ, while a†~k,λ is the corresponding

59

creation operator. The Hamiltonian is written in terms of the modes as

H =1

2

∫d3~r(~E2 + ~B2

)=

1

2

∑k,λ

~ωk(a†k,λak,λ + ak,λa†k,λ)

=∑k,λ

~ωk(a†k,λak,λ +1

2).

(3.28)

The operator Nk,λ = a†k,λak,λ counts the number of photons carrying momentum ~k and

polarization ek,λ. The 12

shift is the ground state energy associated with this mode.

While the sum of the ground state energy may seem to diverge and lead to infinite

ground state energy for the vacuum state, the realistic physical system is subject to

a cutoff at high energies, where the physics is no longer described by say free electro-

magnetic fields in a box. For the moment we will ignore this ground state energy since

it does not affect physical processes, though we will revisit this issue later.

The Hilbert space of free electromagnetic field in a periodic box can then be con-

structed by acting on the vacuum state |0〉 with creation operators. For instance, a

state with nk,λ photons of momentum k and polarization λ is written

|nk,λ〉 =∏k,λ

(a†k,λ)nk,λ√

nk,λ!|0〉. (3.29)

This state has energy E =∑

k,λ ~ωk(nk,λ+ 12) and momentum ~P =

∑k,λ ~~knk,λ (verify-

ing this is left as an exercise). An important consequence which is obvious in retrospect

is that the photons are identical bosons. It followed from the canonical commutation

relation for ~A(x) (as opposed to say canonical anticommutation relations for fermionic

fields).

The angular momentum operator ~J is the generator of rotational symmetry. Under

an infinitesimal spatial rotation R(δ~θ) = exp(iδ~θ · ~J/~) = 1 + i~δ~θ · ~J + · · · ,

~r → ~r′ = ~r + δ~θ × ~r + · · · (3.30)

~A transforms as a vector, meaning ~A→ ~A′ with

~A′(x′) = ~A(x) + δ~θ × ~A(x) + · · · . (3.31)

In terms of operators, the rotational symmetry acts on A(x) by

R(δ~θ) ~A(x)R(δ~θ)−1 = ~A′(x) = ~A(x) + δ~θ × ~A(x)− (δ~r · ∇) ~A(x), (3.32)

60

or

[i

~δ~θ · ~J, ~A(x)] = δ~θ × ~A(x)−

[(δ~θ × ~r) · ∇

]~A(x) (3.33)

It follows that

[i

~δ~θ · ~J,~a~ke

ik·x] = δ~θ × ~a~keik·x − i

[(δ~θ × ~r) · ~k

]~a~ke

ik·x, (3.34)

or in components,

[J i, aj~keik·x] = i~εijmam~k e

ikx − ~εimnxmknaj~keik·x. (3.35)

The last term corresponds to (minus) the orbital angular momentum carried by the

modes of momentum ~k, which is destroyed by the annihilation operator a~k. Writing

~J =

∫d3~r ~r × ~Π + ~S, (3.36)

where ~Π is the momentum density (the spatial component of the Neother current for

translational symmetry), and ~S is the spin operator, then

[Si, aj~k] = i~εijmam~k . (3.37)

We can write Si in terms of the creation and annihilation operators as

~S = −i~∑k

~a†~k × ~a~k

= ~∑k

k(nk,+ − nk,−)(3.38)

This means that the photon created by a†~k,+ carries one unit of angular momentum

along the direction of its momentum k (helicity +1), and the photon created by a†~k,−carries minus one unit of angular momentum along k (helicity −1).

Having established the quantization of free photon fields, it is useful to go back to

delta function normalized states, for instance for one photon,

L32

(2π)32

a†~k,λ|0〉 → |~k, λ〉 = a†λ(

~k)|0〉 (3.39)

where aλ(~k) and its conjugate now obey the canonical commutation relation

[aλ(~k), a†λ′(~k′)] = δλλ′δ

3(~k − ~k′). (3.40)

We will work in ~ = 1 units again from now. The free field mode expansion is written

in term sof the delta function normalized oscillators as

~A(x) =

∫d3~k

(2π)32

√2ωk

∑λ

[aλ(~k)e~k,λe

ik·x + c.c.]

(3.41)

61

While |~k, λ〉 are one photon states of definite momentum and helicity, it will be useful

to also work with states of definite total angular momentum. Such states are labeled

by |kjmλ〉, where (j,m) are the quantum numbers of the total angular momentum and

Jz, and λ is the eigenvalue of the helicity operator ~J · p. They are normalized by

〈kjmλ|k′j′m′λ′〉 =δ(k − k′)

k2δjj′δmm′δλλ′ . (3.42)

Let us begin with the state |kz, λ〉, i.e. a photon moving along z direction with mo-

mentum k and helicity λ. It is an eigenstate with respect to Jz, with eigenvalue λ, and

therefore can be decomposed into angular momentum eigenstates with total angular

momentum j ≥ |λ| = 1. So we can write

|kz, λ〉 =∞∑j=1

Nj|kjλλ〉. (3.43)

The more general helicity state |~k, λ〉 can be obtained by performing a rotation from

z to k. Denote the matrix element of this rotation symmetry on Jz basis Djmm′(k),

namely

Djmm′(

~k) = 〈j,m|R(k, z)|j,m′〉. (3.44)

Then

|~k, λ〉 =∞∑j=1

Nj

j∑m=−j

Djmλ(k)|kjmλ〉 (3.45)

Using the orthogonality relation of Dj-matrix,∫dkDj

mλ(k)Dj′

m′λ′(k)∗ =4π

2j + 1δjj′δmm′δλλ′ , (3.46)

we have ∫d2kDj

mλ(k)∗|~k, λ〉 =4π

2j + 1Nj|kjmλ〉 (3.47)

Taking inner products of states on the LHS∫d2kd2k′Dj

mλ(k)Djm′λ′(k

′)∗〈~k, λ|~k′, λ′〉 =

∫d2kd2k′Dj

mλ(k)Djm′λ(k

′)∗δ3(~k − ~k′)δλλ′

=

∫d2kDj

mλ(k)Djm′λ(k)∗

δ(k − k′)δλλ′k2

=4π

2j + 1

δ(k − k′)δmm′δλλ′k2

(3.48)

and equating it to the inner product on the RHS,(4π

2j + 1Nj

)2δ(k − k′)

k2δmm′δλλ′ , (3.49)

62

we find

Nj =

√2j + 1

4π. (3.50)

So we find the desired change of basis,

|~k, λ〉 =∞∑j=1

j∑m=−j

√2j + 1

4πDjmλ(k)|kjmλ〉,

|kjmλ〉 =

√2j + 1

4π

∫d2kDj

mλ(k)∗|~k, λ〉.

(3.51)

Note in particular that there is no one photon state of total angular momentum zero.

Under parity symmetry, the vector potential ~A acquires a minus sign. While the

spin is invariant under parity, the helicity changes sign. Thus under parity,

Is|~k, λ〉 = −| − ~k,−λ〉. (3.52)

This is the statement that a photon has odd intrinsic parity.

Let us investigate the causality of the theory by examining commutators of opera-

tors at two points x1 and x2. Consider the vacuum two point function

〈0|Ai(x1)Aj(x2)|0〉 =

∫d3~kd3~k′

(2π)32√ωkωk′

∑λ

〈0|aλ(~k)aλ(~k′)†|0〉(e~k,λ)i(e

∗~k′,λ

)jeik·x1−ik′·x2

=

∫d3~k

(2π)32ωk

∑λ

(e~k,λ)i(e∗~k,λ

)jeik·x12

=

∫d3~k

(2π)32ωk

(δij − kikj

)eik·x12

(3.53)

and since the commutator of Ai(x1) with Aj(x2) is a c-number, it is given by

[Ai(x1), Aj(x2)] = −i∫

d3~k

(2π)3k

(δij − kikj

)ei~k·~r12 sin(kt12) (3.54)

The observables are constructed out of gauge invariant operators, in this case field

strengths. For instance,

[Ei(x1), Ej(x2)] = ∂t1∂t2 [Ai(x1), Aj(x2)] = −i∫

d3~k

(2π)3k

(~k2δij − kikj

)ei~k·~r12 sin(kt12)

= −2i(∇2δij −∇i∇j)D(x12),

[Ei(x1), Bj(x2)] = εjmn∂t1∂xm2 [Ai(x1), An(x2)] = −εijm∫

d3~k

(2π)3kmei

~k·~r12 cos(kt12)

= 2iεijm∇m∂tD(x12).(3.55)

63

Here we defined the function

D(x) = −∫

d3~k

(2π)32kei~k·~r sin(kt)

= − 1

8π2r

∫ ∞−∞

dk sin(kr) sin(kt)

=δ(r + t)− δ(r − t)

8πr.

(3.56)

We see that the commutators between the field strengths vanish unless the two points

x1 and x2 are light-like separated. At equal time, while the electric fields commute

with themselves (and so do the magnetic fields with themselves), the electric field and

the magnetic field do not commute.

3.2 Casimir effect

We have seen that each mode of the photon contributes 12~ωk to the ground state

energy. While this energy is unobservable for photon field in a fixed box, it does

become meaningful if we somehow allow the size of the box to vary. Let us consider

two parallel conducting plates of size L× L separated at distance ` along z-direction.

L is taken to be large so that the effects due to the boundary of the plates can be

ignored. The x, y components of the electric field are subject to Dirichlet boundary

condition at the plates, namely

Ex|z=0 = Ex|z=` = Ey|z=0 = Ey|z=` = 0. (3.57)

In the radiation gauge, this implies that Ax, Ay vanish at z = 0, `. The vector potential

has mode expansion of the form

Ai(~r, t) = 2`−12L−1

∞∑n=1

∑kx,ky

√~

2ωk

[ai(kx, ky, n) sin

(πnz`

)eikxx+ikyy−iωkt + c.c.

], i = x, y,

Az(~r, t) = 2`−12L−1

∞∑n=1

∑kx,ky

√~

2ωk

[i`

πnkiai(kx, ky, n) cos

(πnz`

)eikxx+ikyy−iωkt + c.c.

]

+ `−12L−1

∑kx,ky

√~

2ωk

[az(kx, ky)e

ikxx+ikyy−iωkt + c.c.].

(3.58)

The frequency of the modes are

ωk =

√k2x + k2

y + (πn

`)2 (3.59)

64

At given kx, ky, there are two polarizations for each n ≥ 1, and one polarization for

n = 0. The ground state energy is then

E0 = L2

∫dkxdky(2π)2

∞∑n=−∞

1

2~ωk (3.60)

where we have turned the sum over n = 0 and n ≥ 1 into a sum over n ∈ Z. This sum

is obviously divergent. In reality, there is a cutoff at a high energy scale Λ (say size

of atoms) where the conducting plate is no longer a valid approximation. We should

therefore cut off the sum/integral at energy or frequency ω ∼ Λ. A convenient way to

implement this cut off is to multiply the summand by a factor e−εωk , with ε ∼ 1/Λ,

and take ε→ 0 at the end. The “regularized” ground state energy is then

E0 =~L2

4π

∫ ∞0

kdk

∞∑n=−∞

√k2 + (πn/`)2e−ε

√k2+(πn/`)2

=~L2

4π

∞∑n=−∞

e−επ|n|/`(

2

ε3+

2π|n|ε2`

+π2n2

ε`2

)=

~L2

4π

[12`

πε4− π3

180`3+O(ε)

] (3.61)

where in the last step we have performed the sum over n and expanded near small

ε. In taking ε → 0, there is a divergent term of order ε−4. It is proportional to L2`,

the volume of space between the two plates. There would be similar contributions to

the zero point energy from modes in the space outside the two plates, and their total

contribution is unaffected by the positions of the two plates, and hence do not induce

any force between the plates. The term of physical interest is then

E0(`) = −π2~L2

720`3. (3.62)

This zero point energy increases as ` increases, giving rise to an attractive force between

the two plates. This is known as the Casimir force. The pressure on the plate due to

the Casimir force is

P = − 1

L2

dE0(`)

d`= − π

2~c240`4

. (3.63)

where we restored c which was set to 1 previously. This force has indeed been measured

and verified in experiments.

3.3 Radiative transitions

The coupling of charged particles to electromagnetic field is described by a Hamiltonian

of the form

H = HM +Hγ +Hint, (3.64)

65

where HM stands for the matter Hamiltonian in the absence of EM field, Hγ is the

Hamiltonian of the free EM field as we have described, and Hint is the interaction

Hamiltonian that couples the EM field to the charged matter. The systems we will be

mostly considering are atoms coupled to EM fields. We have

HM +Hint =∑a

(~pa − ea ~A(~ra))2

2ma

−∑a

~µa · ~B(~ra) +1

8π

∑a6=b

eaeb|~rab|

+ VM (3.65)

where ~ra and ~pa are the positions and canonical momenta of the a-th particle, ea and

~µa are the electric charge and magnetic moment of the particle, and VM stands for

non-electromagnetic interactions among the particles. Hint can be separated out as

Hint =∑a

[− eama

~pa · ~A(~ra) +e2a

2ma

~A(~ra)2 − ~µa · ~B(~ra)

](3.66)

For now, we shall ignore the coupling of the magnetic moment, whose coupling to the

EM field is a small effect in the nonrelativistic limit. Working to first order perturbation

theory in the EM coupling strength (or the electric charge), we only need to consider

the first term. Let us analyze it for a single particle, coupled to EM field in a box of

size L subject to periodic boundary condition,

H1 = − e

m~p · ~A(~r)

= − e

mL−

32

∑~k,λ

1√2ωk

(~p · e~k,λa~k,λe

i~k·~r + c.c.) (3.67)

To first order in perturbation theory (Born approximation), the transition rate is re-

lated to the matrix element of H1 via the formula

Γi→f = 2π|〈f |H1|i〉|2δ(Ef − Ei) (3.68)

which we derived earlier.

Let us consider the process of spontaneous emission. The initial state has no pho-

tons, whereas the final state has one photon of some momentum ~k and polarization λ.

Let us write|i〉 = |i〉m ⊗ |0〉, |f〉 = |f〉m ⊗ a†~k,λ|0〉, (3.69)

where now |i〉m and |f〉m denote the electron state. The emission rate of a photon into

solid angle dΩ is (k = |~k| is determined by energy conservation)

dΓ = 2π|〈f |H1|i〉|2L3

(2π)3

d3k

dωk

= 2π|〈f |H1|i〉|2L3

(2π)3k2dΩ.

(3.70)

66

The first order transition amplitude is

〈f |H1|i〉 = −L−32

∑k′,λ′

e

m√

2k′m〈f |~p · e∗~k′,λ′e

−i~k′·~r|i〉m〈0|a~k,λa†~k′,λ′|0〉

= −L−32

e

m√

2km〈f |~p · e∗~k,λe

−i~k·~r|i〉m(3.71)

The emission rate within a solid angle is then given by

dΓ

dΩ=

e2

8π2m2k∣∣∣m〈f |~p · e∗~k,λe−i~k·~r|i〉m∣∣∣2 (3.72)

The spontaneous decay width of the atom from state |i〉m to |f〉m is given by the sum

of the emission rate over all final photon states

Γif =e2

8π2m2k∑λ

∫dΩ∣∣∣m〈f |~p · e∗~k,λe−i~k·~r|i〉m∣∣∣2 (3.73)

where k = ωk = Emi − Em

f .

Next, let us consider stimulated emission. The only difference from the case of

spontaneous emission is that instead of having no photons in the initial states, suppose

we have nk,λ photons with momentum k and polarization λ. The transition amplitude

is multiplied by a factor

〈n~k,λ + 1|a†~k,λ|n~k,λ〉 =√n~k,λ + 1. (3.74)

The differential emission rate makes sense in this case only if n~k,λ varies continuously

with ~k in the infinite size L limit. The stimulated emission rate for emission of photon

into solid angle dΩ is then multiplied by the factor

N~k,λ + 1 (3.75)

where N~k,λ is the (average) number of photons in each state of momentum ~k within

the angle dΩ. In infinite space, this is sensible only when there is a nonzero density of

photons, or more precisely, a nonzero energy flux I~k,λ per energy interval dω and per

solid angle dΩ. It is related to the photon occupation number by (exhibiting ~ and c

explicitly)

I~k,λdωdΩ =c

L3~ωN~k,λ

L3

(2π)3k2dkdΩ (3.76)

and so we can related N~k,λ to the flux by

N~k,λ =(2π)3

k3I~k,λ. (3.77)

67

The absorption rate of a photon by the atom can be analyzed similarly. The

transition amplitude is

〈f |~p · e~k,λei~k·~r|i〉〈n~k,λ − 1|a~k,λ|n~k,λ〉 =

√n~k,λ〈f |~p · e~k,λe

i~k·~r|i〉 (3.78)

It follows thatdΓabs,f→idΓem,i→f

=N~k,λ

N~k,λ + 1(3.79)

This relation was first derived by Einstein as a consequence of thermodynamic equi-

librium.

Let us take a closer look at the transition matrix elements

〈f |~p · e∗~k,λe−i~k·~r|i〉 (3.80)

Suppose |i〉 and |f〉 are atom bound states, and that the wave length 1/k is large

compared to the characteristic length scale of the atom. Then we can approximate the

transition matrix element with

〈f |~p · e∗~k,λ|i〉 = −im〈f |[~r · e∗~k,λ, HM ]|i〉

= −im(Ei − Ef )〈f |~r · e∗~k,λ|i〉 = ∓imk〈f |~r · e∗~k,λ|i〉.(3.81)

where the − sign is for the case of emission and + sign for absorption. Generally, with

multiple charged particles in the system, we can replace this matrix element by

∓imk〈f | ~D · e∗~k,λ|i〉 (3.82)

where ~D is the dipole moment operator. This leading contribution to the transition

rate is called dipole transition, given by

dΓifdΩ

=e2

8π2k3|〈f | ~D · e∗~k,λ|i〉|

2 (3.83)

Suppose |i〉 has (J, Jz) angular momentum quantum numbers (j,m). Acting on it with~D can only change j by 0,±1 and change m by 0,±1. Furthermore, the action of ~D

changes parity, and the transition is possible only between states of opposite parity.

An exceptional case is j = 0, which by emitting a photon can only go to a j = 1 atomic

state, since there is no photon state of zero angular momentum.

Going to higher order in perturbation theory, transition by the coupling to mag-

netic dipole moment or higher electric multiple moments are possible, thereby more

general selection rules, accordingly. The transition between j = 0 states is still strictly

forbidden in one-photon emission.

68

3.4 Scattering of photons

The scattering of a photon with say an electron is an effect that first occurs at second

order in perturbation theory. The transition amplitude is given by matrix elements of

the second order T operator

T = H2 +H11

E −H0 + iεH1, (3.84)

where

H1 = − e

m~p · ~A(~r) = − e

mL−

32

∑~k,λ

1√2k

(~p · e~k,λa~k,λe

i~k·~r + c.c.),

H2 =e2

2m~A(~r)2 =

e2

2mL−3

∑k1,k2,λ1,λ2

ek1,λ1 · e∗k2,λ2√4k1k2

ei~k12·~rak1,λ1 , a

†k2,λ2+O(a2, (a†)2).

(3.85)

In the expression for H2, we omitted the terms that involves two a’s or two a†’s, because

they do not contribute to the photon scattering amplitude at the leading nontrivial

order. To study the transition rate or cross section, we simply need to compute the

matrix element

〈f, k′λ′|T |i, kλ〉 = 〈f, k′λ′|H2|i, kλ〉+∑a

〈f, k′λ′|H1|ψa〉〈ψa|H1|i, kλ〉E − Ea + iε

(3.86)

where|i, kλ〉 = |i〉 ⊗ a†kλ|0〉,|f, k′λ′〉 = |f〉 ⊗ a†k′λ′|0〉,

(3.87)

|i〉 and |f〉 stand for the initial and final electron states. |ψa〉 is a complete orthonormal

energy eigen-basis of intermediate states. Since H1 either creates or annihilates a

photon, we only need to consider |ψa〉 that involves zero or two photons.

First let us compute the matrix element of H2,

〈f, k′λ′|H2|i, kλ〉 =e2

2L3m√kk′

e∗k′,λ′ · ek,λ〈f |ei(~k−~k′)·~r|i〉

=e2

2L3m√kk′

e∗k′,λ′ · ek,λFfi(~q).(3.88)

where Ffi(~q) (~q = ~k−~k′) is the form factor. The generalization to the case of a photon

interacting with multiple charged particles in straightforward.

Next let us examine the intermediate state |ψa〉. There are two cases to consider:

|ψa〉 = |n〉 ⊗ |0〉, and |ψa〉 = |n〉 ⊗ |kλ, k′λ′〉, (3.89)

69

where |n〉 runs through a complete set of electron energy eigenstates. Putting these

together, we have

〈f, k′λ′|T |i, kλ〉 =e2

2L3m√kk′

[e∗k′,λ′ · ek,λFfi(~q) +

1

m

∑n

〈f |~p · e∗k′λ′e−i~k′·~r|n〉〈n|~p · ek,λei

~k·~r|i〉Ei − En + k + iε

+1

m

∑n

〈f |~p · ekλei~k·~r|n〉〈n|~p · e∗k′,λ′e−i

~k′·~r|i〉Ei − En − k′ + iε

](3.90)

Our previous derivation of the inelastic cross section is easily generalized to the rela-

tivistic case. The collision rate is

dΓ = 2π |〈f, k′λ′|T |i, kλ〉|2 L3

(2π)3k′2dΩ. (3.91)

The flux of the incident photon is c/L3, and we have set c = 1. The scattering cross

section is thus

dσ

dΩ= L3 dΓ

dΩ= |〈f, k′λ′|T |i, kλ〉|2 L6

(2π)2k′2

=e4

16π2m2

k′

k

∣∣e∗k′,λ′ · ek,λFfi(~q)+

1

m

∑n

(〈f |~p · e∗k′λ′e−i

~k′·~r|n〉〈n|~p · ek,λei~k·~r|i〉

Ei − En + k + iε+〈f |~p · ekλei

~k·~r|n〉〈n|~p · e∗k′,λ′e−i~k′·~r|i〉

Ei − En − k′ + iε

)∣∣∣∣∣2

(3.92)

This is known as the Kramers-Heisenberg formula. To generalize to the case of multiple

charged particles in the atom, we simply need to replace ~p by the sum over the momenta

of each charged particle times its number of units of charge.

Note that this result diverges when k = En − Ei for some excitation energy level

n. This is an artifact of the second order perturbation theory we have been using. We

have already seen that the coupling of the atom to the photon induces spontaneous

decay. As a consequence, |n〉 will only be a metastable state with some decay width,

and its energy effectively acquires a negative imaginary part. We will analyze this in

detail in the next section.

Consider the long wavelength limit, with |i〉 = |f〉 a rotationally invariant atomic

70

state. For instance, by rotational symmetry, we have∑n

〈i|~p · e∗k′λ′ |n〉〈n|~p · ek,λ|i〉Ei − En + k + iε

= e∗k′,λ′ · ek,λ∑n

〈i|pz|n〉〈n|pz|i〉Ei − En + k + iε

= −m2k2e∗k′,λ′ · ek,λ∑n

〈i|Dz|n〉〈n|Dz|i〉Ei − En + k + iε

' −m2k2e∗k′,λ′ · ek,λ∑n6=i

〈i|Dz|n〉〈n|Dz|i〉Ei − En

.

(3.93)

Note that the sum does not involve n = i because 〈i|Dz|i〉 = 0 by rotational invariance

of the state |i〉. Define

D =∑n6=i

〈i|Dz|n〉〈n|Dz|i〉Ei − En

, (3.94)

which depends only on the property of the atomic state |i〉. For the neutral atom, the

form factor Ffi(~q) is of order O(q2) at low frequency, and is small compared to the

term proportional to D in the nonrelativistic limit. The cross section in this limit is

that of Rayleigh scattering,

dσ

dΩ=

e4

16π2m2

∣∣(ef · ei)2mk2D∣∣2 = 4α2ω4 |(ef · ei)D|2 (3.95)

3.5 Atom resonance

Due to the coupling of electrons in the atom to photons, the atomic states are no

longer exact energy eigenstates but are metastable resonance states with some decay

width. Let us examine this phenomenon in some detail. We will consider the example

of radiative transition between 1s and 2p states of the hydrogen atom. For simplicity,

the spin and magnetic moment of the electron will be ignored. 2p has three degenerate

states, with the z-component of orbital angular momentum m = 0,±1. So the atomic

states of interest are |1s〉 and |2p,m〉. It will be convenient to write the transition

operator between the states as

b†m = |2p,m〉〈1s|, bm = |1s〉〈2p,m|. (3.96)

The operators b†mbm = |2p,m〉〈2p,m| and bmb†m = |1s〉〈1s| are projections onto the

respective states. The first order interaction Hamiltonian has matrix elements for the

absorption of a photon

〈2p,m|H1|1s; kλ〉 = −L−32ie√2k〈2p,m|ek,λ · ~pei

~k·~r|1s〉

' −L−32ie(E2 − E1)√

2k〈2p,m|ek,λ · ~r|1s〉

(3.97)

71

where in the second step we have taken the long wavelength approximation. We write

further write the dipole moment matrix elements as

〈2p,m|~r|1s〉 = e∗mD (3.98)

where D = 〈1s|z|2p, 0〉 and

e0 = (0, 0, 1), e±1 =1√2

(∓1, i, 0). (3.99)

We will only keep the first order interactions in the Hamiltonian. Further, we will

simplify things by keeping only the terms that correspond to the absorption γ+1s→ 2p,

and ignore γ + 2p→ 1s because the latter does not have much effect on the resonance

absorption. One should keep in mind, however, that this approximation spoils causality

of the photon field. Now our toy model Hamiltonian is written as

H = Hγ + (E2 − E1)∑m

b†mbm − igL−32

∑k,λ,m

1√2k

[(ek,λ · e∗m)b†mak,λ − (e∗k,λ · em)bma

†k,λ

](3.100)

where g ≡ e(E2 − E1)D. This model is exactly soluble due to the extra constant of

motion N =∑b†mbm +

∑a†k,λak,λ. Let us restrict to the N = 1 sector, which contains

the state of the atom in its ground state, with a photon, and the state of the atom

excited, without the photon.

Let us begin with the state of the atom in its ground state, with one photon,

|Φi〉 = |~k, λ〉 = a†~k,λ|0〉. (3.101)

Here |0〉 includes the ground state of the atom. The scattering state obeys Lippman-

Schwinger equation,

|Ψ+i 〉 = |~k, λ〉+

1

E −H0 + iεHint|Ψ+

i 〉, (3.102)

where|Ψ+

i 〉 =∑k′,λ′

χk′,λ′a†k′,λ′|0〉+

∑m

Cmb†m|0〉. (3.103)

We then have simply a linear equation on the coefficients χk,λ and Cm. First, compute

Hint|Ψ+i 〉 = −igL−

32

∑k′,λ′,m

χk′,λ′ek′,λ′ · e∗m√

2k′b†m|0〉+ igL−

32

∑k′,λ′,m

Cme∗k′,λ′ · em√

2k′a†k′,λ′ |0〉.

(3.104)

72

Then, we derive the equations on the coefficients,

χk′,λ′ = δkk′δλλ′ + igL−32

∑m

1√2k′

Cm(e∗k′,λ′ · em)

k − k′ + iε,

Cm = −igL−32

∑k′,λ′

χk′,λ′ek′,λ′ · e∗m√

2k′1

k + E1 − E2 + iε.

(3.105)

From these we obtain an equation for Cm,

Cm = −igL−32ek,λ · e∗m√

2k

1

k + E1 − E2 + iε+ g2L−3

∑k′,λ′,m′

Cm′

2k′(ek′,λ′ · e∗m)(e∗k′,λ′ · em′)

(k + E1 − E2 + iε)(k − k′ + iε),

(3.106)

whose solution is of the form Cm = (ek,λ · e∗m)C in the L→∞ limit, with C obeying

C =1

k + E1 − E2 + iε

[−igL

− 32

√2k

+ g2C∑λ′

∫d3~k′

(2π)3

1

2k′|e∗k′,λ′ · ek,λ|2

k − k′ + iε

]. (3.107)

So we obtained the solution

C = −igL− 3

2

√2k

1

k + E1 − E2 − Σ(k) + iε,

Σ(k) = g2∑λ′

∫d3~k′

(2π)3

1

2k′|e∗k′,λ′ · ek,λ|2

k − k′ + iε,

(3.108)

and so

|Ψ+i 〉 = a†~k,λ|0〉+

g2L−3

k + E1 − E2 − Σ(k) + iε

∑k′,λ′

1

2√kk′

e∗k′,λ′ · ek,λk − k′ + iε

a†~k′,λ′ |0〉

− igL−32

√2k

1

k + E1 − E2 − Σ(k) + iε

∑m

(ek,λ · e∗m)b†m|0〉.(3.109)

Let us examine the “self-energy” Σ(k). An immediate trouble is that it appears to be

the result of a linearly divergent integral. In fact, while its imaginary part is finite,

ImΣ(k) = −πg2∑λ′

∫d3~k′

(2π)3

|e∗k′,λ′ · ek,λ|2

2k′δ(k − k′)

= − πg2k

2(2π)3

∑λ′

∫d2k′|e∗k′,λ′ · ek,λ|2

= − πg2k

2(2π)3

∫d2k′

[1− |k′ · ek,λ|2

]= −g

2k

6π,

(3.110)

73

its real part diverges linearly,

ReΣ(k) =g2

6π2

∫ ∞0

dk′k′

2

(1

k − k′ + iε+

1

k − k′ − iε

)=

g2

6π2P

∫ ∞0

dk′k′

k − k′.

(3.111)

Here P stands for the principal part of the integral. Note however that divergence comes

from large k. We used the dipole approximation to derive our toy model Hamiltonian,

which is no longer valid at large k. For now we will sweep this difficult under the rug,

by assuming an ultraviolet cutoff at energy/frequency Λ that is large compared to the

resonance energy E2−E1, but still small when multiplied by g2 and does not shift the

resonance energy significantly. A more systematic treatment of this problem will be

given in the next section.

Let us then write

Σ(k) = ∆− i

2Γ, (3.112)

with ∆ = ReΣ(k) and Γ = −2ImΣ(k) = g2k3π

. The scattering amplitude is

Tfi = 〈Φf |Hint|Ψ+i 〉

= C〈0|ak′,λ′Hint

∑m

(ek,λ · e∗m)b†m|0〉

=g2

2kL3

e∗k′,λ′ · ek,λk + E1 − E2 − Σ(k)

.

(3.113)

The differential cross section is

dσ

dΩ=

g4

16π2

∣∣∣∣ e∗k′,λ′ · ek,λk + E1 − E2 − Σ(k)

∣∣∣∣2=

g4

16π2

|e∗k′,λ′ · ek,λ|2

(k + E1 − E2 −∆)2 + 14Γ2.

(3.114)

At the resonance frequency k = k∗ = E2 − E1 + ∆, the cross section is enhanced to

dσresdΩ

=9|e∗k′,λ′ · ek,λ|2

4k2∗

,

σtotres =6π

k2∗,

(3.115)

where in the total cross section we have summed over final helicity λ′. Note that

the coupling parameter g has dropped out in this expression, and the result is much

larger than the cross section away from the resonance energy. This may be compared

to the unitarity bound on the `-wave cross section in non-relativistic scattering, σ` =πk2 (2` + 1)|η` − 1|2 ≤ 4π

k2 (2` + 1). In our toy model with the truncated Hamiltonian,

74

only the ` = 1 photon scatters. While the non-relativistic bound on the cross section

doesn’t apply to the scattering of relativistic photons, we see that it is of the same

order as the resonance cross section.

Let us examine the decay of the excited state |2p,m〉 = b†m|0〉. Suppose the system

is in this state at time t = 0. After time t, the amplitude for the atom to remain in

this state isA(t) = 〈0|bme−iHtb†m|0〉

=∑f

〈0|bm|Ψ+f 〉e

−ikt〈Ψ+f |b†m|0〉

= g2∑λ

∫d3~k

(2π)32k

|e∗m · ek,λ|2e−ikt

|k + E1 − E2 − Σ(k)|2

=g2

6π2

∫ ∞0

kdke−ikt

|k + E1 − E2 − Σ(k)|2

(3.116)

Recall that

Σ(k) =g2

6π2

∫ Λ

0

dk′k′

k − k′ + iε, (3.117)

where Λ is a UV cutoff, which we assume to be independent of k for now.

If we assume the resonance is sharp and approximate Σ(k) = ∆−iΓ/2 by a constant

(its value at resonance energy), we may write

A(t) ' Γ

2π

∫ ∞0

dke−ikt

(k − k∗)2 + Γ2

4

' Γ

2π

∫ ∞−∞

dke−ikt

(k − k∗)2 + Γ2

4

= e−ik∗t−12

Γt

(3.118)

where in the second step we have extended the integration range to k ∈ (−∞,∞).

This gives the classical exponential decay behavior of the excited atom state.

In making the above approximation, we have thrown away terms that would con-

tribute to a “power-law tail” that modifies the exponential decay behavior at late times.

A better treatment is as follows. Define

G(z) = z + E1 − E2 −g2

6π2

∫ Λ

0

dk′k′

z − k′. (3.119)

G(z) is an analytic function in z away from the positive real z-axis (as Λ → ∞). We

may write A(t) as

A(t) =i

2π

∫ ∞0

dke−ikt[

1

G(k + iε)− 1

G(k − iε)

]=

∫C

dz

2πi

e−izt

G(z)

(3.120)

75

where C is a contour in the complex z-plane that encircles the positive real z-axis in

counterclockwise direction. Let us divide it into two parts, C+ which lies along R+ + iε

(going to the left), and C− which lies along R+ − iε (going to the right).

For t > 0, the integrand falls off exponentially as Imz → −∞. It is useful to deform

the contour C to a contour that begins and ends at z = −i∞, say by deforming C−and C+ to iR−. In order to do so, however, we need to deform the C+ part of C into a

second Riemann sheet of the function G(z). Let us denote the first Riemann sheet by

R1 and the second Riemann sheet by R2. On R1, we have

ImG(z) = Imz

[1 +

g2

6π2

∫ Λ

0

dk′k′

(Rez − k′)2 + (Imz)2

](3.121)

In particular we see that 1/G(z) has no poles on R1. So we are free to deform C− to

the contour running from z = 0 to z = −i∞ on R1, while no poles are crossed. The

resonance pole z = z∗ = E1−E2 + ∆− i2Γ is in fact on the second Riemann sheet R2.

At weak coupling g, this is the only pole of relevance. When C+ moves pass z = z∗,

the integral picks up the residue. So we can write

A(t) =e−iz∗t

G′2(z∗)+

∫ ∞0

dy

2πe−yt

[1

G1(−iy)− 1

G2(−iy)

], (3.122)

where G1(z) and G2(z) stand for the function G(z) on the two sheets R1 and R2. The

contribution from the residue at z∗ to the survival amplitude is

1

1− Σ′(z∗)e−i(E2−E1+∆)te−

12

Γt (3.123)

where Σ′(z∗) is of order g2 ln Λ. This gives an exponentially decaying survival proba-

bility P (t) = |A(t)|2 ' e−Γt. There is, however, a second contribution to A(t). In the

weak coupling limit, it goes like∫ ∞0

dy

2πe−yt

G2(−iy)−G1(−iy)

(iy + E2 − E1 + Σ)2

=g2

6π2

∫ ∞0

dyye−yt

(iy + E2 − E1 + Σ)2

' g2

6π2(E2 − E1 + Σ)2

1

t2, t 1

E2 − E1

.

(3.124)

This term modifies the exponential decay law at late time. Note that the 1/t2 behavior

is a consequence of the masslessness of the photon, and does not hold for decay into

massive particles.

76

3.6 Comment on oscillation vs decay

In quantum mechanical systems with discrete energy levels, perturbing the Hamiltonian

by interactions that allow transitions between the levels lead to oscillatory behavior. In

systems with a continuous spectrum, typically due to asymptotic free particle states,

the states can decay as we have seen. Let us explore the two situations. We will

consider a simple model that consists one “excited state” |1〉, and a number of states

|0, k〉 it can decay into. Think of |1〉 as an atomic state with no photons, and |0, k〉states with one photon, for instance. With respect to the unperturbed Hamiltonian,

|1〉 has energy zero and |0, k〉 has energy νk. Now introduce transition matrix elements

〈0, k|Hint|1〉 = gk, and similarly its Hermitian conjugate. A general state

|ψ〉 = C(t)|1〉+ χk(t)|0, k〉 (3.125)

obeys the time evolution

iC(t) =∑k

g∗kχk(t),

iχk(t) = νkχk(t) + gkC(t).

(3.126)

A useful way to find and represent the solution is through Laplace transform

C(s) =

∫ ∞0

e−stC(t)dt, (3.127)

and similarly χ(s). For t > 0, the integral converges absolutely if Re(s) > 0, and

defines an analytic function in s at least when Re(s) > 0. The inverse transform is

given by

C(t) =

∫I

ds

2πiestC(s). (3.128)

where the contour I goes along ε+ iR, where ε is taken to be an infinitesimal positive

parameter.

By Laplace transforming the time evolution equations for C(t) and χk(t), we hae∫ ∞0

e−stiC(t)dt = ie−stC(t)

∣∣∣∣∞0

+ is

∫ ∞0

e−stC(t)dt

= −iC(0) + isC(s)

=∑k

g∗kχk(s).

(3.129)

and similarly

−iχk(0) + isχk(s) = νkχk(s) + gkC(s). (3.130)

77

Now let us study the evolution of |1〉 by setting C(0) = 1, χk(0) = 0 for all k. Then

sC(s) = 1− i∑

g∗kχk(s) = 1−∑ |gk|2

s+ iνkC(s), (3.131)

and we can solve

C(s) =1

s+Q(s), Q(s) ≡

∑k

|gk|2

s+ iνk. (3.132)

Laplace transforming back, we find

C(t) =

∫I

ds

2πi

est

s+Q(s). (3.133)

Let us consider two situations.

I) Suppose |0, k〉 are a set of discrete energy levels (of the unperturbed Hamiltonian).

We can try to deform the contour I to s→ −∞, where the integrand goes to zero. In

doing this deformation, the contour integral picks up contributions from the residues

at poles, located at solutions of

s+Q(s) = s+∑k

|gk|2

s+ iνk= 0. (3.134)

The solutions are at purely imaginary values of s. For example, if there is only one

|0, k〉 state of energy ν, then the pole is located at

s+|g|2

s+ iν= 0, s = −i

(ν

2±√ν2

4+ g2

). (3.135)

The resulting survival amplitude C(t) is the superposition of a periodic functions with

a discrete set of periodicity in time. If the periodicities are integer fractions of some

common period, then C(t) would be a period function in time. Even if the periodicities

are not integer fractions of some common period, as long as they are a discrete set,

C(t) will be a quasi-periodic function in time, in the sense that it will repeatedly come

back to close to 1 at late times.

II) Suppose |0, k〉 are a continuum of energy levels, with some density of states ρ(k).

Q(s) is now expressed as an integral

Q(s) =

∫|g(k)|2ρ(k)dk

s+ iν(k). (3.136)

Generally, this function will have branch cuts along the imaginary s-axis. It is not

straightward to deform the contour I to the left complex s-plane due to the branch

cut. As we have seen in the previous section, one may often deform the contour I pass

78

the imaginary s-axis, but to different analytic sheets, and pick up residues at poles as

well as possibly some remaining integration along cuts.

In the case of very weak coupling g, Q(s) is small and we may assume that the

poles occur at small s. Since Q(s) is defined by analytic continuation from Re(s) > 0,

for very small s, Q(s) acquires a real part

ReQ(ε) = π

∫δ(ν(k))|g(k)|2ρ(k)dk (3.137)

The residue at the pole s∗ ' −Q(ε) then gives the exponentially decaying behavior

C(t) ' es∗t (3.138)

Note that s∗ has negative real part, and 2ReQ(ε) is the decay width.

This exponential decay behavior is typically modified at very late times, as we seen

in the previous section, and also at very early times. In fact, if we expand the time

evolution equations around t = 0 (with the same initial condition), we find

iχk(t) = gk +O(t), iC(t) = g∗kχk(t) +O(t2), (3.139)

and so

C(t) = 1− 1

2

∑k

|gk|2t2 +O(t3). (3.140)

If the decay were exactly exponential, C(t) would be a linear function in t near t = 0,

which is not the case. This is what gives rise to the “quantum Zeno effect”.

3.7 Lamb shift

We have seen in the previous section that the frequency of the photon emitted in the

spontaneous decay of the atom is shifted by ReΣ(k) = ∆. This correction to the atomic

energy levels is due to emission and absorption of virtual photons. Given the interaction

Hamiltonian Hint that couples the atom to EM field, the shift of energy levels up to

second order in perturbation theory is given by (in the case of nondegenerate levels)

∆En = 〈n|Hint|n〉+∑i 6=n

〈n|Hint|i〉〈i|Hint|n〉En − Ei (3.141)

with

Hint = H1 +H2 = − e

m~p · ~A(~r) +

e2

2m~A(~r)2. (3.142)

For an atom bound state |n〉 (with no external photons), 〈n|Hint|n〉 = 〈n|H2|n〉 =e2

2m〈n| ~A2(~r)|n〉. The only contribution comes from a commutator of a with a†, which

79

would be a zero point energy. This zero point energy is unphysical and can be removed

by a redefinition of the Hamiltonian of electron-photon coupling. We will use from now

on

Hint = − e

m~p · ~A(~r) +

e2

2m: ~A(~r)2 : (3.143)

where the normal ordering removes the zero point energy. Now 〈n|Hint|n〉 = 0, and we

only need to consider the contribution at second order in perturbation theory. Note that

though (3.141) was derived assuming nondegenerate energy levels, the same conclusion

(i.e. only the second term contributes) holds when there are degenerate atomic energy

levels.

At order O(e2), the correction to the energy level comes from intermediate state

|i;~kλ〉 (3.144)

where i labels an atomic energy level and |~kλ〉 denotes a one-photon state of momentum~k and polarization λ. We have the matrix element

〈i,~kλ|Hint|n〉 = − e

m

L−32

√2ωk〈i|~p · e∗k,λe−i

~k·~r|n〉 ≡ L−32fi,n(k) (3.145)

and therefore

∆En = L−3∑i

∑k,λ

|fi,n(k)|2

En − Ei − ωk. (3.146)

Note that the sum over i now includes the case i = n, as i denotes only the atom state

here.

In the continuum limit, this is

∆En =∑i

∫d3~k

(2π)32ωk

e2

m2

∑λ |〈i|e∗k,λ · ~pe−i

~k·~r|n〉|2

En − Ei − ωk

=∑i

∫ ∞0

dk

En − Ei − ke2k

(2π)32m2

∑λ

∫dΩk|〈i|e∗k,λ · ~pe−i

~k·~r|n〉|2(3.147)

As we have seen before in scattering theory, the integration over k should be regularized

using the iε prescription, and the resulting ∆En is complex, with its real part giving the

shift of the energy level and its imaginary part the decay width. Here we are interested

in the real part of ∆En, which is given by the principal part of the integral.

The low energy contribution can be approximated using the dipole approximation,

as before. Using ∑λ

∫dΩk|〈i|e∗k,λ · ~p|n〉|2 =

8π

3|〈i|~p|n〉|2, (3.148)

80

we obtain (in the dipole approximation)

∆En =2α

3πm2

∑i

|〈i|~p|n〉|2P∫ ∞

0

kdk

En − Ei − k (3.149)

where α = e2/4π. While the integral appears to diverge linearly, note that the correc-

tion is present even for a free electron! For a free electron with nonrelativistic dispersion

relation Efree(p) = p2/2m, we would have (the dipole approximation is not valid in

this case)

∆Efree(p) =α

4π2m2P

∫ ∞0

kdk∑~q

∑λ

∫dΩk

|〈~q|e∗k,λ · ~qe−i~k·~r|~p〉|2

p2

2m− q2

2m− k

=α

4π2m2P

∫ ∞0

kdk∑~q

∑λ

∫dΩk

|e∗k,λ · ~p δ~q,~p−~k|2p2

2m− q2

2m− k

=α

4π2m2P

∫ ∞0

kdk∑λ

∫dΩk

|e∗k,λ · ~p|2

p2

2m− (~p−~k)2

2m− k

(3.150)

The integration over k is valid only in the nonrelativistic regime, k m. For p

comparable to the typical momentum of the electron in the atomic bound state, we

also have p m. The denominator is approximated by

p2

2m− (~p− ~k)2

2m− k =

~p · ~km− k2

2m− k ' −k (3.151)

We have then

∆Efree(p) ' −2α

3πm2p2

∫ Λ

0

dk = − 4αΛ

3πm

p2

2m(3.152)

where the cutoff scale Λ should be taken to be of order the electron rest energy m.

This energy shift amounts to a renormalization of the electron mass.

The observable shift of atomic energy level therefore is given by the difference

between ∆En and the expectation value of ∆Efree in the state |n〉, at leading nontrivial

order in perturbation theory. We have

∆En =2α

3πm2P

∫ ∞0

dk

[∑i

k|〈i|~p|n〉|2

En − Ei − k+ 〈n|~p2|n〉

]

=2α

3πm2P

∫ ∞0

dk∑i

(En − Ei)|〈i|~p|n〉|2

En − Ei − k

(3.153)

We see that the linear divergence is canceled, leaving a logarithmically divergent inte-

gral. Using

P

∫ Λ

0

dk

x− k= − ln

Λ

|x|, (3.154)

81

we have

∆En = − 2α

3πm2

∑i

(En − Ei)|〈i|~p|n〉|2 lnΛ

|En − Ei| (3.155)

Now we make use of the identity∑i

(En − Ei)|〈i|~p|n〉|2 = 〈n|~p · [~p,HM ]|n〉 =1

2〈n|[pi, [pi, HM ]]|n〉

= −1

2〈n|∇2V |n〉 = −1

2Ze2〈n|δ3(~r)|n〉 = −1

2Ze2|ψn(0)|2.

(3.156)

Since ln |En − Ei| varies slowly with the energy levels, we see that the shift of energy

∆En occurs primarily for s-wave states and is much smaller for ` 6= 0 states (whose

wave function vanishes at the origin). To compute the energy difference between 2s1/2

and 2p1/2, we mainly need to compute the shift of energy of 2s1/2 due to radiative

corrections. The quantity

ln〈|En − Ei|〉 =

∑i (En − Ei)|〈i|~p|n〉|2 ln |En − Ei|∑

i (En − Ei)|〈i|~p|n〉|2(3.157)

depends only on the energy spectrum of the non-relativistic atom. It can be evaluated

by summing over many atomic states, and the result for n = 2s is

〈|E2s − Ei|〉 ' 16.64Ry = 226.3eV (3.158)

The cutoff scale should be of order Λ ∼ m ' 0.511MeV, which is significantly larger.

To determine the precise contribution from energies near m requires a fully second

quantized theory of relativistic electrons. Nonetheless, we can make a good estimate

since the cutoff dependence is logarithmic. We have

lnm

〈|E2s − Ei|〉' 7.72. (3.159)

The resulting energy shift of 2s level is

∆E(2s) = − α

3πm2e2|ψ2s(0)|2 × 7.72 (3.160)

Using

ψ2s(r) =a−3/20

2√

2π(1− r

2a0

)e− r

2a0 , (3.161)

we arrive at

∆E(2s) = − α2

3πm24π(

a−3/20

2√

2π)2 × 7.72 = 7.72

mα5

6π' 1051MHz. (3.162)

This rough estimate is remarkably close to the experimental value 1057MHz.

82

4 Multi-particle system

We have already seen systems of identical particles in the context of scattering of two

identical particles, and in the quantization of electromagnetic field. Now we will give

a systematic treatment of (non-relativistic) system of identical particles. Consider

a system of N identical particles, with some Hamiltonian H. The wave function is

written Ψ(x1, · · · , xN). To say that the particles labelled by i and j are identical

means that there is a symmetry Pij interchanging i with j that commutes with the

Hamiltonian H, such that P 2ij = 1. Pij can have eigenvalute +1 or −1, corresponding

to Bose-Einstein or Fermi-Dirac statistics, and the particles obeying such statistics

are bosons or fermions. In relativistic quantum field theory, it is a consequence of

Lorentz invariance, causality, and unitarity that all particles with integer spins are

bosons, while all particles with half integer spins are fermions. Within the framework

of nonrelativistic quantum mechanics, we can only take this spin-statics relation as

given.

4.1 Bose-Einstein statistics

We will enlarge the Hilbert space by including not just states with N particles at

given N , but all non-negative integer N . Beginning with the vacuum state |0〉 with

no particles, we act with creation operator a†p to generate states with N particles of

momenta p1, p2, · · · , pN . The creation and annihilation operators a†p and ap obey the

commutation relation

[ap, aq] = [a†p, a†q] = 0, [ap, a

†q] = δpq. (4.1)

The vacuum is annihilated by all ap’s. In the case of discrete momenta, we need to

include appropriate normalization factors so that the resulting state has unit norm.

The state with np particles at momentum p is

|np〉 =∏p

(a†p)np√np!|0〉. (4.2)

To see that this description is equivalent to a completely symmetric wave function,

consider a state

|p1, p2, · · · , pN〉 = a†p1a†p2· · · a†pN |0〉. (4.3)

Here the pi’s may or may not be equal, and so the state vector is not necessarily nor-

malized. Given some N -particle state |Ψ〉, the N -particle wave function in momentum

space is

Ψ(p1, · · · , pN) = 〈p1, p2, · · · , pN |Ψ〉 (4.4)

83

which is, of course, completely symmetric in p1 through pN .

We now define the analogous annihilation and creation operators in position space,

by Fourier transforming ap and a†p,

ψ(x) =1√V

∑p

eip·xap, ψ†(x) =1√V

∑p

e−ip·xa†p. (4.5)

Here V is the volume of the space. The volume factor is included so that the state

|x〉 = ψ†(x)|0〉 (4.6)

is delta function normalized. ψ and ψ† obey the commutation relation

[ψ, ψ] = [ψ†, ψ†] = 0, [ψ(x), ψ†(x′)] = δ(~x− ~x′). (4.7)

The particle number operator is

N =∑p

a†pap =

∫d~xψ†(x)ψ(x). (4.8)

The position space wave function for an N -particle state |Ψ〉 is

Ψ(x1, · · · , xN) =1√N !〈0|ψ(x1) · · ·ψ(xN)|Ψ〉. (4.9)

The normalization is such that

〈Ψ|Ψ〉 =

∫d~x1 · · · d~xN |Ψ(x1, · · · , xN)|2. (4.10)

The position eigenstates can be defined (with appropriate normalization) as

|x1, · · · , xn〉 =1√n!ψ†(x1) · · ·ψ†(xn)|0〉. (4.11)

Be aware that with this normalization, the action of position space creation and anni-

hilation operators are

ψ†(y)|x1, · · · , xn〉 =√n+ 1|y, x1, · · · , xN〉,

ψ(y)|x1, · · · , xn〉 =1√n

n∑i=1

δ(~xi − ~y)|x1, · · · , xi−1, xi+1, · · ·xN〉(4.12)

As an example, consider an N -particle system, with each particle obeying non-

relativistic dispersion relation E = p2/2m, and pairwise interaction potential energy

V (|~x|) where ~x is the separation between the two particles. The total Hamiltonian in

terms of the position space creation and annihilation operators is written as

H =

∫dxψ†(x)〈x| p

2

2m|x′〉ψ(x′) +

1

2

∫dx1dx2dx

′1dx

′2ψ†(x1)ψ†(x2)〈x1, x2|V |x′1, x′2〉ψ(x′1)ψ(x′2)

=

∫dxψ†(x)

−∇2

2mψ(x) +

1

2

∫dx1dx2ψ

†(x1)ψ†(x2)V (|x1 − x2|)ψ(x2)ψ(x1).

(4.13)

84

4.2 Fermi-Dirac statistics

The wave function of identical fermions is antisymmetric under the exchange of two

particles. Take a basis of states in which each particle has a definite momentum p and

spin s. Denote them by ν = (p; s). The vacuum (no particles) is annihilated by all

annihilation operators bν , i.e. bν |0〉 = 0 for all ν. For the N -particle state

|ν1, · · · , νN〉 = b†ν1· · · b†νN |0〉 (4.14)

to be antisymmetric under the exchange of a pair of momenta, the creation operators

should anti-commute among themselves. And similarly, the annihilation operators

anti-commute with themselves. For the state b†ν |0〉 to have unit norm, we need

〈0|bνb†ν′|0〉 = δν,ν′ . (4.15)

One may assume a commutation relation of the form bνb†ν′ + cb†ν′bν = δν,ν′ , for some

number c, but this relation would be consistent with the anti-commuting property of

bν (or b†ν) among themselves only for c = 1. So the commutation relation involving a

creation and an annihilation operator for the fermion is

bν , b†ν′ = δν,ν′ . (4.16)

The fermion number operator (for particles of momentum ν) is

Nν = b†νbν = 1− bνb†ν . (4.17)

It obeys the standard relations

[Nν , bν′ ] = −δν,ν′bν′ , [Nν , b†ν′ ] = δν,ν′b

†ν′ . (4.18)

Note that N2ν = b†νbνb

†νbν = b†νbν , b†νbν = Nν , and so the eigenvalues of Nν are 0 and

1. As in the boson case, we can introduce the position space annihilation and creation

operators (field operators) by

ψs(x) =∑ν

bν〈x; s|ν〉, ψ†s(x) =∑ν

b†ν〈ν|x; s〉. (4.19)

They obey anti-commutation relations

ψs(x), ψs′(x′) = 0, ψs(x), ψ†s′(x

′) = δss′δ(~x− ~x′). (4.20)

The Hamiltonian governing N fermions with say pairwise potential interaction can be

constructed analogously to the bosonic case.

85

4.3 The equation of motion for the field operator

Let us consider the equation of motion for the field operator in Heisenberg picture,

ψ(x, t) = eiHtψ(x)e−iHt. (4.21)

The equal time commutator between the field operators are the same as in the case of

Schrodinger picture operators. Suppose the Hamiltonian is of the form

H = K + U + V

=

∫dxψ†(x)

[−∇2

2m+ U(x)

]ψ(x) +

1

2

∫dx1dx2ψ

†(x1)ψ†(x2)V (|x1 − x2|)ψ(x2)ψ(x1).

(4.22)

where K, U , V are the kinetic energy, the one-body potential, and the two-body

interactions. The equation of motion is

iψ(x, t) = [ψ(x, t), H]

=

[−∇2

2m+ U(x)

]ψ(x, t) +

∫dyψ†(y)V (|x− y|)ψ(y)ψ(x).

(4.23)

This formula applies to both the bosonic and fermionic case. Superficially, it takes

the form of a nonlinear generalization of the Schrodinger equation. The last term

represents the effective potential due to interactions between pairs of particles.

4.4 Ideal gas

A quantity of interest in thermodynamics is the partition function

Z = Tr e−β(H−µN) (4.24)

where β = (kT )−1, k being Boltzmann’s constant and T the temperature, and µ is

the chemical potential. In the grand canonical ensemble, the thermal density matrix

is given by

ρ =1

Ze−β(H−µN) (4.25)

For a system of free particles, the Hamiltonian is diagonalized by states of definite

particle numbers and each particle having a definite energy. We will label the one-

particle states by ν. ν may stand for the momentum and possibly spin/polarization in

both bosonic and fermionic cases. The energy of a particle in the state ν is ε(ν). The

partition function for free particles then factorizes into products of partition functions

of particles occupying states labelled by ν, namely

ZB(β, µ) =∏ν

∞∑n=0

e−βn(ε(ν)−µ) =∏ν

1

1− e−β(ε(ν)−µ)(4.26)

86

for bosons, and

ZF (β, µ) =∏ν

1∑n=0

e−βn(ε(ν)−µ) =∏ν

[1 + e−β(ε(ν)−µ)

](4.27)

for fermions.

The average particle number in the state ν is easily obtained from the partition

function,

〈Nν〉 =TrNνe

−β(H−µN)

Tr e−β(H−µN)

= ∓ 1

β

∂

∂µln(1∓ e−β(ε(ν)−µ))

=1

eβ(ε(ν)−µ) ∓ 1

(4.28)

where the − and + sign are for the boson and fermion cases, respectively. A crucial

difference is that the ground state of the bosonic system has all particles in the one-

particle ground state, whereas the ground state of the fermionic system has fermions

filling up one-particle states up to some energy εF , the Fermi energy.

4.5 Mean field approximation

The system of many identical particles becomes very complicated when interactions

between the particles are introduced. In this course we will only explore the mean field

method. Let us begin with the bosonic case. The ground state of a non-interacting Bose

gas has all particles in their respective ground states, i.e. they form the Bose-Einstein

condensate. Such a state is very sensitive to perturbations that introduce interactions

between particles, for the following reason. Suppose the interaction is attractive and

has range R. Then the ground state should correspond to the system of N -particles

collapsing into a volume of size R. With a sufficiently large number of particles, even

with weak interactions, the interaction energy scales like N2, dominating the kinetic

energy which scales like N . In the opposite situation, suppose the interaction repulsive,

and the system is confined in a box of size R so that the wave function vanishes at the

boundary of the box. If one could ignore the interactions, the ground state would be

described by the product of N one-particle ground state wave function which varies

over distance R, independently of N . When a repulsive interaction is present, this

cannot be the case as the particles should tend to a uniform distribution, at least away

from the boundary of the box.

This is in contrast to the case of identical fermions. Consider a simple example of

N fermions in a one-dimensional box of size L. We will ignore the spin here, as the

87

effect of spin or any other internal quantum numbers is simply to introduce further

degeneracy of energy eigenstates. The one-particle energy eigenstates are described by

the wave function

ψn(x) =

√2

Lsin

nπx

L, n = 1, 2, 3, · · · . (4.29)

The ground state of N fermions has the wave function

Ψ(x1, · · · , xN) =1√N !

∑σ∈SN

(−)σN∏n=1

ψn(xσ(n)) (4.30)

corresponding to each fermion occupying one of the first N energy levels. The density

of fermions in space is

N∑n=1

|ψn(x)|2 =N + 1

2

L− 1

2L

sin( (2N+1)πxL

)

sin(πxL

)(4.31)

In the limit of large N,L with the ratio N/L fixed, this is close to being a uniform

density (away from the boundary of the box). As opposed to the bosonic case, for

the fermions adding weak interactions does not drastically change the behavior of the

ground state.

We will now investigate the ground state of a system of many identical particles

with interactions. Let us first consider the bosonic case, and consider a system of

bosons with delta function interaction potential,

V (r12) =4πa

mδ(~r1 − ~r2) (4.32)

where a is the scattering length and m the mass of the particle. Note that if we

consider scattering between two particles in the center of mass frame, m/2 would be

the reduced mass. In the Born approximation, the scattering amplitude would be a

constant fB = −a in the center of mass frame and the total cross section 4πa2. Note

however that V is a pseudopotential in the sense that it can only be used in the first

order Born approximation, and would lead to divergent results at the next order in

perturbation theory.

In the dilute gas limit, namely the limit in which the mean distance between the

particles d is much greater than the scattering length a, we can work with the pseu-

dopotential and write the interaction term in the second quantized Hamiltonian as

V =2πa

m

∫d~xψ†(x)ψ†(x)ψ(x)ψ(x). (4.33)

This approximation is consistent only for a > 0, which we assume for now. Let φ(x) be

the one-particle ground state wave function, and uν(x) an orthogonal basis of excited

88

one-particle wave functions. The Bose field operator can be expanded on this basis as

ψ(x) = φ(x)c+∑ν

uν(x)bν . (4.34)

The mean field approximation assumes that the ground state can be described by each

particle moving in a mean field due to the interaction with the collection of other

particles. The ground state by hypothesis takes the form

|G〉 =1√N !

(c†)N |0〉. (4.35)

The ground state energy in this approximation is

〈G|H|G〉 =

∫d~x

[− N

2mφ∗(x)∇2φ(x) +NU(x)|φ(x)|2 +

2πa

mN(N − 1)|φ(x)|4

](4.36)

where we have used 〈G|c†c|G〉 = N and 〈G|(c†)2c2|G〉 = N(N − 1). We will minimize

this energy expectation value by varying the trial wave function φ(x). Define the

condensate wave function

Ψ(x) =√Nφ(x), (4.37)

and we can write the “ground state” energy functional in the large N limit as

E [Ψ] =

∫d~x

[− 1

2mΨ∗(x)∇2Ψ(x) + U(x)|Ψ(x)|2 +

2πa

m|Ψ(x)|4

](4.38)

To find the ground state wave function, we minimize E [Ψ] with respect to Ψ (and Ψ∗),

while keeping∫d~x|Ψ|2 = N . This is achieved by extremizing

E [Ψ]− µ[∫

d~x|Ψ(x)|2 −N]

(4.39)

with respect to an unconstrained Ψ(x), and with respect the Lagrangian multiplier

µ which imposes the constraint the total particle number (norm of Ψ). Varying this

functional with respect to Ψ∗ yields the Gross-Pitaevskii equation

− 1

2m∇2Ψ(x) + U(x)Ψ(x) +

4πa

m|Ψ(x)|2Ψ(x) = µΨ(x). (4.40)

Suppose the potential U(x) vanishes. There may be walls where we impose certain

boundary conditions on Ψ(x). Far away from the boundary of U = 0 region, there is

a constant (modulus) solution

|Ψ(x)|2 =mµ

4πa= n. (4.41)

89

where n is the particle density. To understand the behavior near the boundary (say a

wall), it suffices to consider the one-dimensional equation

Ψ′′(x) + 8πaΨ(n− |Ψ|2) = 0, (4.42)

subject to the boundary condition Ψ(0) = 0, Ψ(∞) =√n. The solution is given by

Ψ(x) =√n tanh

(√4πanx

). (4.43)

We see that it approaches the constant value√n for x 1/

√4πan = d

32/√

4πa, where

d is the mean distance between particles. Recall that the dilute gas approximation is

valid when d a.

Now let us consider the fermionic version of the mean field approximation. We will

assume that the ground state can be expressed as the antisymmetric product of N

one-particle wave functions,

φ1(ν), φ2(ν), · · · , φN(ν), (4.44)

where ν denotes the position (or momentum), spin, and possibly other quantum num-

bers of the fermion. The φi’s are assumed to be orthonormal, and we want to minimize

the energy functional with respect to the φi’s. Let us expand the fermion field operator

in the formψ(ν) =

∑i

φi(ν)bi +∑α

uα(ν)bα. (4.45)

where the b’s are the annihilation operators. The trial ground state takes the form

|Φ〉 =N∏i=1

b†i |0〉. (4.46)

The Hamiltonian can be written as

H =

∫dνψ†(ν)Fψ(ν) +

1

2

∫dµdνψ†(µ)ψ†(ν)V ψ(ν)ψ(µ) (4.47)

where F is an operator acting on one-body wave functions and V is an operator acting

on two-body wave functions. The expectation value of H in the state |Φ〉 is

〈Φ|H|Φ〉 =∑i,j

〈i|F |j〉〈Φ|b†ibj|Φ〉+1

2

∑i,j,k,`

〈i, j|V |k, `〉〈Φ|b†jb†ibkb`|Φ〉. (4.48)

The notation here requires some explanation. |i〉, |j〉 etc. stand for one-particle states

given by the wave functions φi, φj, etc. For instance,

〈i|F |j〉 =

∫dνφ∗i (ν)F φj(ν). (4.49)

90

|Φ〉 on the other hand is a vector in the Fock space built out of the fermion creation

operators. In the second quantized formalism, the Fock space is the physical Hilbert

space whereas the space of |i〉’s is only an auxiliary space of one particle wave functions.

Using〈Φ|b†ibj|Φ〉 = δij,

〈Φ|b†jb†ibkb`|Φ〉 = δikδj` − δi`δjk,

(4.50)

we can write the energy functional as

〈Φ|H|Φ〉 =∑i

〈i|F |i〉+1

2

∑i,j

[〈i, j|V |i, j〉 − 〈i, j|V |j, i〉

](4.51)

We now want to minimize this energy functional with respect to |1〉, · · · , |N〉, or their

one-body wave functions φ1, · · · , φN subject to the constraint that each φi is normalized

with∫dν|φi(ν)|2 = 1. This is achieved by considering an energy functional with N

Lagrangian multipliers,

E [Φ] = 〈Φ|H|Φ〉 −N∑i=1

εi

[∫dν|φi(ν)|2 − 1

](4.52)

Extremizing E [Φ] gives the Hartree-Fock equations

δ

δφ∗k(ν)〈Φ|H|Φ〉 = εkφk(ν). (4.53)

Using ν-eigenbasis |ν〉, we can write 〈Φ|H|Φ〉 as

〈Φ|H|Φ〉 =∑i

∫dνφ∗i (ν)F φi(ν) +

1

2

∑i,j

∫dµdν

[φ∗i (µ)φ∗j(ν)V φj(ν)φi(µ)− φ∗i (µ)φ∗j(ν)V φi(ν)φj(µ)

](4.54)

And so the Hartree-Fock equations takes the form

F φk(ν) +∑i

∫dµ[φ∗i (µ)V φk(ν)φi(µ)− φ∗i (µ)V φi(ν)φk(µ)

]= εkφk(ν) (4.55)

As an application, take the system of Z-electron atom. ν = (~r, s), and F and V are

given by (in atomic units)

F =p2

2− Z

r, V =

1

r12

. (4.56)

These are understood to be operators that act on functions of ~r and functions of (~r1, ~r2),

respectively. The Hartree-Fock equations is now written as[−1

2∇2 − Z

r

]φk(~r, s) +

∫d3~r′

1

|~r − ~r′|

[∑i,s′

φ∗i (~r′, s′)φi(~r

′, s′)φk(~r, s)

−∑i,s′

φ∗i (~r′, s′)φi(~r, s)φk(~r

′, s′)

]= εkφk(~r, s).

(4.57)

91

One may then try to solve this nonlinear integral equation numerically.

5 Lattice systems

We now turn to a new subject: the study of quantum mechanical systems on a lattice.

This is an important subject with close connection to solid state physics and to quantum

field theory. A key insight and theme will be the collective excitations (quasi-particles)

arising from complex systems that often admit simpler and universal descriptions.

5.1 1D Ising Model

Let us begin with the simplest spin system, the 1D Ising model. This is a 1-dimensional

chain of particles whose only degrees of freedom are their spins, | ↑〉 and | ↓〉. The

Hamiltonian involves nearest neighbor interactions,

H = −JL∑i=1

sisi+1 (5.1)

where si acts on the i-th site as σz. The spectrum of this Hamiltonian is somewhat

trivial since H is already diagonalized by states of definite sz on each site. This

spectrum can also be described in terms of the partition function

Z = Tre−βH =∑si

eβJ∑i sisi+1

= 2∑

t1,··· ,tL−1=±1

eβJ∑i ti = 2(cosh βJ)L−1.

(5.2)

5.2 Heisenberg spin chain

Now let us turn to a much less trivial example, the Heisenberg XXX 12

spin chain.

The basis of states is the same as in the case of 1D Ising model, namely one spin-1/2

particle on each site, the spin being its only degree of freedom. For later convenience

we will consider periodic boundary condition, i.e. identify the (n+L)-th site with the

n-th site, and so in particular the L-th site interacts with both the (L− 1)-th site and

the 1st site. The Hamiltonian is

H = −L∑n=1

(~Sn · ~Sn+1 −

1

4

)(5.3)

92

where ~Sn acts on the n-th site as 12~σ. We have dropped the overall coupling constant

here, which is obviously unimportant in finding the energy spectrum. Though, what

we call the ground state will depend on the sign of the Hamiltonian. A useful fact is

that

Pn,n+1 =1

2+

1

2~σn · ~σn+1 =

1

2+ 2~Sn · ~Sn+1 (5.4)

is the permutation operator on the spins at site n and n + 1. For instance, one can

easily checkP12| ↑, ↑〉 = | ↑, ↑〉,P12| ↑, ↓〉 = | ↓, ↑〉.

(5.5)

So we can write the Hamiltonian as

H =1

2

L∑n=1

(1− Pn,n+1). (5.6)

The lowest energy states have eigenvalue 1 under Pn,n+1. In other words, the ground

states are completely symmetric under permutations of sites. One such state is

|Ω〉 = | ↓↓ · · · ↓〉. (5.7)

It has total spin S = L/2. All other ground states are obtained by SO(3) rotations of

|Ω〉. There are 2S + 1 = L+ 1 such states.

To obtain an excited state, we may begin with |Ω〉 and flip one of the spins from

| ↓〉 to | ↑〉, say consider

| ↓↓ · · · ↓↓↓↑↓↓↓ · · · ↓〉 (5.8)

This is not an energy eigenstate, however, since by acting on it with the Hamiltonian

the up spin gets moved to the left or right adjacent site. We shall make use of the

translational symmetry of the system. The operator that shifts every site to the right

by one spot can be written as

U = P1,2P2,3 · · ·PL−1,L ≡ e−iP (5.9)

where P stands for the momentum operator, not to be confused with permutation

operator Pn,n+1. Since UL = 1, the momentum takes discrete value 2πk/L, k =

0, 1, · · · , L− 1, and has periodicity 2π. Since

[U,H] = 0, (5.10)

we can simultaneously diagonalize H and U . This means that we only need to look for

energy eigenstates of definite momentum P . Another conserved quantity is the total

spin ~S, in particular Sz. Each time a spin down is flipped to spin up, a “magnon” is

93

created. The conservation of Sz implies that the number of magnons is conserved. So

now let us look for one magnon energy eigenstates. Such a state should have definite

momentum p, which then must take the form

|p〉 ≡ 1√L

L∑n=1

eipn| ↓〉1 ⊗ | ↓〉2 ⊗ · · · ⊗ | ↓〉n−1 ⊗ | ↑〉n ⊗ | ↓〉n+1 ⊗ · · · | ↓〉L. (5.11)

Indeed, we see that

U |p〉 =1√L

L∑n=1

eipn| ↓〉1 ⊗ | ↓〉2 ⊗ · · · ⊗ | ↓〉n ⊗ | ↑〉n+1 ⊗ | ↓〉n+2 ⊗ · · · | ↓〉L

= e−ip|p〉.(5.12)

Note that the momentum takes the values p = 2πk/L, k = 0, 1, · · · , L − 1. It is then

straightforward to check that |p〉 is in fact an energy eigenstate,

H|p〉 =1

2(2− eip − e−ip)|p〉 = 2 sin2(

p

2)|p〉. (5.13)

So we have learned that a single magnon obeys dispersion relation

E(p) = 2 sin2(p

2). (5.14)

In the infinite chain limit L→∞, for small momentum this looks like

E(p) ≈ p2

2, (5.15)

i.e. the magnon looks like a non-relativistic particle of mass = 1! The non-relativistic

dispersion is corrected to (5.14) at finite momentum, as it must because p is periodically

valued with periodicity 2π.

Next, let us consider excitations that involve a pair of magnons. Far away from

each other, the magnons “move” along the spin chain according to the dispersion

relation (5.14). When they come nearby, there will be scattering due to the interacting

Hamiltonian. This is in contrast with the 1D Ising model, where the energy does not

depend on the momentum of the magnon, but only the number of magnon excitations.

In that case there was no interaction between the magnons, whereas in the XXX 12

spin chain the magnons interact nontrivially.

It will be convenient to characterize the magnon states using the latticized position

space wave function, defined by

Ψ(n) =

(1〈↓ | ⊗ 2〈| ↓ | ⊗ · · · ⊗ n−1〈↓ | ⊗ n〈↑ | ⊗ n+1〈↓ | ⊗ · · · L〈↓ |

)|Ψ〉 (5.16)

94

So the one magnon state of momentum p is characterized by the wave function

Ψ(x) =1√Leipx. (5.17)

The two-magnon state can then be described in terms of a two-particle wave function

Ψ(x1, x2), which is, by definition, symmetric under the exchange of x1 with x2. As is

almost obvious, the magnons are identical bosons. A two-magnon energy eigenstate

should be described by a scattering wave function, of the form

Ψ(x1, x2) = eip1x1+ip2x2 + S(p2, p1)eip2x1+ip1x2 , x1 < x2. (5.18)

For x1 > x2, the wave function is related by symmetry, i.e. x1 ↔ x2. Since the

number of magnons is conserved, the scattering is elastic due to momentum and energy

conservation. So the two magnons simply cross each other with a phase shift. Note

that while the total momentum p1 + p2 must be an integer multiple of 2π/L, this need

not be the case for p1 and p2 separately. In fact, requiring the wave function to be

periodic, namely for x1 < x2, Ψ(x1, x2) = Ψ(x2, x1 + L), means that

eip1x1+ip2x2 + S(p2, p1)eip2x1+ip1x2 = eip2Leip1x2+ip2x1 + eip1LS(p2, p1)eip2x2+ip1x1 (5.19)

It follows that for the two-magnon scattering state, p1 and p2 must obey

S(p2, p1) = eip2L = e−ip1L. (5.20)

Let us now compute S(p2, p1). The Schrodinger equation H|Ψ〉 = E|Ψ〉 is written

explicitly on the wave function as

EΨ(x1, x2) = 2Ψ(x1, x2)− 1

2Ψ(x1 − 1, x2)− 1

2Ψ(x1 + 1, x2)− 1

2Ψ(x1, x2 − 1)− 1

2Ψ(x1, x2 + 1),

x1 ≤ x2 − 2,

EΨ(x, x+ 1) = Ψ(x, x+ 1)− 1

2Ψ(x− 1, x+ 1)− 1

2Ψ(x, x+ 2).

(5.21)

The first equation simply says that the energy of the scattering state is the sum of the

would-be energy of two separate magnons with momenta p1, p2,

E = E(p1) + E(p2) = 2 sin2(p1

2) + 2 sin2(

p2

2). (5.22)

Note however that p1, p2 do not necessarily obey the momentum quantization condition

for a single magnon. In fact, they need not be real!

It suffices to consider the second equation at x = 0, which reads

2(1− E)eip2 + 2(1− E)S(p2, p1)eip1 = e−ip1+ip2 + S(p2, p1)e−ip2+ip1 + e2ip2 + S(p2, p1)e2ip1 .

(5.23)

95

From this we solve

S(p2, p1) = −1− 2eip2 + eip1+ip2

1− 2eip1 + eip1+ip2. (5.24)

Note that S(p2, p1) is a phase when p1, p2 are real, as required by unitarity. But also

note that S(p2, p1) is no longer a phase when p1, p2 are complex. This will become

important shortly.

It is convenient and conventional to introduce the variable λ, related to the mo-

mentum p by

eip =λ+ i

2

λ− i2

, λ =1

2cot(

p

2). (5.25)

In terms of λ, the scattering phase takes a very simple form

S(p2, p1) =λ1 − λ2 − iλ1 − λ2 + i

. (5.26)

The dispersion relation is expressed in terms of λ as

E(λ) =1

2

1

λ2 + 14

. (5.27)

To determine the two-magnon spectrum on a length N chain, we need to use the

relation (5.20) (which is the two body version of the Bethe ansatz):

λ1 − λ2 + i

λ1 − λ2 − i=

(λ1 + i

2

λ1 − i2

)L=

(λ2 + i

2

λ2 − i2

)−L. (5.28)

There are two qualitatively distinct cases. First, suppose p1, p2 are real, hence λ1, λ2

are real. We have

p1 =2πk1

L− i

LlnS(p1, p2),

p2 =2πk2

L+i

LlnS(p1, p2).

(5.29)

The solution may be found perturbatively in 1/L:

p1 =2πk1

L− i

LlnS(

2πk1

L,2πk2

L) +O(

1

L2),

p2 =2πk2

L+i

LlnS(

2πk1

L,2πk2

L) +O(

1

L2).

(5.30)

What we see here are scattering states of a pair of unbound magnons, whose momentum

quantization condition is shifted due to the scattering phase.

One may think that these are all the solutions of two-magnon states, but that isn’t

the case. Suppose λ1 is complex, say with Imλ1 > 0. Then |eip1| > 1, and eip1L grows

96

exponentially in L. In the infinite chain limit L → ∞, this demands the solution to

(5.28) to obey

λ1 − λ2 = i. (5.31)

By the same reason, we also need Imλ2 = −Imλ1 in the infinite L limit, which requires

λ1 = λ 12

+i

2, λ2 = λ 1

2− i

2, (5.32)

for some real λ 12. The total momentum is

p 12

= p1 + p2 = −i lnλ 1

2+ i

λ 12− i

, (5.33)

and the energy

E = E 12

= E(λ1) + E(λ2) =1

λ212

+ 1 (5.34)

In terms of p = p 12, we have a new dispersion relation for such two-magnon states

E 12(p) = sin2(

p

2). (5.35)

In the small momentum limit, this looks like the non-relativistic dispersion relation for

a particle of mass = 2! Compared to two separate magnons carrying momentum p1, p2,

however, E 12(p) is smaller:

E 12(p1 + p2) = sin2(

p1 + p2

2) < E(

p1

2) + E(

p2

2) = 4 sin2(

p1

4) + 4 sin2(

p2

4). (5.36)

So the scattering wave function with complex p1, p2 we have found actually describes

a bound state of two magnons. The binding energy goes to zero at zero momentum.

Note that the bound state is found precisely at the pole of the (analytic continuation

of) two-body S-matrix

S(p1, p2) =λ1 − λ2 + i

λ1 − λ2 − i, (5.37)

as anticipated.

5.3 Algebraic Bethe ansatz

Now let us turn to the states of ` magnons. One may derive the Bethe ansatz equation

from the `-body wave function we have done in the case of particles scattering with

a delta function potential before. Note that one must be careful here with the cases

97

where the positions of magnons are adjacent, but everything works out as they should.

The resulting Bethe equation is

eipkL =∏j 6=k

S(pk, pj), k = 1, · · · , `, (5.38)

or in terms of the “Bethe roots” λk,(λk + i

2

λk − i2

)L=∏j 6=k

λk − λj + i

λk − λj − i. (5.39)

Guided by the intuition from the two-magnon case, we expect the ` magnons to group

into a number of bound particles scattering off one another; each of the bound particles

is a bound state of several magnons. Such a bound state may be described as follows.

Let M be an integer or a half integer. Take 2M + 1 Bethe roots λi of the form

λm = λ+ im, m = −M,−M + 1, · · · ,M. (5.40)

This would solve the Bethe equation for 2M + 1 magnons in the L → ∞ limit. The

total momentum of this state is

pM(λ) = −iM∑

m=−M

lnλm + i

2

λm − i2

= −i lnλ+ i(M + 1

2)

λ− i(M + 12), (5.41)

and the energy is

EM(λ) =1

2

M∑m=−M

1

λ2m + 1

4

=1

2

2M + 1

λ2 + (M + 12)2

=2

2M + 1sin2(

pM2

).

(5.42)

In the small momentum limit, this bound state has mass 2M + 1, i.e. the binding

energy goes to zero.

Now we would to determine the S-matrix of one magnon scattering with a type

M magnon bound state. The claim, which we have not quite justified, is that this

S-matrix (a scattering phase) is simply the product of the S-matrix with each of the

2M + 1 successive magnon “components” in the bound state. This gives

S0,M(λ, µ) =M∏

m=−M

S(λ, µ+ im)

=M∏

m=−M

λ− µ− i(m− 1)

λ− µ− i(m+ 1)

=λ− µ+ iM

λ− µ− iM· λ− µ+ i(M + 1)

λ− µ− i(M + 1).

(5.43)

98

Furthermore, we can calculate the scattering phase of a type M bound state with a

type N bound state,

SM,N(λ, µ) =M∏

m=−M

S0,N(λ+ im, µ)

=M∏

m=−M

λ− µ+ i(N +m)

λ− µ− i(N −m)· λ− µ+ i(N +m+ 1)

λ− µ− i(N −m+ 1)

=M+N∏

L=|M−N |

S0,L(λ, µ).

(5.44)

A general excited state can be characterized by νM, M = 0, 12, 1, · · · , where νM

is the number of type M magnon bound states. The total magnon charge (i.e. Szrelative to that of |Ω〉) is ` =

∑M νM(2M + 1). Each bound particle of type M has

its momentum pM,j (j = 1, · · · , νM), or corresponding Bethe root λM,j. The Bethe

equation for the ` magnons can be rewritten in terms of the momenta for each bound

state aseipM (λM,j)L =

∏M ′

∏(M ′,k)6=(M,j)

SM,M ′(λM,j, λM ′,k) (5.45)

We shall take log of both sides, and make use the relation

−i lnλ+ ia

λ− ia= π − 2 arctan

λ

a. (5.46)

Here a choice of branch is assumed, and a is assumed to be nonzero. We then obtain

2L arctanλM,j

M + 12

= 2πQM,j +∑M ′

∑(M ′,k)6=(M,j)

ΦM,M ′(λM,j − λM ′,k) (5.47)

where ΦM,M ′ the scattering phase (that depends only on the difference between Bethe

roots) is given by

ΦM,M ′(λ) = 2M+M ′∑

K=|M−M ′|

(arctan

λ

K+ arctan

λ

K + 1

), (5.48)

and QM,j (j = 1, · · · , νM) are a set of integers or half integers depending on whether

L+νM−1 is even or odd (this is because SM,M = S0,0S0,1 · · ·S0,2M , and S0,0 is special).

The QM,j’s are a set of numbers that characterize the Bethe roots. While the

magnons were introduced as identical bosons, due to interactions they behave rather

like fermions: the solution cannot contain a pair of identical Bethe roots, namely λM,j

are all different for given M . For a given solution, it turns out that the QM,j’s increase

99

as λM,j increase, and for given M , there are no coinciding QM,j’s for different j. For

now we take it as a hypothesis. Nontrivial consistency checks will be given shortly.

For real momentum p ranging from 0 to 2π, λ ranges from ∞ to −∞ along the

real axis. Note a peculiarity of our parameterization: as p is periodic, when it varies

from −ε to ε (ε > 0), λ jumps from −∞ to ∞. A quantity like arctan(λ/a) (for

a > 0) would jump from −π/2 to π/2. Also, if we restrict λ to finite values, we have

implicitly excluded zero momentum magnons. These are taken into account by overall

SO(3) rotation on the states that are described by finite Bethe roots.

According to our hypothesis, λM,j takes value in a range on the real axis, and

accordingly QM,j. Suppose we put a root λM,j = ∞ (call the corresponding QM,j

QM,∞), then we obtain from the Bethe equation

πL = 2πQM,∞ +∑M ′ 6=M

νM′∑k′=1

M+M ′∑K=|M−M ′|

2π +

νM∑k 6=j

(π +

2M∑K=1

2π

)

⇒ QM,∞ =L

2−∑M ′ 6=M

νM ′(2min(M,M ′) + 1)− (2M +1

2)(νM − 1).

(5.49)

It turns out that the maximal admissible root QM,max is related by

QM,max = QM,∞ − (2M + 1). (5.50)

This can be understood in the following way. At QM,∞, all 2M + 1 Bethe roots of the

type M bound state go to infinity. Each time we decrease QM,j by one, one of the

Bethe roots moves back to a finite value. Of course, this picture is only heuristic and

so far unjustified. But we will see that this gives a consistent result, namely we will

recover all states of the spin chain in the end.

So we have now

QM,max =L− 1

2−∑M ′

J(M,M ′)νM ′ , (5.51)

where

J(M,M ′) ≡

2min(M,M ′) + 1, M 6= M ′,

2M + 12, M = M ′.

(5.52)

Since arctan is an odd function, we expect the solutions to be symmetric under λ→ −λ,

and QM,j → −QM,j. The QM,j’s range from QM,min = −QM,max to QM,max. A general

solution would be described in terms of distributing the νM QM,j’s among PM allowed

vacancies,

PM = 2QM,max + 1 = L− 2∑M ′

J(M,M ′)νM ′ . (5.53)

100

This is much like filling energy levels with identical fermions. Given νM, the number

of ways of distribute the QM,j’s is given by

Z(L, νM) =∏M

(PMνM

)(5.54)

The total number of Bethe roots ` =∑

M(2M + 1)νM goes up to L/2. Such a state in

fact has spin L2− ` and has the lowest Sz in its SO(3) multiplet, namely Sz = ` − L

2.

The reason is that an SO(3) rotation on the state characterized by finite Bethe roots

would introduce a Bethe root at λ = −∞. For example, in the ` = 1 case, there are

L− 1 = 2(L2− 1) + 1 states of different momenta.

Now let us count the total number of states describe by Bethe root configurations.

Firstly, the total number of magnons ` ranges from 0 to N/2. Given `, we consider

all possible partitions of the form ` =∑

(2M + 1)νM , and for each partition there are

Z(L, νM) states. So in total, the number of states are

Z =

L/2∑`=0

(L− 2`+ 1)∑

νM:∑

(2M+1)νM=`

Z(L, νM) (5.55)

It is an exercise of combinatorics to show that (see http://arxiv.org/abs/hep-th/9605187)

in fact

Z = 2L, (5.56)

precisely recovering all states of the spin chain.

5.4 The antiferromagnetic case and the thermodynamic limit

We have seen that the ground state |Ω〉 of the Heisenberg spin chain Hamiltonian H

has maximal spin, i.e. all spins aligned. If we are to consider a Hamiltonian with

the opposite sign of coupling, e.g. H ′ = −H, then the ground state would be one

described by a maximal number of Bethe roots, namely ` = L/2. This describes a

one-dimensional antiferromagnetic system. The state with highest energy with respect

to H (i.e. ground state w.r.t. −H) has L/2 type 0 magnons and no bound states, i.e.

ν0 = L/2, and νM = 0 for all M ≥ 1/2. As mentioned earlier, this state has spin 0 and

is therefore a singlet under SO(3) rotational symmetry. We have

Q0,max =L− 1

2− 1

2ν0 =

L

4− 1

2. (5.57)

The L/2 roots then fill up the range Q0,j = j = −Q0,max,−Q0,max + 1, · · · , Q0,max.

Here we are identifying the label “j” with the value of Q0,j. We will write λj = λ0,j.

101

The Bethe ansatz equation for this state is

arctan(2λj) =πj

L+

1

L

∑k

arctan(λj − λk) (5.58)

for all j, where the sum over k also ranges from −Q0,max to Q0,max. We are interested

in the large L limit, in which case

x ≡ j

L(5.59)

becomes a continuous parameter that takes value on the interval [−14, 1

4]. λj in the

continuum limit is replaced by a function of x, λ(x). The Bethe equation now becomes

an integral equation

arctan(2λ(x)) = πx+

∫ 1/4

−1/4

arctan(λ(x)− λ(y))dy. (5.60)

Since we are interested in the distribution of Bethe roots, we would like to map λ(x)

back to x. A good way to characterize this is the density of Bethe roots, ρ(λ), defined

to be such that ρ(λ)dλ is the number of roots in the range λ to λ+ dλ, divided by the

total number of roots L/2. So

ρ(λ) =1

L/2

dj

dλ= 2

dx

dλ=

2

λ′(x). (5.61)

It obeys ∫ ∞−∞

ρ(λ)dλ = 1. (5.62)

Differentiating (5.60) with respect to x, we obtain

2λ′(x)

4λ2 + 1= π +

∫ 1/4

−1/4

λ′(x)

(λ(x)− λ(y))2 + 1dy

= π +1

2λ′(x)

∫ ∞−∞

ρ(µ)dµ

(λ(x)− µ)2 + 1

(5.63)

where we made the change of variable µ = λ(y). Dividing both sides by λ′(x), we have

1

λ2 + 14

= πρ(λ) +

∫ ∞−∞

ρ(µ)dµ

(λ− µ)2 + 1(5.64)

This equation can be solved by Fourier transforming both sides, ρ(ω) =∫ρ(λ)eiωλdλ,

since the convolution product turns into ordinary product under Fourier transform.

The Fourier transform of the above equation reads

2πe−|ω|/2 = πρ(ω) + πe−|ω|ρ(ω), (5.65)

102

from which we solve

ρ(ω) =1

cosh (ω/2), (5.66)

and hence the Bethe root distribution function

ρ(λ) =1

cosh(πλ). (5.67)

The momentum of this state is

P =L

2

∫(π − 2 arctan(2λ))ρ(λ)dλ =

πL

2. (5.68)

In the antiferromagnetic case, it is natural to redefine the momentum by shifting this

to zero, and count only the contribution from “hole excitations” which we will discuss

shortly. The energy is given by

E =L

2

∫1

2

ρ(λ)

λ2 + 14

dλ = L ln 2. (5.69)

Now let us consider the next to highest energy state, or w.r.t the antiferromagnetic

Hamiltonian H ′ = −H, the first excited state. There are cases to consider: (I) ν0 =

L/2− 1, νM = 0 for M ≥ 1/2, and (II) ν0 = L/2− 2, ν 12

= 1, νM = 0 for M > 1.

First consider case (I). Since ν0 has decreased by 1, Q0,max increases by 1/2, and

there are 2Q0,max+1 = L2

+1 vacancies and L2−1 roots to fill these vacancies. It means

that there are two “holes”, say at level j = j1 and j2. Equivalently, we may write Q0,j

as

Q0j = j + θ(j − j1) + θ(j − j2), (5.70)

where θ is the step function: θ(j) = 1, j ≥ 0 and θ(j) = 0, j < 0. The continuum limit

of the Bethe equation now takes the form

arctan(2λ(x)) = πx+π

L(θ(x− x1) + θ(x− x2)) +

∫ 1/4

−1/4

arctan(λ(x)− λ(y))dy,

(5.71)

or after taking derivative on both sides,

1

λ2 + 14

= πρ(λ) +

∫ ∞−∞

ρ(µ)dµ

(λ− µ)2 + 1+

2π

L(δ(λ− λ1) + δ(λ− λ2)) (5.72)

Let ρ0(λ) = 1/ cosh(πλ) be the solution for the fully filled state we saw previously.

Then ρ(λ) takes the form

ρ(λ) = ρ0(λ) +2

L(σ(λ− λ1) + σ(λ− λ2)) (5.73)

103

where σ(λ) solves the equation

σ(λ) +1

π

∫ ∞−∞

σ(µ)dµ

(λ− µ)2 + 1+ δ(λ) = 0. (5.74)

Fourier transforming both sides, we obtain

σ(ω) = − 1

1 + e−|ω|. (5.75)

We can now calculate the momentum and energy due to the “holes”,

∆P =2∑i=1

∫(π − 2 arctan(2λ))σ(λ− λi)dλ

= π −2∑i=1

∫dλ(2 arctan(2λ))

∫dω

2πe−i(λ−λi)ωσ(ω)

= π − i2∑i=1

∫dωe−|ω|/2

ωσ(ω)eiλiω = π +

i

2

2∑i=1

∫dω

eiλiω

ω cosh(ω/2)

= k(λ1) + k(λ2),

(5.76)

where

k(λ) =π

2− arctan sinh(πλ). (5.77)

k(λ) can be thought of as the momentum of an individual “hole excitation” on top

of the ground state of the antiferromagnetic Hamiltonian H ′ = −H. These holes can

only appear in pairs. The energy of the two holes is given by ∆E = −ε(λ1) − ε(λ2),

where

ε(λ) =

∫1

2

σ(λ)

λ2 + 14

dλ =π

2 cosh(πλ). (5.78)

In terms of k(λ), we find the dispersion relation for a hole excitation (this energy is

with respect to H ′ = −H),

ε(k) =π

2sin k. (5.79)

Here the range of momentum for a single hole is k ∈ [0, π]. The two-hole state describe

here has total spin 1. It is thus referred to as a triplet state.

Now let us turn to the case (II): ν0 = L/2 − 2, ν 12

= 1, νM = 0 for M > 1. The

range of Q0,j is bounded by

Q0,max =L− 1

2− ν0

2− ν 1

2=L

4− 1

2. (5.80)

There are now 2Q0,max + 1 = L/2 vacancies for type 0 particles, and once again two

holes, say at levels j1 and j2. There is in additional a type 12

bound particle, whose

104

corresponding root λ 12

is to be determined by the Bethe equation. The Bethe equation

for type 0 roots in the continuum limit gives

1

λ2 + 14

= πρ(λ) +

∫ ∞−∞

ρ(µ)

(λ− µ)2 + 1dµ+

2π

L[δ(λ− λ1) + δ(λ− λ2)] +

1

LΦ′

0, 12(λ− λ 1

2)

(5.81)

where Φ0, 12(λ) is the scattering phase of type 0 with type 1

2magnons, and Φ′

0, 12

(λ) its

derivative in λ. There is another Bethe equation for the type 12

root,

arctanλ 12

=1

L

∑j

Φ 12,0(λ 1

2− λ0,j). (5.82)

Here Q 12,j is bounded by Q 1

2,max = L−1

2−ν0− 3

2ν 1

2= 0, hence there is only one admissible

value Q 12,j = 0 which is why we did not write the term πQ 1

2,j in above equation.

The density ρ(λ) can be written in terms of ρ0 and σ as

ρ(λ) = ρ0(λ) +2

L

[σ(λ− λ1) + σ(λ− λ2) + Ω(λ− λ 1

2)]

(5.83)

where Ω(λ) obeys the equation

Ω(λ) +1

π

∫ ∞−∞

Ω(µ)

(λ− µ)2 + 1dµ+

1

2πΦ′

0, 12(λ) = 0. (5.84)

Recall that

Φ0, 12(λ) = 2

[arctan(2λ) + arctan(

2λ

3)

],

Φ′0, 1

2(λ) =

1

λ2 + 14

+3

λ2 + 94

.(5.85)

Fourier transforming (5.84) gives

Ω(ω) + e−|ω|Ω(ω) +(e−|ω|2 + e−

3|ω|2

)= 0, (5.86)

from which we solve

Ω(ω) = −e−|ω|2 , Ω(λ) = − 1

2π

1

λ2 + 14

. (5.87)

Next, let us solve λ 12

from (5.82); the latter is written in the continuum limit as

arctanλ 12

=

∫Φ 1

2,0(λ 1

2− λ)ρ(λ)dλ

=

∫Φ 1

2,0(λ 1

2− λ)ρ0(λ)dλ+

2

L

2∑i=1

∫Φ 1

2,0(λ 1

2− λ)σ(λ− λi)dλ+

2

L

∫Φ 1

2,0(λ 1

2− λ)Ω(λ− λ 1

2)dλ.

(5.88)

105

The last term is zero because Φ 12,0(λ) is an odd function in λ, whereas Ω(λ) is even.

The first term in the second line is∫Φ 1

2,0(λ 1

2− λ)ρ0(λ)dλ =

∫dωe−λω

[2ie−|ω|

ωcosh

ω

2

]ρ0(ω) =

∫dωe−λω

2ie−|ω|

ω

= arctanλ 12,

(5.89)

which cancels the LHS of (5.88). This leaves

2∑i=1

∫Φ 1

2,0(λ 1

2− λ)σ(λ− λi)dλ = 0. (5.90)

Using σ(ω) derived earlier, this equation is

2∑i=1

arctan 2(λ 12− λi) = 0, (5.91)

from which we solve

λ 12

=λ1 + λ2

2. (5.92)

Now we have all the necessary information to solve for the dispersion relation of the

hole states. Taking into account the contribution from type 0 and type 12

Bethe roots,

the momentum is

∆P =

∫(π − 2 arctan(2λ))

[2∑i=1

σ(λ− λi) + Ω(λ− λ 12)

]dλ+

(π − 2 arctanλ 1

2

)= k(λ1) + k(λ2)− π + 2 arctanλ 1

2+(π − 2 arctanλ 1

2

),

= k(λ1) + k(λ2).(5.93)

and the energy is

∆E = −ε(λ1)− ε(λ2) +

∫1

2

Ω(µ− λ 12)

µ2 + 14

dµ+1

λ212

+ 1

= −ε(λ1)− ε(λ2).

(5.94)

Note that the contribution due to the shift of type 0 root distribution function by Ω(λ)

exactly cancels the contribution due to the type 1/2 root, and we end up with exactly

the same dispersion relation of two holes as in the triplet case.

Thus we have found that, remarkably, the two-hole excitations behave like two

separate spin 1/2 particles. The spin triplet and singlet two-hole state however have

different interpretations in terms of Bethe roots: the former is described by L2−1 Bethe

roots filling all but two vacancies, while the latter is described by L2− 2 type 0 Bethe

roots together with one type 12

bound particle.

106

5.5 Conserved charges

We shall now uncover the reason for factorized S-matrix of magnon scattering in the

Heisenberg spin chain which allows for the solution in terms of algebraic Bethe ansatz:

the existence of “sufficiently many” conserved charges, i.e. Hermitian operators that

commute with the spin chain Hamiltonian H.

One begins by introducing an auxiliary space Va which is a two-dimensional complex

vector space. Let hn be the two-dimensional space of spin-12

states on the n-th site of

the spin chain. We then define a set of “Lax operators” that acts on hn ⊗ Va,

Ln,a(λ) = λIn ⊗ I + i~Sn ⊗ ~σ, (5.95)

where the second identity matrix I and Pauli matrices ~σ are understood to act on Va.

λ is a spectral parameter (which will turn out to be the Bethe root!). Later on we

will make use of multiple auxiliary spaces Va1 , Va2 , etc., and correspondingly the Lax

operators will be denoted Ln,a1 , Ln,a2 , etc.

As seen before, we can write the Lax operator in terms of permutation operator

Pn,a that exchanges hn with Va,

Ln,a(λ) = (λ− i

2)In,a + iPn,a. (5.96)

Next, one defines the T -matrix for the entire spin chain, as the ordered product of

Lax operators associated with each site,

Ta(λ) = LL,a(λ) · · ·L1,a(λ). (5.97)

It acts on H⊗Va, H being the 2L-dimensional quantum Hilbert space of the spin chain.

More explicitly, one may write

Ta(λ) =

(A(λ) B(λ)

C(λ) D(λ)

)(5.98)

where A,B,C,D are operators acting on H. Note that Ta(λ) is in fact a degree L

polynomial in λ with operator coefficients.

Finally, define F (λ) as the trace of the T -matrix,

F (λ) = TrT (λ) = A(λ) +D(λ). (5.99)

We claim[F (λ), F (µ)] = 0, ∀λ, µ;

H =L

2− i

2

d

dλlnF (λ)

∣∣∣∣λ= i

2

.(5.100)

107

The two relations together establish the set of commuting charges which are the co-

efficients of F (λ) as a polynomial in λ, which includes the Hamiltonian H and hence

the charges are conserved. What makes these relations possible is the fundamental

commutation relation

Ra1,a2(λ− µ)Ln,a1(λ)Ln,a2(µ) = Ln,a2(µ)Ln,a1(λ)Ra1,a2(λ− µ), (5.101)

where Ra1,a2(λ) (called the “R-matrix”) is defined much like the Lax operator,

Ra1,a2(λ) = λIa1,a2 + iPa1,a2 . (5.102)

The verification of (5.101) is simple and is left as an exercise.

By repeatedly applying (5.101), one finds the commutation relation

Ra1,a2(λ− µ)Ta1(λ)Ta2(µ) = Ta2(µ)Ta1(λ)Ra1,a2(λ− µ) (5.103)

Taking the trace of

Ta1(λ)Ta2(µ) = (Ra1,a2(λ− µ))−1Ta2(µ)Ta1(λ)Ra1,a2(λ− µ) (5.104)

over Va1 ⊗ Va2 then establishes that F (λ) and F (µ) commute for arbitrary λ and µ.

To derive the relation between F (λ) and the spin chain Hamiltonian, we only need

to expand F (λ) near λ = i/2. Firstly, note that

Ln,a(i

2) = iPn,a,

d

dλLn,a(λ)|λ= i

2= In,a. (5.105)

It follows that

Ta(i

2) = iLPL,a · · ·P1,a,

d

dλTa(λ)|λ= i

2= iL−1

L∑n=1

PL,a · · · Pn,a · · ·P1,a. (5.106)

Note that PL,a · · ·P1,a shifts (12 · · ·n) to the left and replaces a by 1 and n by a. Taking

trace over Va, it gives the shift operator by 1 to the left. We obtain

F (i

2) = iLP1,2P2,3 · · ·PL−1,L,

d

dλF (λ)|λ= i

2= iL−1

L∑n=1

P1,2 · · ·Pn−1,n+1 · · ·PL−1,L.

(5.107)

Finally,

d

dλlnF (λ)|λ= i

2=dF (λ)

dλF (λ)−1|λ= i

2= −i

L∑n=1

Pn,n+1, (5.108)

and hence (5.100) follows.

108

To get a better feeling of what (5.103) means, let us expand it out in terms of

operators acting on the physical Hilbert space H only. We shall write Ra1,a2 , Ta1 and

Ta2 as 4× 4 matrices that act on the 4-dimensional vector space Va1 ⊗ Va2 . We have

Ia1,a2 =

1 0 0 0

0 1 0 0

0 0 1 0

0 0 0 1

, Pa1,a2 =

1 0 0 0

0 0 1 0

0 1 0 0

0 0 0 1

, Ra1,a2(λ) =

λ+ i 0 0 0

0 λ i 0

0 i λ 0

0 0 0 λ+ i

,

Ta1(λ) =

A(λ) 0 B(λ) 0

0 A(λ) 0 B(λ)

C(λ) 0 D(λ) 0

0 C(λ) 0 D(λ)

, Ta2(µ) =

A(µ) B(µ) 0 0

C(µ) D(µ) 0 0

0 0 A(µ) B(µ)

0 0 C(µ) D(µ)

.

(5.109)

(5.103) implies commutation relations such as

[B(λ), B(µ)] = 0,

(λ− µ)A(λ)B(µ) = (λ− µ− i)B(µ)A(λ) + iB(λ)A(µ),

(λ− µ)D(λ)B(µ) = (λ− µ+ i)B(µ)D(λ)− iB(λ)D(µ).

(5.110)

Let |Ω〉 be the spin chain state with all spins up. The Lax operator acts on it as

Ln,a(λ)|Ω〉 =

(λ+ i

2iS−n

0 λ− i2

)|Ω〉 (5.111)

The T -matrix then acts on |Ω〉 in the upper triangular form

T (λ)|Ω〉 =

((λ+ i

2)L ∗

0 (λ− i2)L

)|Ω〉 (5.112)

From this, we know

C(λ)|Ω〉 = 0, A(λ)|Ω〉 = (λ+i

2)L|Ω〉, D(λ)|Ω〉 = (λ− i

2)L|Ω〉. (5.113)

The operator B(λ) acts nontrivially on |Ω〉. In fact, it is the creation operator of

a magnon of rapidity λ. To see this, first note that the fundamental commutation

relation implies

[B(λ), B(µ)] = 0. (5.114)

A state with ` magnons turns out to be of the form

|Ψ〉 = B(λ1)B(λ2) · · ·B(λ`)|Ω〉. (5.115)

109

Requiring it to be an eigenstate of F (λ), and repeatedly applying (5.110) then give the

Bethe equations.

If we write the scattering state in the form

|Ψ〉 = B(λ1) · · ·B(λ`)|Ω〉 = exp

[∑j=1

lnB(λj)

]|Ω〉, (5.116)

it is then straightforward to take the continuum limit. For instance, the antiferromag-

netic ground state can be represented as

|Ψ0〉 = exp

[L

2

∫ ∞−∞

lnB(λ)ρ0(λ)dλ

]|Ω〉 (5.117)

The spin triplet two-hole state can be expressed as

|Ψ(λ1, λ2)〉 = Z(λ1)Z(λ2)|Ψ0〉, (5.118)

where

Z(λ) = exp

[∫ ∞−∞

lnB(µ)σ(λ− µ)dµ

]. (5.119)

5.6 Factorized S-matrices

In the continuum limit, one can talk about the S-matrix of asymptotic states. In

situations where infinitely many conservation laws exist, such as in the continuum

limit of the XXX 12

spin chain, it generally follows that the n-body S-matrix factorizes

into a product of 2-body S-matrices. The argument goes as follows. Suppose Qk are

an infinite set of conserved charges, k = 1, 2, 3, · · · , and suppose they are diagonalized

on one-particle momentum eigenstates

Qk|p(a)〉 = ω(a)k (p)|p(a)〉, (5.120)

where a labels the type of particle and p(a) its momentum, and assuming that the

interactions are of finite range, then on an n-particle asymptotic state (described by

far separated wave packets), or the corresponding scattering states, we have

Qk|p(a1)1 , p

(a2)2 , · · · , p(an)

n ;±〉 =n∑i=1

ω(ai)k (pi) |p(a1)

1 , p(a2)2 , · · · , p(an)

n ;±〉 (5.121)

where the ± labels the scattering state associated with incoming asymptotic wave

packets (+) or outgoing asymptotic wave packets (−).

With sufficiently many conserved charges Qk, ω(a)k (p) form a complete basis of

functions of p, say all polynomials in p. Then∑n

i=1 ω(ai)k (pi) form a complete basis of

110

symmetric functions of (p1, · · · , pn). This means that the set of momenta (p1, · · · , pn)

(as well as the total number of particles of given mass) is conserved in a scattering

process, though the particle type ai can change among particles of the same mass.

Consider the n-particle wave function Ψ(x1, x2, · · · , xn). In the domain x1 < x2 <

· · · < xn and |xi − xj| R where R is the range of interactions, then the conserved

charges Qk (5.120) are simultaneously diagonalized on a momentum basis,

Ψ(x1, x2, · · · , xn) =∑σ∈Sn

S(pi; pσ(i)) exp(ipσ(1)x1 + · · · ipσ(n)xn

)(5.122)

This form of the n-particle wave function as superposition of plane waves, of course,

does not hold when xi and xj come close. As xi crosses xj and back to the asymptotic

region where the x’s are far separated again, the wave functions are related by the

2-body S-matrix (between particles of momenta pi and pj). The n-particle S-matrix

is then given by the composition of 2-body S-matrices corresponding to successive

exchanges of pairs of particles. In each 2→ 2 scattering process, the two momenta are

unchanged, while the particle type can change only among particles of the same mass,

due to energy conservation.

The consistency of the factorization of n-particle S-matrix requires, for three par-

ticles labeled by momentum and particle type p(ai)i , i = 1, 2, 3, that

S12S13S23 = S23S13S12 (5.123)

where the 2-body S-matrix Sij ≡ S(pi, pj) for given momenta pi, pj is a matrix that

acts on the internal indices that label the particle type. (5.123) is known as the Yang-

Baxter equation. While it is trivially satisfied in the case of one species of particle

(in which case Sij are simply scattering phases), it gives nontrivial constraints on the

S-matrix involving different types of particles of the same mass (typically related by

symmetry).

As an example of factorized S-matrix, let us consider the scattering of two “hole

excitations” in the antiferromagnetic XXX 12

spin chain. We have already established

that the hole excitations come in a spin-12

doublet, and so by SO(3) symmetry the

2-body S-matrix of holes with rapidity λ and µ takes the form

S(λ, µ) = S(λ− µ) = a(λ− µ) [h(λ− µ)I + P ] , (5.124)

where P is the permutation operator on the two holes. Just as in the case of magnons

in the ferromagnetic spin chain, the S-matrix of a pair of holes depend only on the

difference in the rapidity variables, λ− µ. The Yang-Baxter equation then reads

S12(λ)S13(λ+ µ)S23(µ) = S23(µ)S13(λ+ µ)S12(λ) (5.125)

111

or more explicitly,

[h(λ) + P12] [h(λ+ µ) + P13] [h(µ) + P23] = [h(µ) + P23] [h(λ+ µ) + P13] [h(λ) + P12] .

(5.126)

Expanding this out, and using the relations

P12P23 = P13P12 = P23P13,

P13P23 = P12P13 = P23P12,

P12P13P23 = P13 = P23P13P12,

(5.127)

the only resulting nontrivial condition on h(λ) is

h(λ+ µ) = h(λ) + h(µ). (5.128)

It follows that h(λ) is proportional to λ. Acting on triplet and singlet two-hole states,

the 2-body S-matrix takes the form

St(λ) = a(λ)(h(λ) + 1), Ss(λ) = a(λ)(h(λ)− 1) (5.129)

Unitarity requires that St(λ) and Ss(λ) are phases for real λ. In particular, h(λ) = iγλ,

for a real constant γ. It will turn out that γ = −1. In fact, this 2-body S-matrix

is proportional to the R-matrix Ra1,a2(λ) we wrote earlier, which obeys the Yang-

Baxter equation. It takes more work to determine St(λ). This can be done using the

representation (5.118) (See Faddeev’s lecture for details). We only quote the result

here

St(λ) = −iΓ(1+iλ

2)Γ(1− iλ

2)

Γ(1−iλ2

)Γ(1 + iλ2

). (5.130)

It has infinitely many resonance poles.

Next, let us consider a somewhat richer example: a factorized S-matrix of two

types of non-relativistic particles with O(2) symmetry in one dimension. The 2-body

S-matrix takes the form

Sij→kl(p1, p2) = δijδklSA(p12) + δikδjlST (p12) + δilδjkSR(p12). (5.131)

Here i, j, k, l = 1, 2 labels the type of the particle. ij → kl denotes the scattering

process in which particle i carries momentum p1, colliding with particle j carrying

momentum p2, and the result of the scattering is a particle of type k leaving with mo-

mentum p1, while a particle of type l leaving with momentum p2. The index structure of

the S-matrix is constrained by the O(2) symmetry. The components SA, ST , SR stand

for the annihilation, transmission, and reflection amplitudes. Write S = SA +ST +SR.

Some components of Yang-Baxter equations are trivially satisfied. For exam-

ple, if we consider the scattering amplitude of three particles of type 1, 1, 1, of mo-

menta p1, p2, p3, into 1, 1, 1, of momenta p3, p2, p1, the Yang-Baxter equation would

112

be trivial in this case. Some other components of the Yang-Baxter equation are

(112) → (112), (112) → (121), (112) → (211). From now on we will write SaSbScfor Sa(p12)Sb(p13)Sc(p23). The first case gives

SRSSR + STSRST = SSRS + SASSA. (5.132)

The second case (112)→ (121) gives

STSSR + SRSRST = SSTSR + SASAST . (5.133)

The third case (112)→ (211) gives

SASASR + SSTST = SSTST + SASASR (5.134)

which is in fact trivial.

(5.133) can be written in terms of SA, ST , SR as (after canceling a term on both

sides)

STSASR + STSRSR + SRSRST = SASTSR + SRSTSR + SASAST . (5.135)

(5.132) can be written as

SRSASR + SRSTSR = SASRSA + (ST + SR)SRSA + SASR(ST + SR) + STSRSR + SRSRST + SASSA

= 2SASRSA + STSRSA + SRSRSA + SASRST + SASRSR + STSRSR + SRSRST + SASTSA + SASASA.(5.136)

These two equations turn out to be the only nontrivial components of the O(2) invariant

Yang-Baxter equation. We won’t be able to solve the overall phase of the S-matrix,

which drops out of the Yang-Baxter equations. This phase must be fixed by other

physical requirements, e.g. on the pole structure of the S-matrix. Let us consider the

ratios

h(p) =ST (p)

SR(p), g(p) =

SA(p)

SR(p). (5.137)

Dividing the two equations above by SRSRSR on both sides, and after a bit of algebra,

we can write them as

h(p) + h(q)− h(p+ q) = g(p)h(p+ q)− h(p)g(p+ q) + g(p)g(p+ q)h(q),[1 + g(p+ q) + h(p+ q)

][1− g(p)g(q)

]+ h(p)h(q) =

[1 + g(p) + h(p)

][1 + g(q) + h(q)

].

(5.138)

We also need to implement the constraints due to unitarity of the S-matrix. Assuming

time reversal symmetry, S∗(p) = S(−p), unitarity amounts to S(p)S(−p) = I, or in

components,

ST (p)ST (−p) + SR(p)SR(−p) = 1,

ST (p)SR(−p) + SR(p)ST (−p) = 0,

2SA(p)SA(−p) + SA(p)ST (−p) + SA(p)SR(−p) + ST (p)SA(−p) + SR(p)SA(−p) = 0.(5.139)

113

In terms of g and h, these equations can be expressed as

h(p)h(−p) + 1 =1

SR(p)SR(−p),

h(p) + h(−p) = 0,

2g(p)g(−p) + (1 + h(p))g(−p) + (1 + h(−p))g(p) = 0.

(5.140)

Setting p = q = 0, the Yang-Baxter equations imply

h(0) = g(0)2h(0),

(1 + g(0) + h(0))(1− g(0)2) + h(0)2 = (1 + g(0) + h(0))2.(5.141)

while the unitarity condition implies

h(0) = 0, g(0)2 + g(0) = 0, SR(0) = 1. (5.142)

The solution to these equations are either h(0) = g(0) = 0, or h(0) = 0, g(0) = −1.

We will consider the first possibility, h(0) = g(0) = 0.

Taking derivative of (5.138) with respect to p, and then setting p = 0, we obtain

the first order differential equations

h′(q) = (1 + g(q))[h′(0)− g′(0)h(q)

],

h′(q) + g′(q) = (1 + g(q))[h′(0) + g′(0)(1 + h(q) + g(q))

] (5.143)

Define α = h′(0), β = g′(0). We can solve g in terms of h from the first equation,

g(p) =h′(p)

α− βh(p)− 1 (5.144)

and plug into the second equation, which can be simplified to

h′′(p) = 2βh(p)h′(p)⇒ h′(p) = βh(p)2 + α. (5.145)

The solution ish(p) = (α/β)

12 tan

[(αβ)

12p],

g(p) =(α/β)

12 tan

[(αβ)

12p]

+ 1

(α/β)12 cot

[(αβ)

12p]− 1

.(5.146)

Finally, using the unitarity relation again, we obtain

SR(p)SR(−p) =1

h(p)h(−p) + 1=

1

1− αβ

tan2((αβ)12p)

(5.147)

114

We have labeled the two types of particles by the index i = 1, 2. Alternatively, we

could define their linear combinations (say in terms of annihilation operators ai(p))

a(p) = a1(p) + ia2(p), a(p) = a1(p)− ia2(p). (5.148)

a(p) could be the annihilation operator of a charged particle, and a(p) the annihila-

tion operator of its “anti-particle”. Note that in our integrable model the charged

particle and anti-particle cannot simply annihilate; rather, they scatter according to

the factorized S-matrix. Let SI , ST , and SR be the scattering amplitude for iden-

tical particles (or anti-particles), the transmission, and the reflection amplitudes for

particle-anti-particle, respectively. They are related to SA, ST , SR by

SI = ST + SR,

ST = SA + ST ,

SR = SA + SR.

(5.149)

We then derive for instance,

SR(p)SR(−p) =sin2(πκ)

sin2(πκ) + sinh2(πγp), etc. (5.150)

where we redefined the constants

sin(πκ) =2√αβ

α + β, γ =

2i

π

√αβ. (5.151)

A solution to the functional equation (5.150) is

SR(p) =sin(πκ)

i sinh(πγp)

Γ(−iγp− κ)Γ(−iγp+ κ+ 1)

Γ(−iγp)Γ(1− iγp). (5.152)

Of course, there are many other solutions, differing by factors of the form∏k

αk − ipαk + ip

. (5.153)

Note that SR(p) should be an analytic function that approaches 1 as p → 0. Factors

like (5.153) would introduce extra poles on the upper half complex momentum plane,

corresponding to bound states. For κ > 0, (5.152) turns out to be the solution with a

minimal set of singularities. It has bound state poles at (positive imaginary momentum)

p =i(κ− n)

γ, (5.154)

where n is an integer in the range 0 ≤ n < κ.

115

(5.152) turns out to be the scattering amplitude of a particle and “anti-particle”

interaction through the attractive potential (setting mass to 1)

U(x) = − g

γ2

1

cosh2(x/γ), g = κ2 − κ+

3

4. (5.155)

The scattering between identical particles turns out to be that of the repulsive potential

(note that SI(p) does not have bound state poles)

V (x) =g

γ2

1

sinh2(x/γ). (5.156)

The Hamiltonian for such a system of N identical particles and M identical “anti”-

particles is

H = −N∑i=1

∂2xi

2−

M∑j=1

∂2yj

2+ 2

N∑i<i′

V (xi − xi′) + 2M∑j<j′

V (yj − yj′) + 2N∑i=1

M∑j=1

U(xi − yj).

(5.157)

This is in the fact the non-relativistic limit of solitons and anti-solitons in the so-called

sine-Gordon model - a well understood example of an integrable quantum field theory.

6 Path integral revisited

6.1 Derivation of path integral

Begin with a quantum mechanical system with canonical momenta and coordinates

(p, q), and Hamiltonian H. Let |q〉 be the eigenstate of q with eigenvalue q. Consider

the time evolution amplitude

A = 〈qf |e−i~ HT |qi〉 (6.1)

We will work in Heisenberg picture now, writing

|q, t〉 = ei~ Ht|q〉, (6.2)

and so

A = 〈qf , T |qi, 0〉. (6.3)

Now divide the time period T into N short intervals, separated by T = tN > tN−1 >

· · · > t1 > t0 = 0, and inserting a complete q(t)-eigenbasis at each time tk, we can

rewrite the evolution amplitude as

A =

∫ N−1∏k=1

dqk

N−1∏m=0

〈qm+1, tm+1|qm, tm〉, (6.4)

116

where

(q0, t0) ≡ (qi, 0), (qN , tN) ≡ (qf , T ). (6.5)

Each matrix element involved in (6.4) can be written as

〈qm+1, tm+1|qm, tm〉 = 〈qm+1|e−i~ H(tm+1−tm)|qm〉

=

∫dpm〈qm+1|pm〉〈pm|e−

i~ H(tm+1−tm)|qm〉

=

∫dpm2π~

exp

[i

~(pm∆qm −H(pm, qm)∆tm) +O(∆t2)

].

(6.6)

In the last step, we have expanded e−iH∆t/~ in powers of ∆tm = tm+1 − tm, and kept

only the linear terms in ∆t in the exponent. As we will see shortly, the O(∆t2) terms

in the exponent become negligible in the limit where N is taken to infinity and ∆t

going to zero. H(p, q) here is defined by taking H(p, q), moving all the p’s to the left

of the q’s using the canonical commutation relation, and then replacing p and q by p

and q. This is so that

〈pm|H|qm〉 = H(pm, qm)〈pm|qm〉. (6.7)

Putting this back in (6.4), we have

A =

∫ N−1∏k=1

dpmdqm2π~

dp0

2π~exp

[i

~

N−1∑m=0

(pm∆qm −H(pm, qm)∆tm) +O(N−1)

]. (6.8)

Taking the continuum limit, N →∞, ∆t→ 0, we obtain

A =

∫[dp dq] exp

[i

~

∫ T

0

dt (pq −H(p, q))

]. (6.9)

The functional integral measure [dp dq] is formally defined as

[dp dq] = limN→∞

N−1∏k=1

dpmdqm2π~

dp0

2π~. (6.10)

Note that the exponent in (6.9) does not depend on the time derivative of p, and so the

integration over p(t) can be done separately at each time interval. If the Hamiltonian

is of the form p2

2+V (q), as is typically the case in non-relativistic quantum mechanics,

then the integration over p(t) is a Gaussian integral, giving

A =

∫[Dq] exp

[i

~

∫ T

0

dtL(q, q)

], (6.11)

where we have used the familiar property that the Lagrangian L(q, q) is the Lendre

transform of the Hamiltonian,

L(q, q) =[pq −H(p, q)

]∣∣∂pH(p,q)=q

. (6.12)

117

The measure [Dq] is rescaled from [dq] in (6.9) due to the Gaussian integrals. With

a Hamiltonian like p2

2+ V (q), this only changes the normalization of the measure

by a constant and is not important. One should note that for H(p, q) with more

general p-dependence, however, there can be additional factors from the integral over

p which are functions of (q, q), that enter the measure [Dq]. The continuum limit from

which we obtained the path integral measure is also subtle. In practice, the measure

[Dq] is often taken as the integration over Fourier modes of q(t). The resulting path

integral may suffer from short distance divergences, which requires regularization and

renormalization. We will see an example of this later.

Next, let us consider the insertion of q(t) into the path integral. For instance,∫[Dq]

qf ,Tqi,0

q(t) eiS/~ =

∫[Dq]q,tqi,0dq(t)[Dq]

qf ,Tq,t q(t) eiS/~

=

∫dq〈qf , T |q, t〉q〈q, t|qi, 0〉

= 〈qf , T |q(t)|qi, 0〉.

(6.13)

So the insertion of the operator q(t) into the amplitude amounts to the insertion of the

corresponding classical canonical coordinate q(t) into the path integral. If we insert

two (or more) q(t)’s into the path integral, however, following the same derivation,

the result is the amplitude with the insertion of the time-ordered product of the q(t)’s.

Namely, ∫[Dq]

qf ,Tqi,0

q(t)q(t′) eiS/~ = 〈qf , T |T [q(t)q(t′)] |qi, 0〉 (6.14)

Let us emphasize that while on the LHS there is no ordering between q(t) and q(t′) as

these are classical variables, the corresponding operators on the RHS are automatically

time ordered. This is an important property of the path integral. Similarly, if we insert

more q(t)’s into the path integral, the result gives the amplitude with the insertion of

the time ordered product of all the q(t)’s.

6.2 Euclidean path integral

The path integral described in the previous section is given by a functional integration

over phase factors, which does not obviously converge. The Euclidean version of it, on

the other hand, involves an exponentially damped integrand and has better convergence

properties. To begin, we define the Euclidean time evolution amplitude as

AE = 〈qf |e−HTE/~|qi〉, (6.15)

118

obtained from the ordinary (“Lorentzian”) amplitude by the Wick rotation T → −iTE.

The corresponding path integral representation is∫[Dq]

qf ,TEqi,0

exp

[1

~

∫ TE

0

dτ L(q, i∂τq)

](6.16)

The Euclidean Lagrangian and action are defined as

LE(q, ∂τq) = −L(q, i∂τq),

SE =

∫dτLE(q, ∂τq),

(6.17)

so that the Euclidean path integral takes the form∫[Dq]

qf ,TEqi,0

e−SE/~. (6.18)

For instance, if L(q, q) = 12q2 − V (q), then LE(q, ∂τq) = 1

2(∂τq)

2 + V (q). In this case,

SE is positive definite.

One may also consider the Euclidean path integral with periodic boundary condition

over [0, T ], which amounts to identifying qf with qi and integrating over qi. The result

is the trace of the Euclidean evolution operator, also known as the partition function.

Namely

Tr e−1~ HTE =

∫dq〈q|e−

1~ HTE |q〉 =

∫[Dq]q(0)=q(TE)e

−SE/~. (6.19)

More generally, suppose there is a symmetry Ω that act on the states |q〉 by

Ω|q〉 = |qΩ〉, (6.20)

where |qΩ〉 is another q-eigenstate with eigenvalue qΩ, then we can work with the path

integral with periodic boundary condition twisted by Ω,

Tr Ω e−1~ HTE =

∫dq〈q|e−

1~ HTE Ω|q〉 =

∫[Dq]qΩ(0)=q(TE)e

−SE/~. (6.21)

One should be aware that a Euclidean operator A in the Euclidean Heisenberg

picture turns into

A(τ) = eHτ Ae−Hτ , (6.22)

which is no longer Hermitian. Rather, it obeys

A(τ)† = A(−τ). (6.23)

119

6.3 The free path integral

As an explicit and useful example, consider the path integral of a harmonic oscillator,

H =1

2p2 +

1

2ω2q2,

L =1

2q2 − 1

2ω2q2,

LE =1

2∂τq

2 +1

2ω2q2.

(6.24)

We will work in ~ = 1 units in this section. We would like to compute the Euclidean

transition amplitude

〈qf , T |qi, 0〉E =

∫[Dq]

qf ,Tqi,0

e−SE . (6.25)

First, we shall separate q(τ) into the classical path plus quantum fluctuations,

q(τ) = qcl(τ) + q(τ), (6.26)

where qcl obeys the classical equation of motion and the path integral boundary con-

ditions

−∂2τ qcl(τ) + ω2qcl(τ) = 0, qcl(0) = qi, qcl(T ) = qf . (6.27)

The measure can be written as [Dq(τ)] by a trivial shift of integration variable, while

q(τ) now vanishes on the boundary of the path, τ = 0 and τ = T . Since qcl solves the

classical equation of motion, the action SE[q] separates into the classical piece SE[qcl]

and an action for quantum fluctuation S ′. Generally, S ′ will involve quadratic and

higher order terms in q and may depend on qcl as well. In the simple example of a

harmonic oscillator, it is only quadratic and is independent of qcl.

It is easy to derive the classical action,

Scl(qi, qf ) = SE[qcl] =1

2

∫ T

0

dτ[(∂τqcl)

2 + ω2q2cl

]= ω

(q2i + q2

f ) cosh(ωT )− 2qiqf

2 sinh(ωT ).

(6.28)

The action for the quantum fluctuation is

S ′ =1

2

∫ T

0

dτ[(∂τ q)

2 + ω2q2]. (6.29)

So we can write the transition amplitude as

〈qf , T |qi, 0〉E = e−Scl∫

[Dq]0,T0,0 e−S′[q] (6.30)

120

Now, we will perform the functional integration by decomposing q(τ) on a eigenbasis

with respect to the Euclidean kinetic operator

∆ = −∂2τ + ω2. (6.31)

Namely, expand

q(τ) =∑k

φkfk(τ),

∆fk(τ) = λkfk(τ),

(6.32)

and integrate over the coefficients φk,

[Dq]→∏k

dφk. (6.33)

In making this substitution on the functional integration measure, a potentially diver-

gent normalization factor is dropped. For now, we assume that this only introduces

an overall constant factor which will be fixed later. The eigenfunctions fk(τ) and

eigenvalues λk subject to Dirichlet boundary condition at τ = 0, T are

fk(τ) =

√2

Tsin

kπτ

T, λk =

k2π2

T 2+ ω2. (6.34)

We have then∫[Dq]0,T0,0 e

−S′[q] →∫ ∏

k

dφk exp

(−1

2

∑λkφ

2k

)=∏k

√2π

λk(6.35)

More generally, the RHS can be written in terms of the functional determinant,

1√det ∆

2π

, det ∆ =∏k

λk. (6.36)

Using the explicit expression for λk, we can compute the determinant as

det∆

2π=∞∏k=1

k2π2 + ω2T 2

2πT 2. (6.37)

This infinite product is obviously divergent. On the other hand, we have not been care-

ful in keeping track of the normalization factor of the functional integration measure.

The divergence has to do with the high frequency modes, i.e. large k. This divergence

can be regularized by considering

det∆

2π−→

∞∏k=1

k2π2 + ω2T 2

k2π2 + Λ2T 2, (6.38)

121

where Λ is a large “cutoff” frequency. The difference between (6.37) and (6.38), over

finite k, becomes a constant normalization factor in the Λ→∞ limit (and in particular

independent of ω and T ). This justifies the consistency of the regularization as Λ is

taken to infinity. This regularization scheme is known as Pauli-Villars regularization.

The product on the RHS of (6.38) now converges, and the result is

Λ sinh(ωT )

ω sinh(ΛT ). (6.39)

One can easily see that this expression has the same set of zeros and poles as (6.38) and

the same T → 0 limit. The regularized functional integral for the transition amplitude

(in the Λ→∞ limit) is

〈qf , T |qi, 0〉E →√

ω

2 sinh(ωT )exp

[−Scl +

1

2(ΛT − ln Λ)

]. (6.40)

The term linear in Λ in the exponential can be canceled by a constant shift of the

Lagrangian. The ln Λ term gives what is known as “wave function renormalization”.

To cancel these, we need to include a counterterm to the action, Sct, which depends on

Λ but is independent of q(t) in this case. Sct can be determined by demanding in the

T →∞ limit,

〈qf , T |qi, 0〉 → e−E0T 〈qf |0〉〈0|qi〉 =

√ω

πe−

ω2

(q2i+q2

f+T ), (6.41)

where |0〉 is the ground state of the harmonic oscillator, and E0 the ground state energy.

The regularized path integral, together with the counterterm, gives

〈qf , T |qi, 0〉 −→T→∞√ω

2e−ωT/2 exp

[−ω

q2i + q2

f

2+

1

2(ΛT − ln Λ)− Sct

]. (6.42)

We see that the counterterm is

Sct =1

2

∫ T

0

dτΛ− 1

2ln

Λ

π. (6.43)

After this is taken into account, (6.40) gives

〈qf , T |qi, 0〉 =

√ω

2π sinh(ωT )e−Scl(qi,qf ). (6.44)

This is indeed the correct transition amplitude between position eigenstates for the

harmonic oscillator.

122

6.4 Diagrammatic expansion

Now we would like to develop a systematic way of computing transition amplitudes with

insertions of time ordered operators, or (time ordered) “correlators”, using the path

integral. We will slightly simplify our task by considering Euclidean path integral in

the T →∞ limit, with insertions of operators at finite time. Since the Euclidean path

integral is exponentially damped by the positive definite Euclidean action SE, we may

relax the boundary conditions as T is taken to infinity. This makes sense because the

Euclidean time evolution operator e−HT becomes e−E0T times the projection operator

onto the ground state in the T →∞ limit. For instance,

〈qf , T |qi, 0〉E → 〈qf |0〉e−E0T 〈0|qi〉. (6.45)

Given some set of operators Oi(ti), we have

〈0|TO1(τ1) · · · On(τn)|0〉 =

∫[Dq]O1(τ1) · · · On(τn) e−SE∫

[Dq]e−SE(6.46)

Let us specialize to the case where the inserted operators are q(τ)’s. To compute the

path integral with the insertion of a string of q(τ)’s, it suffices to consider the generating

functional

Z[J ] =

⟨exp

[i

∫dτJ(τ)q(τ)

]⟩=

∫[Dq] exp

[−SE + i

∫dτJ(τ)q(τ)

] (6.47)

We have for instance,

〈0|Tq(τ1) · · · q(τn)|0〉 =(−i)n

Z[0]

δ

δJ(τ1)· · · δ

δJ(τn)Z[J ]. (6.48)

Suppose the action is quadratic, as in the case of a harmonic oscillator. It can be

written in the form

SE =1

2

∫dτq(τ)∆q(τ), (6.49)

where ∆ is some differential operator in τ . We shall assume that ∆ is an invertible

Hermitian operator and that SE is positive definite. Then we have

Z[J ] = Z[0] exp

[−1

2

∫dτdτ ′J(τ)G(τ, τ ′)J(τ ′)

], (6.50)

where G(τ, τ ′) is the kernel for the inverse operator of ∆, i.e. the Green’s function. It

obeys

∆τG(τ, τ ′) = δ(τ − τ ′), (6.51)

123

so that

∆τ

∫dτ ′G(τ, τ ′)J(τ ′) = J(τ). (6.52)

Note that G(τ, τ ′) = G(τ ′, τ). Z[0] is given by the functional determinant

Z[0] =

(det

∆

2π

)− 12

. (6.53)

It is useful to write Z[J ] in the form

Z[J ] = Z[0] exp

[1

2

∫dτdτ ′G(τ, τ ′)

δ

δq(τ)

δ

δq(τ ′)

]ei

∫dτJ(τ)q(τ)

∣∣∣∣q=0

. (6.54)

By multiplying both sides with an arbitrary functional of J(τ), and performing the

functional integration over J(τ), we arrive at the formula

〈F [q]〉 = Z[0] exp

[1

2

∫dτdτ ′G(τ, τ ′)

δ

δq(τ)

δ

δq(τ ′)

]F [q]

∣∣∣∣q=0

, (6.55)

for any functional F of q(τ). For example,

Z[0]−1〈q(τ1)q(τ2)q(τ3)q(τ4)〉 = exp

[1

2

∫dτdτ ′G(τ, τ ′)

δ

δq(τ)

δ

δq(τ ′)

]q(τ1)q(τ2)q(τ3)q(τ4)

∣∣∣∣q=0

= G(τ1, τ2)G(τ3, τ4) +G(τ1, τ3)G(τ2, τ4) +G(τ1, τ4)G(τ2, τ3).(6.56)

For the harmonic oscillator,

∆τ = −∂2τ + ω2. (6.57)

The Green’s function, or propagator, is given by

G(τ, τ ′) =1

2ωe−ω|τ−τ

′|. (6.58)

Sometimes it is convenient to work with Its Fourier transform, G(E) = 1/(E2 + ω2).

G(τ, τ ′) =

∫dE

2π

eiE(τ−τ ′)

E2 + ω2. (6.59)

Now consider the following example of an anharmonic oscillator,

H =1

2p2 +

1

2ω2q2 +

1

4gq4, (6.60)

where g is a coupling constant. The path integral

Z =

∫[Dq]e−SE =

∫[Dq] exp

[−∫dτ(q∆q +

1

4gq4)

](6.61)

124

with periodic boundary condition, in the limit T → ∞ gives e−E0T where E0 is the

ground state energy. When g = 0, we know that E0 = 12ω. An interesting question is

to compute E0 as a function of g. Writing E0(g) = E0(0) + δE0(g), we have

e−E0(g)T = e−E0(0)T

∞∑n=0

1

n!(−δE0(g)T )n (6.62)

To compute δE0(g), we only need to compute the term linear in T that multiplies

Z0 = e−E0(0)T . Expanding e−SE perturbatively in powers of g, we have

Z(g) ≈ Z0

∞∑n=0

1

n!(−g

4)n⟨[∫

dτq(τ)4

]n⟩(6.63)

where we wrote “≈” indicating that the RHS is only a perturbative expansion of Z(g).

The significance of this point will be explained in the next section. The linear term in

T multiplying Z0 is given by the “connected diagrams”, where all q(τ)’s are contracted

via propagators. At order O(g), we have

−g4

∫dτ3G(0)2 = −3g

4(

1

2ω)2T (6.64)

At the next order, O(g2), we have

1

2!(−g

4)2

∫dτdτ ′

[(4

2

)(4

2

)2G(τ, τ ′)2G(0)2 + 4!G(τ, τ ′)4

]=

1

2!(−g

4)2(

1

2ω)4

∫dτdτ ′

[72e−2ω|τ−τ ′| + 24e−4ω|τ−τ ′|

]=

1

2!(−g

4)2(

1

2ω)4

(72

2

2ω+ 24

2

4ω

)T,

(6.65)

and so on and so forth. We obtain an expansion of the form

E0(g) =∞∑n=0

angnω1−3n, (6.66)

with

a0 =1

2, a1 = − 3

16, a2 =

21

128, · · · (6.67)

While this expansion can give approximate answers for small g, it suffers from the

following problem: the sum (6.66) has zero radius of convergence in g! The simplest

way to understand this is the argument due to Dyson: if the sum had finite radius

of convergence, then Z(g) should be an analytic function in g at g = 0, and should

behave perfectly regularly for sufficient small negative g. But this cannot be the case

since the Hamiltonian is not bounded from below for negative g.

125

Concretely, what happens is that the absolute values of the coefficients an grow

like n! at large n, and the sum (6.66) thus has zero radius of convergence in g. This

growth of |an| can be roughly understood from the fact that the number of connected

diagrams of n quartic vertices grows like n!. A simple way to see this is to consider the

following integral (related to Bessel function)

I(g) =

∫dx exp

(−1

2x2 − g

4x4

)(6.68)

which involves a similar counting problem. It has the perturbative expansion

∞∑n=0

1

n!(−g

4)n∫dx x4ne−x

2/2 =√

2∞∑n=0

(−)nΓ(2n+ 1

2)

n!gn (6.69)

We see that the coefficients up to the sign (−)n indeed grows like n! times the n-th

power of a constant, and hence the sum has zero radius of convergence, as expect from

the non-analyticity at g = 0.

To understand the physical origin of this problem and how it can be fixed, let

us try to find the “optimal” perturbative approximation to Z(g), by truncating the

perturbative series at the order where |an|gn is minimized. Suppose g is small and

|an| ∼ n!c−n, then

|an|gn ∼ exp[n ln(g/c) + n(lnn− 1)

](6.70)

at large n. As a function of n, it is minimized at

ln(g/c) + lnn = 0 ⇒ n =c

g. (6.71)

At this value of n,

|an|gn ∼ e−cg . (6.72)

If we simply cut off the sum at this order, one expects that the “error” is of order

exp(−c/g). Note that the latter has trivial Taylor series when expanded in small

positive g. Had we made such an error in the computation of Z(g) or E0(g), we would

never know it from the perturbative expansion. Such terms are thus referred to as

“non-perturbative”.

Note that if we restore ~ dependence, an will be multiplied by ~n, and the non-

perturbative contribution would go like exp(−c/(g~)).

6.5 Instantons

Generally it is highly nontrivial to compute the non-perturbative contributions to the

path integral. Sometimes, it is easy to identify the leading non-perturbative correction

126

as coming from non-trivial saddle points of the Euclidean path integral. If there is

a classical solution to the Euclidean classical equation of motion, xcl(τ) obeying the

prescribed boundary condition of the path integral, then the integrand of the path

integral evaluated at this solution is

exp

(−1

~SE[xcl]

)(6.73)

Such a contribution precisely has the ~ dependence to account for non-perturbative

effects. One should be cautious that not all non-perturbative effects are accounted for

by the contribution from such saddle points of the classical action. However, sometimes

certain effects does not receive any perturbative contribution, and the contribution from

a nontrivial Euclidean classical solution dominates the path integral.

These classical solutions of the Euclidean equation of motion are called “instan-

tons”. The terminology refers to the property that the nontrivial part of the solution

is typically localized in Euclidean time (assuming that the action is finite). Consider

the Lagrangian L = 12x2−V (x). The Euclidean Lagrangian is LE = 1

2x2 +V (x), where

the dot now stands for derivative with respect to the Euclidean time. The Euclidean

equation of motion in this case is therefore exactly the same as the ordinary equation

of motion with V (x) replaced by −V (x). If we consider the path integral with periodic

boundary condition on the Euclidean time interval [0, T ], then the instanton solutions

describe the motion of a particle bouncing back and forth around a local minimum of

−V (x) with periodicity T .

The classic example that exhibits the use of instantons is the double well potential,

V (x) = V (−x) with two local minima at x = −a and a, as well as a local maximum at

x = 0. For instance, we may take V (x) = 14g(x2 − a2)2. Classically, the lowest energy

configuration of the particle is at either x = −a or x = a. Quantum mechanically, the

tunneling between the two minima of the potential yields a splitting of the two lowest

energy states. This splitting is a non-perturbative effect, and would not be visible

if we consider only the perturbative contribution to the path integral by expanding

around the two classically solutions x(τ) ≡ −a and x(τ) ≡ a. Suppose we compute

the partition function using the latter approach, we would see two copies of identical

contributions from paths near each minimum, and would have incorrectly concluded

that there are two degenerate ground states of the same energy.

To be concrete, consider the position eigenstates |a〉 and | − a〉. We would like to

compute

〈±a|e−HT | ± a〉, 〈±a|e−HT | ∓ a〉 (6.74)

from the path integral. The ground state energy can be read off from these, in the

large T limit. Apart from x ≡ −a and x ≡ a, the obvious nontrivial classical solutions

127

are the ones that rolls from −a to a, which we will refer to as an instanton, and the one

that rolls from a to −a, which we refer to as an anti-instanton. With large but finite

T , such an instanton solution starts with a small velocity at the top x = −a of the

potential −V (x), and ends on the other top of the potential x = a with small velocity.

In the infinite T limit, the solution starts at x = −a with zero velocity, and takes an

infinite amount of time to roll to x = a. Most of the action of the instanton, however,

is given by the contribution from the solution at finite Euclidean time. This is seen by

noting that near say x = a,

x(τ) ≈ ω(x− a), ω = 2ga2. (6.75)

At late time, x(τ) − a ∼ e−ωτ , and its contribution to the action is exponentially

suppressed. We also see from this that the instanton extends to a region in Euclidean

time of order 1/ω.

Because the instanton action receives little contribution from very early and very

late times, there are also many approximate solutions in the large T limit, consisting

of the composition of alternating instantons and anti-instantons centered at t1 < t2 <

· · · < tn. t1, · · · , tn are assumed to be widely separated so that the solutions are a

good approximations to stationary points of the action. We must then also integrate

over the positions ti of the instantons and anti-instantons. The approximation would

be consistent only if the contribution is dominated by widely separated instantons. We

will see that this is indeed the case.

We have chosen our potential with V (a) = V (−a) = 0, and the instanton solutions

of interest correspond to zero energy rolling solutions of the Euclidean Lagrangian.

The solution then obeys

x(τ) =√

2V (x) (6.76)

The classical action of an instanton (or an anti-instanton) is

S0 =

∫dτ

[1

2x2 + V (x)

]=

∫dτx√

2V =

∫ a

−adx√

2V (x). (6.77)

The approximate solution involving n instantons and anti-instantons therefore has

action nS0. Note that n is even for solutions starting at ±a and ending at ±a, while

n is odd for solutions starting at ±a, ending at ∓a.

We now expand around the instanton solution

x(τ) = xcl(τ) + y(τ) (6.78)

and write the action as S[x] = nS0 + S ′ where S ′ is the action for the quantum

fluctuations around the classical solution. S ′ involves quadratic and higher order terms.

128

In the semi-classical approximation, we simply integrate out the fluctuations using the

quadratic part of S ′, while ignoring the higher order terms in S ′. This can be done as

before by integrating over all modes of y(τ) on [0, T ]. An important subtlety here is

that, there is a “zero mode” of y(τ) that does not contribute to the action S ′. In other

words, there is an y0(τ) such that

S[xcl + y0] = S[xcl]. (6.79)

This is only possible if xcl(τ) + y0(τ) is another classical solution, for small y0(τ). And

of course we know there is such a y0(τ), because we can simply translate the instanton

in time. This is a general lesson: whenever the classical solution comes in a continuous

family, there are always zero modes associated to the first order variation of the classical

solution along this family. In this case, we have

y0(τ) ∝ dxcl(τ)

dτ(6.80)

The functional integration is performed by expanding y(τ) on an orthonormal basis

yn(τ) of functions on [0, T ],

y(τ) =∑n

cnyn(τ) (6.81)

with

yn(0) = yn(T ) = 0,

∫ T

0

yn(τ)ym(τ)dτ = δn,m, (6.82)

and then integrating over all cn’s. To normalize y0(τ) in this way, we need

y0(τ) = S− 1

20

dx(τ)

dτ, (6.83)

in the case of a single instanton. Here x(τ) stands for the one-instanton solution. To

integrate over c0 is equivalent to integrating over the center of the instanton (call it

t1). The integration measures are related by

dx(τ)

dτdt1 = y0(τ)dc0, (6.84)

and so

dc0 = S120 dt1. (6.85)

Doing the remaining Gaussian integral gives the factor∫ ∏n

dcn√2πe−S

′=

(S0

2π

) 12∫dt1(det ′

[−∂2

τ + V ′′(x(τ))])− 1

2 (6.86)

129

for one instanton (or anti-instanton), where det′ stands for the functional determinant

over all “non-zero modes” yn(τ) with n 6= 0 (the kinetic operator −∂2τ + V ′′(x(τ)) has

vanishing eigenvalue on the zero mode).

With n instantons and anti-instantons, the integration over the n center positions

(equivalent, over n zero modes), is∫ T

0

dtn

∫ tn

0

dtn−1 · · ·∫ t2

0

dt1 →T n

n!(6.87)

since the alternating instantons and anti-instantons are ordered in time by our con-

struction of the approximate solution. The integration over nonzero modes in the

n-instanton case can be treated by separating out the functional determinant of the

harmonic oscillator (i.e. that of the operator −∂2τ +ω2) from the one-instanton contri-

bution,(S0

2π

) 12 (

det ′[−∂2

τ + V ′′(x(τ))])− 1

2 ≡ K(det[−∂2

τ + ω2)])− 1

2 = K

√ω

πe−ωT/2,

(6.88)

in the large T limit, as seen previously, and then take the n-th power of the “one-

instanton correction factor” K.

The resulting n-instanton contribution is√ω

πe−ωT/2

(Ke−S0T )n

n!(6.89)

Summing up n (even and odd respectively in two cases), we obtain

〈±a|e−HT | ± a〉 ≈√ω

πe−ωT/2 cosh

(Ke−S0T

)(6.90)

and

〈±a|e−HT | ∓ a〉 ≈√ω

πe−ωT/2 sinh

(Ke−S0T

). (6.91)

Note that the sum over n dies off quickly for n greater than Ke−S0T . For n less

than Ke−S0T , the average separation between the instantons and anti-instantons is

T/n = eS0/K. This is very large in the semi-classical limit, and hence the dilute

“instanton gas” approximation is valid.

By the decomposition into energy eigenstates,

〈x|e−HT |x′〉 =∑m

〈x|m〉e−EmT 〈m|x′〉, (6.92)

130

we see that there are two energy eigenstates states that contribute to the transition

amplitudes from |a〉 to | ± a〉 in the large T limit, with energies

E± =ω

2±Ke−S0 (6.93)

The split between the two energy is non-perturbative (goes like exp(−S0~) when ~ is

restored), and would be invisible in the perturbative path integral. We also see from

(6.91) that Ke−S0 is the tunneling rate from a to −a (or −a to a).

131