1

ATOMIC PHYSICS(PHYS4011) LECTURE NOTES

Lecture notes based on a course given by Tom Kirchner.The emphasis of the course is on solving atomic systems,

in particular the Hydrogen atom through perturbation theory.Some relativistic quantum mechanics is introduced at the end

York University, 2011

Presented by: TOM KIRCHNER

LATEXNotes by: JEFF ASAF DROR

2011

YORK UNIVERSITY

2

CONTENTS

I. Introduction: the field-free Schrodinger hydrogen atom 3

A. Reduction to an effective one-body problem 3

B. Central-field problem for relative motion 3

C. Solutions of the Coloumb problem 3

D. Assorted Remarks 5

II. Atoms in electric fields: the Stark effect 7

A. Non-Degenerate perturbation theory (PT) 8

1. Comments 10

B. Degenerate Perturbation Theory 12

C. Effect on excited states: the linear Stark effect 14

III. Interaction of Atoms with Radiation 18

A. The semiclassical Hamiltonian 18

B. Time-Dependent Perturbation Theory 21

1. General Formulation 21

2. Comments 23

3. Discussion of 1st order result 23

4. Example: Slowly varying perturbation 24

5. Solution of TDSE up to 1st order 25

6. Comments 25

7. Example: Sudden Perturbation 26

8. Example:Periodic perturbation 26

C. Photoionization 28

1. Transitions into the continuum: Fermi’s golden rule (FGR) 28

2. Dipole Approximation 30

D. Outlook on Field Quantization 31

1. Construction of HF 32

2. Creation and Annihilation Operators 34

3. Interaction Between Photon Field and Electrons 35

4. The Transition Matrix Elements 36

5. Spontaneous Emission 38

IV. Brief Introduction to Relativistic Quantum Mechanics 43

A. Klien-Gordon Equation 43

1. Setting up a relativistic wave equation 43

B. Discussion of KG equation 43

C. Dirac Equation 44

1. Free Particles 44

2. Solutions of the free Dirac equation 47

3. Add Electromagnetic Potentials 49

4. The relativistic hydrogen problem 49

5. Nonrelativistic limit of the Dirac equations 51

Lecture 3 - Jan 09, 2011

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I. INTRODUCTION: THE FIELD-FREE SCHRODINGER HYDROGEN ATOM

A. Reduction to an effective one-body problem

B. Central-field problem for relative motion

Hrel =p2

2µ+ V (r); Hrelφrel(r) = Erelφrel(r) (I.1)

Note: omit hats from now on. We use the ansatz:

φrel(r) = R(r)Y`,m(φ, θ) (I.2)

=y(r)

rY`,m(φ, θ) (I.3)

where y(r) ≡ rR(r). This gives the radial Schrodinger equation which given by

y′′` (r) +

[ε− U(r)− ` (`+ 1)

r2

]y`(r) = 0 (I.4)

where ε ≡ 2µ~2Erel, U(r) = 2µ

~2 V (r).

C. Solutions of the Coloumb problem

Consider the potential

V (r) = − Ze2

4πεor(I.5)

For now on we will only consider Erel < 0 (we don’t consider the scattering problems of Erel > 0, only the bound

r

FIG. 1. The Hydrogen potential energy

states). The energy levels as well as the energy states are quantized.

ε→ εn

Erel → En = ~2

2µεn = − ~2

2µa2Z2

n2 ≈ −13.6eV(Z2

n2

)ψrel → ψn,`,m =

yn,`(r)r Y`,m

4

TABLE I. Some Hydrogen States and Their Properties. R ≡ ~22µa

≈ 13.6eV

n ` m nr = n− `− 1 state −En1 0 0 0 1s RZ2

2 0 0 1 2s RZ2

4

2 1 −1 0 2p−1RZ2

4

2 1 0 0 2p0RZ2

4

2 1 1 0 2p+1RZ2

4

where yn,` = An,`r`+1e−κnrL2`+1

n−`−1 (2κnr) and κn =√−εn = Z

na , and L is the associated Laguerre polynomials. n is

an integer and it’s called the principle quantum number. Lastly, a ≡ 4πε0~2

µe2 ≈ 0.53A is called the Bohr radius.

There is a degeneracy in the energy since the energies are independent on `,m. For a given `, there are 2` + 1states. For a given n we have the states ` = 0, 1, ..., n− 1. Hence the number of states is

n−1∑`=0

2`+ 1 = n2 (I.6)

Some states are summarized in table I The energy spectrum is shown in figure 2. Probability density is defined as

n=1

E

0

n=22p2s

1s

n=33p3s 3d

FIG. 2. The energy levels of Hydrogen

ρn,`,m(r) |φn,`,m|2 = R2n,`(r) |Yn,`(φ, θ)|

2(I.7)

If

∫r

ρn,`,m(r)dr3 =

∫ ∞0

r2R2n,`(r)dr

1︷ ︸︸ ︷∫|Y`,m|2 dΩ (I.8)

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then ρn,`,mdr3 is the probability to find the electron in the interval [r, r + dr]. We define the radial probability density

by

ρn,`(r) = r2R2n,`(r) |Y`,m(φ, θ)|2 dΩ (I.9)

= r2R2n,`(r) (I.10)

Lecture 4 - Jan 11, 2012Next we consider the momentum space representation of the Hydrogen atom. Recall

ψn,`,m(r) = 〈r|n, `m〉 (I.11)

with Hrel |n, `m〉 = En |n, `,m〉. Alternatively we can project our states onto momentum space

ψn,`,m(p) = 〈p|n, `m〉 (I.12)

= 〈p| I |n, `,m〉 (I.13)

=

∫<〈p|r〉 〈r|n, `,m〉 d3r (I.14)

=1

(2π~)3/2

∫<eip·rψn,`,m(r)d3r (I.15)

Use k = p~ and

e−ik·r = 4π

∞∑L=0

L∑M=−L

(−i)LBessel︷ ︸︸ ︷jL (kr)YLM (Ωk)Y ∗LM (Ωr) (I.16)

where jL is the spherical Bessel function. Note that here the Ωr represents normal θ and φ while Ωk represents θ andφ in spherical momentum coordinates. Using these relations

ψn,`,m(p) =4π

(2π~)3/2

∑L,M

(−i)L∫ ∞

0

r2jL (kr)Rn,`(r)dr

δL,`δM,m︷ ︸︸ ︷∫Y ∗L,M (Ωr)Y`,m(Ωr)dΩr YL,M (Ωk) (I.17)

=4π

(2π~)3/2

(−i)`∫ ∞

0

r2j`(kr)Rn,`(r)drY`,m(Ωk) (I.18)

= Pn,`(p)Y`,m(Ωp) (I.19)

The probability density is

ρn,`,m(p) = |φn,`,m(p)|2 (I.20)

While the radial probability density in momentum space is

ρn,`(p) = p2 |φn,`,m(p)|2 (I.21)

Lecture 13th, 2012

D. Assorted Remarks

1. Review Griffith’s chapters 4.1-4.3

2. Hydrogen-like ions are also solved (for Z ≥ 2). Energy scales like En ∝ Z2.

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3. One can also look at a number of exotic systems using the same results as well. e.g. positronium(e+, e−),muonium (µ+, e−), muonic atom (p, µ−). When considering these systems the energy change is due to En ∝µ = m1m2

m1+m2. For more on exotic systems consider C.T. book, volume I

4. There are corrections to be looked at which we will consider in detail later

5. Thus far we have used SI units. In these units we have the Hamiltonian (for Hydrogen)

HSI = − ~2

2m∇2 − e2

4πε0r(I.22)

To eliminate some constants we introduce atomic units. To get rid of the constants we

• measure mass in me = 1a.u. (atomic unit).

• measure charge in units of e = 1a.u..

• measure angular momentum in units of ~ = 1a.u.

• measure permittivity of 4πεo = 1a.u.

In short atomic units, a.u. are defined by me = e = ~ = 4πεo = 1.As a consequence of this the Hamiltonian in atomic units are given by

Ha.u. = −1

2∇2 − 1

r(I.23)

• For length in atomic units we use Bohr’s radius, ao = 4πεo~2

mee2= 0.53A = 1a.u. (as a consequence of our

earlier definitions).

• For energy in atomic units we consider the Hydrogen ground state. En=1 = − ~2

2mea2o= −13.6eV = − 1

2a.u.

(as a consequence of earlier definitions). Other units of energy may also be used:

1a.u.(of energy) = 27.2eV = 1 Hartree = 2 Rydberg (I.24)

• For time in atomic units we use dimensional analysis in SI:

time =distance

speed=

distance×mass

momentum=

distance2 ×mass

angular momentum(I.25)

Hence we define a unit of time:

to =a2ome

~= 2.4× 10−17s = 1a.u. (I.26)

The Bohr-like revolution time of the electron around the proton in the Hydrogen ground state is

τ = 2πto (I.27)

• We can now infer a velocity in atomic units:

τ = 2πto =2πaovo

(I.28)

⇒vo =~

meao= 2.2× 106m/s =

1

137c = 1a.u. (I.29)

• The fine-structure constant is

α =e2

4πεo~c=

~meaoc

=1

137=

(1

c

)a.u.

(I.30)

Lecture 6: January 16th, 2012

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II. ATOMS IN ELECTRIC FIELDS: THE STARK EFFECT

From classical electromagnetism we know that a uniform electric field in the z direction with field strength F is

E = F k (II.1)

The electrostatic potential is

φ(r) = −E · r (II.2)

= −Fz (II.3)

(II.4)

The potential energy of an electron is

W (r) = −eφ(r) (II.5)

a = Fez (II.6)

We need to solve the stationary Schrodinger equation:

H |φα〉 = E |φα〉 (II.7)

for (in atomic units)

H = −1

2∇2 − 1

r+ Fz (II.8)

= Ho + +W (II.9)

To get a better intuition on the problem we sketch the potential energies for x = y = 0. This is shown in figure 3.In principle there are no stationary state since “bound” electrons can always tunnel out of the potential well. This is

V

(-1/|z|)Coloumb(-1/|z|)Coloumb

z

Fz(F>0)

Eo

Eexcited

FIG. 3. The potential energies of the Stark effect

called ionization. In practice we only have ‘weak’ electric fields:

Flab ∼ 104V/cm

F1s ∼e2

4πεoa2o

= 5× 109V/cm

8

Thus in practice the tunnel effect is unimportant for low-lying states. What does happen is the energy levels shiftand split.

A. Non-Degenerate perturbation theory (PT)

1. Nondegerenate PT:General formulationThere are different types of perturbation theory.Consider the Hamiltonian of the form

H |φα〉 = Eα |φα〉 (II.10)

where we assume H = Ho +W and 〈φα|φβ〉 = δαβ . Assume further that

Ho |φoα〉 = E(0)α |φoα〉 (II.11)

is known and non degenerate with⟨φoα|φoβ

⟩= δαβ . In other words that φα forms a basis. If we assume that W

is small then

W = λw λ 1, λ ∈ < (II.12)

We can now write

Ho + λw |φα(λ)〉 = Eα(λ) |φα(λ)〉 (II.13)

Note that the eigenvalues and eigenvectors should depend on λ since the perturbation changes the system. Wecan now use Taylor expansions:

Eα(λ) = E(0)α +

dEαdλ

∣∣∣∣λ=0

λ+1

2

d2Eαdλ

∣∣∣∣λ=0

λ2 + ... (II.14)

|φα(λ)〉 =∣∣φ0α

⟩+

d

dλ|φα〉

∣∣∣∣λ=0

λ+ ... (II.15)

Consider the derivative of the Schrodinger equation with respect to λ:

d

dλ[(Ho + λw − Eα(λ)) |φα(λ)〉] = 0 (II.16)

(Ho + λw − Eα(λ)) |φ′α(λ)〉+ (w − E′α(λ)) |φα(λ)〉 = 0 (II.17)

(II.18)

where ddλ |φα(λ)〉 ≡ |φ′α(λ)〉. Multiplying both sides of the equation by the bra 〈φβ(λ)|⟨

φβ(λ)|H(λ)− Eα(λ)|φ′β(λ)⟩

+ 〈φβ(λ)|w − E′α(λ)|φ(λ)〉 = 0 (II.19)

Assume that α = β

〈φα(λ)|Eα(λ)︷ ︸︸ ︷〈φα|H(λ) + 〈φα(λ)|w|φα(λ)〉 − E′α(λ)

1︷ ︸︸ ︷〈φα(λ)|φα(λ)〉 = 0 (II.20)

E′α(λ) = 〈φα(λ)|w |φα(λ)〉 (II.21)

E′α (λ = 0) = 〈φoα|w |φoα〉 (II.22)

Next we assume that α 6= β

〈φβ(λ)|Eβ(λ)− Eα(λ) |φ′α(λ)〉+ 〈φβ(λ)|w |φα(λ)〉 = 0 (II.23)

(Eβ(λ)− Eα(λ)) 〈ψβ(λ)|φ′α(λ)〉+ 〈φβ(λ)|w |φα(λ)〉 = 0 (II.24)

〈φβ(λ)|φ′α(λ)〉 =〈φβ(λ)|w |φα(λ)〉Eα(λ)− Eβ(λ)

(II.25)

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Consider the completeness relation and inserting inside the above equation gives

|φ′α(λ)〉 =∑β

|φβ(λ)〉 〈φβ(λ)|φ′α(λ)〉 (II.26)

∑β

|φβ(λ)〉 〈φβ(λ)|w |φα(λ)〉Eα(λ)− Eβ(λ)

(II.27)

but this doesn’t include α = β. What about 〈φα(λ)|φ′α(λ)〉?

d

dλ〈φα|φα(λ)〉 = 0 (II.28)

〈φ′α(λ)|φα(λ)〉+ 〈ψα(λ)|φ′α(λ)〉 = 0 (II.29)

〈φα(λ)|φ′α(λ)〉∗ + 〈ψα(λ)|φ′α(λ)〉 = 0 (II.30)

2 〈φα(λ)|φ′α(λ)〉 = 0 (II.31)

if 〈φα|φ′α〉 ∈ <. However the eigenstate of a Hermitian operator can always be transformed into a basis (bytaking linear combinations of them) such that they are real. Hence these overlaps are real. Which means thatthe overlaps are zero.

|φ′(λ)〉 =∑β 6=α

〈φα(λ)|w |φα(λ)〉Eα(λ)− Eβ(λ)

|φβ(λ)〉 (II.32)

and hence

|φ′α (λ = 0)〉 =∑β 6=α

⟨φ0β

∣∣∣w ∣∣φ0α

⟩E

(0)α − E(0)

β

∣∣φ0β

⟩(II.33)

One can go on to other orders as well

d2

dλ2Eα(λ) =

d

dλE′α(λ) (II.34)

Lecture 7 - January 18th, 2012We can express the perturbed energies by

Eα(λ) = E(0)α +

dEαdλ λ=0

λ+1

2

d2E

dλ2

∣∣∣∣λ=0

λ2 + ... (II.35)

= E(0)α + λ

⟨φ0α

∣∣w ∣∣φ0α

⟩+ λ2

∑β 6=α

∣∣∣⟨φ0α

∣∣w ∣∣∣φ0β

⟩∣∣∣2E

(0)α − E(0)

β

+ ... (II.36)

= E(0)α +

⟨φ0α

∣∣W ∣∣φ0α

⟩+∑β 6=α

∣∣∣⟨φ0α

∣∣W ∣∣∣φ0β

⟩∣∣∣2E

(0)α − E(0)

β

+ ... (II.37)

= E(0)α + ∆E(1)

α + ∆E(2)α + ... (II.38)

(II.39)

|φα(λ)〉 =∣∣φ0α

⟩+∑β 6=α

⟨φ0α

∣∣W ∣∣∣φ0β

⟩E

(0)α − E(0)

β

∣∣φ0β

⟩(II.40)

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1. Comments

(a) These equations are non-defined only when E(0)α 6= E

(0)β ∀α, β. In other words non-degenerate systems.

(b) Convergence is difficult to check in perturbation theory but consistency checks can be done. One can checkthat ∣∣∣∣∣∣

⟨φ0α

∣∣W ∣∣∣φ0β

⟩E

(0)α − E(0)

β

∣∣∣∣∣∣ << 1 (II.41)

If this is not fulfilled then it indicates that perturbation theory will likely fail.

(c) Full calculations for the energies, Eα, beyond first order are in general not possible (due to the infinitesum).

(d) Griffith, 6.1; Liboff, B.1

2. Applications to H(1s) in an electric field

φ01s(r) =

1√πe−r (II.42)

E01s = −1

2(II.43)

W = Fz (II.44)

We can calculate the first order energy correction:

∆E(1)1s =

⟨φ0

1s

∣∣Fz ∣∣φ01s

⟩(II.45)

=F

π

∫e−2rzd3r (II.46)

=F

π

∫r cos θ sin θe−2rr2drdθdφ (II.47)

=F

π

∫ ∞0

r3e−2r

:0∫ π

0

1

2sin 2θdθ

∫dφ (II.48)

= 0 (II.49)

However this doesn’t tell us whether or not the rest of the correction orders are zero or non-zero. To get an ideafor the second order correction we check the consistency criterion. Consider the overlap of the contribution ofthe 2p and 1s states. Note that this gives the smallest possible denominator (greatest correction).∣∣∣∣∣∣

⟨φ

(0)1s

∣∣∣Fz |φ2p〉

E(0)1s − E

(0)n=2

∣∣∣∣∣∣ =

∣∣∣∣∣∣⟨φ

(0)2p0

∣∣∣ z ∣∣φ01s

⟩− 1

2 + 18

∣∣∣∣∣∣ (II.50)

≈ 2F (II.51)

For the Stark effect the F ≈ 10−5 in atomic units. Hence we expect the second order effect to be small but nonzero. Next we estimate the full second order correction

∣∣∣∆E(2)1s

∣∣∣ =

∣∣∣∣∣∣∣∑β 6=1s

∣∣∣⟨φ0β

∣∣∣Fz ∣∣φ01s

⟩∣∣∣2E

(0)1s − E

(0)β

∣∣∣∣∣∣∣ (II.52)

To get an upper bound for this estimate we can replace the denominator by it’s minimum value which correspondsto n = 2. ∣∣∣E(0)

1s − E(0)β

∣∣∣ ≥ ∣∣∣E(0)1s − E

(0)n=2

∣∣∣ =3

8(II.53)

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Thus we can write ∣∣∣∆E(2)1s

∣∣∣ =8

3F 2

∑β 6=1s

∣∣⟨φ0β

∣∣ z ∣∣φ01s

⟩∣∣2 (II.54)

=8

3F 2

∑β 6=1s

∣∣⟨φ01s

∣∣ z ∣∣φ0β

⟩ ⟨φ0β

∣∣ z ∣∣φ01s

⟩∣∣ (II.55)

=8

3F 2

∣∣∣∣∣∣⟨φ01s

∣∣ z ∑β 6=1s

∣∣φ0β

⟩ ⟨φ0β

∣∣ z ∣∣φ01s

⟩∣∣∣∣∣∣ (II.56)

but ∑β 6=2s

∣∣φβ0

⟩ ⟨φ0β

∣∣ = 1−∣∣φ0

1s

⟩ ⟨φ0

1s

∣∣ (II.57)

Hence ∣∣∣∆E(2)1s

∣∣∣ =8

3F 2∣∣⟨φ0

1s

∣∣ z (1− ∣∣φ01s

⟩ ⟨φ0

1s

∣∣) z ∣∣φ01s

⟩∣∣ (II.58)

=8

3F 2

⟨φ0

1s

∣∣ z2∣∣φ0

1s

⟩−⟨φ0

1s

∣∣ z ∣∣φ01s

⟩

:0⟨

φ01s

∣∣ z ∣∣φ01s

⟩(II.59)

=8F 2

3

⟨φ0

1s

∣∣ z2∣∣φ0

1s

⟩(II.60)

It turns out that the integral is easy to do ∣∣∣∆E(2)1s

∣∣∣ ≤ 8F 2

3(II.61)

We also know that the energy correction is negative, ∆E(2)1s < 0. One can also calculate the exact result:

∆E(2)1s = −9

4F 2 (II.62)

This is called the “quadratic Stark effect”. Since F is small this is a very small correction.

3. InterpretationConsider a classical charge distribution in an electric field. The classical energy of a charge distribution is givenby

U =

∫ρ(r)φ(r)d3r (II.63)

= −F∫ρ(r)zd3r (II.64)

= −Fpz (II.65)

where pz ≡ z component of the dipole moment. Connection to QM:

ρ(r) = − |φ1s(r)|2 (II.66)

pz = −∫|φ1s(r)|2 zd3 (II.67)

= −〈φ1s| z |φ1s〉 (II.68)

= −⟨φ

(0)1s + λφ

(1)2s + ...

∣∣∣ z ∣∣∣φ(0)1s + φ

(1)1s + ...

⟩(II.69)

= −⟨φ

(0)1s

∣∣∣ z ∣∣∣φ(0)1s

⟩− λ

⟨φ

(1)1s

∣∣∣ z ∣∣∣φ(0)1s

⟩− λ

⟨φ

(0)1s

∣∣∣ z ∣∣∣φ(1)1s

⟩+O(λ2) + ... (II.70)

12

The first term is zero. By choosing real eigenstates we can

pz = −2∑β 6=1s

⟨φ

(0)1s

∣∣∣ z ∣∣∣φ(0)β

⟩⟨φ

(0)β

∣∣∣Fz ∣∣∣φ(0)1s

⟩E

(0)1s − E

(0)β

(II.71)

= −2F∑β 6=1s

∣∣∣〈φ1s| z∣∣∣φ(0)β

⟩∣∣∣2E

(0)1s − Eβ(0)

+O(λ2) (II.72)

= − 2

F∆E

(2)1s +O(λ2) (II.73)

=9

2F +O(F 2) (II.74)

In summary

(a)

∆E(1)1s = 0 ⇐⇒ p(0)

z = −⟨φ

(0)1s

∣∣∣ z ∣∣∣φ(0)1s

⟩= 0 (II.75)

This is expected since a spherical charge distribution has no static dipole moment (and equivalently for aspherical probability distribution)

(b) ∆E(2)1s 6= 0 reflects that we have a nonzero induced dipole moment. This means that we have a nonzero

polarizability

αD =1

Fp(1)z =

9

2(II.76)

B. Degenerate Perturbation Theory

We have a Hamiltonian given by

H = Ho + λw (II.77)

where

Ho

∣∣φoα,j⟩ = E(0)α

∣∣φoα,j⟩ , j = 1, 2, ..., gα (II.78)

H |φα,j〉 = Eα,j |φα,j〉 (II.79)

Here we have assumed that after the perturbation the energy levels are non-degenerate (this is not always the casebut it is the case for the Stark effect). The states∣∣φoα,j⟩ , j = 1, 2, ...gα

(II.80)

span the subspace (of Hilbert space) associated with E(0)α . Any linear combination is an eigenstate of Ho for E

(0)α .

Effect of perturbation theory is shown in figure 4 Note that as the perturbation is increased we encounter a changein state to some linear combination of states.

|φαj〉 →∣∣φ0α,j

⟩(II.81)

We make the ansatz:

Eα,j(λ) = E(0)α + λE

(1)α,j + λ2E

(2)α,j + ... (II.82)

|φα,j(λ)〉 =∣∣φ0α,j

⟩+ λ

∣∣φ1α,j

⟩+ ... (II.83)

Inserting this into the Schrodinger equation:

(Ho + λw)∣∣φ0

α,j

⟩+ λ

∣∣φ1α,j

⟩+ ...

=E(0)α + λE

(1)α,j

(∣∣φ0α,j

⟩+ λ

∣∣∣φ(1)α,j

⟩+ ...

)(II.84)

Ho

∣∣φ0α,j

⟩+ λ

w∣∣φ0α,j

⟩+Ho

∣∣∣φ(1)α,j

⟩+O

(λ2)

= E(0)α

∣∣φ0α,j

⟩+ λ

E(1)α

∣∣φ0α,j

⟩+ E(0)

α

∣∣∣φ(1)α,j

⟩+O

(λ2)

(II.85)

13

0

E(0)

E

E

E

E

E g

1

2

3

4

FIG. 4. The effects of perturbation theory: degeneracy is lifted

By equation coefficients we know that

λ0 : Ho

∣∣φoα,j⟩ = E(0)α

∣∣φ0α,j

⟩(II.86)

λ1 : Ho

∣∣∣φ(1)α,J

⟩+ w |φα,j(0)〉 − E(0)

α

∣∣∣φ(1)α,j

⟩− E(1)

α,j |φα,j〉 = 0 (II.87)

The first equation just shows consistency. By adding on a bra to the second equation we get⟨φoβ,`

∣∣Ho − E(0)α

∣∣∣φ(1)α,j

⟩+⟨φoβ,`

∣∣w − E(1)α,j

∣∣φoα,j⟩ = 0 (II.88)(E

(0)β − E

(0)α

)⟨φoβ,`|φ

(1)α,j

⟩+⟨φoβ,`

∣∣w ∣∣φoα,j⟩− Eα,jδαβδ`,j = 0 (II.89)

Consider the case of α = β and ` = j

E(1)α,j =

⟨φoα,j

∣∣w ∣∣φoα,j⟩ (II.90)

Consider the case of α = β and ` 6= j ⟨φoβ,`

∣∣w ∣∣φoα,j⟩ = 0 (II.91)

This result is a property of the perturbation and it follows due to the ansatz we chose. However this is not true forall perturbations. For example

〈H(2s)| z |H(2po)〉 6= 0 (II.92)

To ensure that this is always the case we need to diagonalize our perturbations. We modify our ansatz by

Eα,j = E(0)α + λE

(1)α,j + ... (II.93)

|φα,j〉 =∣∣∣φoα,j⟩+ λ

∣∣∣φ(1)α,j

⟩+ ... (II.94)

with∣∣∣φ0α,j

⟩=∑gαk=1 a

αk,j

∣∣∣φ0α,k

⟩such that

⟨φ0α,`

∣∣∣w ∣∣∣φ0α,j

⟩= 0.

Insert the two ansatz into the Schrodinger equation, sort, and project on⟨φ0α,`

∣∣∣⟨φ0α,`

∣∣w − E(1)α,j

∣∣∣φ0α,j

⟩= 0 (II.95)

gα∑k=1

⟨φ0α,`

∣∣w − E(1)α,j

∣∣φ0α,k

⟩aαk,j = 0; ` = 1, 2, ...gα (II.96)

14

This can be rewritten in matrix form aswα11 − Eα,j w12 w13 ...

wα21 wα22 − E(1)α,j ... ...

......

. . ....

... ... ... wgα,gα − E(1)α,j

aα1,jaα2,j

...aαgα,j

00...0

(II.97)

The condition for a nontrivial solution is

det ( ) = 0 (II.98)

and has gα roots E(1)α,j , j = 1, 2, ...gα.

The procedure for using degenerate perturbation theory is as follows

1. Solve “secular” (characteristic) equation and obtain eigenvaluesE

(1)α,j , j = 1, ...gα

(II.99)

2. Insert eigenvalues into the matrix equations and obtain the expansion coefficients.aαk,j ; k, j = 1, ...gα

(II.100)

3. One can go further and check that ⟨˜φ0α,`

∣∣∣w ∣∣φ0α,j

⟩= E

(1)α,jδ`,j (II.101)

4. It’s possible to go on to calculate wavefunctions and 2nd order energy corrections but it’s tedious

C. Effect on excited states: the linear Stark effect

1. Matrix elements

wα`,k =⟨φ0α,`

∣∣w ∣∣φ0α,k

⟩(II.102)

or for the Hydrogen atom (w = Fz): ⟨φ0n`m

∣∣ z ∣∣φ0n`′m′

⟩(II.103)

with φn`m(r) = Rn`(r)Y`m(Ω), z = r cos θ√

4π3 rY10

⟨φ0n`m

∣∣ z ∣∣φ0n`′m′

⟩=

√4π

3

∫ ∞0

r3Rn`(r)Rn`′(r)dr

∫Y`,m(Ω)Y10(Ω)Y`′,m′(Ω)dΩ (II.104)

The radial integral is simple enough. We have done similar ones in the past. The angular integral can be donein general using “Wiger-Eckant-Theorem”:√

4π

2L+ 1

∫Y ∗`m(Ω)YLM (Ω)Y`′m′(Ω)dΩ = (−1)

m√

(2`+ 1) (2`′ + 1)

(` L `′

−m M m′

)(` L `′

0 0 0

)(II.105)

with

Wigner’s 3j symbols︷ ︸︸ ︷(j1 j2 j3m1 m2 m3

)= (−1)

j1−j2−m3 (2j3 + 1)−1/2 ×

Clebsch-Gordan Coefficients︷ ︸︸ ︷〈j1,m1, j2,m2|j3,−m3〉 For more information on these

topics refer to Liboff, chapter 9 or Cohen-T. Chapter 6 and 10.

15

Lecture 9 - January 25th, 2012 The selection rules can be written in terms of the Wigner 3j symbols.

(j1 j2 j3m1 m2 m3

)6=⇐⇒

m1 +m2 +m3 = 0

and

|j1 − j2| ≤ j3 ≤ j1 + j2(triangular condition)

(II.106)

(j1 j2 j30 0 0

)6=⇐⇒

triangular condition

and

j1 + j2 + j3 = even

(II.107)

Applying these relations to our situation our integral is non zero only if

m = m′ (II.108)

∆` = `− `′ = ±1 (II.109)

These are sometimes called the electric dipole selection rules (E1) (special case).

2. Linear Stark effect for H(n = 2)Consider the degenerate stats

φo2s, φo2po , φ

o2p−1

, φo2p1

(II.110)

We consider the matrix eigenvalue problem:∑`′,m′(n=2)

〈φo2`m|w − E(1)n=2

∣∣φo2n,`′,m′⟩ an=2`,m,`′,m′ = 0 (II.111)

Consider

wi,j = 〈φo2`m|w∣∣φo2n,`′m′⟩ (II.112)

If i = j then wi,j = 0 because ∆` = 0. Further we have a symmetric matrix since the states are real. By usingthe selection rules we see that

w =

0 w12 0 0w12 0 0 00 0 0 00 0 0 0

(II.113)

Hence the only potentially nonzero element is the w12. Note that if the radial part is nonzero then the elementmay still be zero. It’s easy to calculate the element explicitly:

w12 = 〈φo2s| z∣∣φo2po⟩ = −3a.u. (II.114)

Hence we have the secular equation ∣∣∣∣∣∣∣∣−E(1) w12 0 0w12 −E(1) 0 00 0 −E(1) 00 0 0 −E(1)

∣∣∣∣∣∣∣∣ = 0 (II.115)

This matrix is block diagonal and it’s easy to find the equation:(E(1)

)2(

E(1))2

− w212

= 0 (II.116)

E(1) = 0, 0, w12,−w12 (II.117)

Hence the first order energy corrections are

∆E(1) = 0, 0,−3F, 3F (II.118)

We are left with three lines. Two lines stay degenerate as represented by the zeroes however the originally oneline splits into three.Next we calculate the mixing coefficients of the expansion, a`,m,`′,m′ . We insert the eigenvalues into our equation.

16

(a) First consider E(1)1 = E

(1)2 = 0 0 w12 0 0

w12 0 0 00 0 0 00 0 0 0

a2s

a2po

a2p−1

a2p+1

=

0000

(II.119)

This is true only if a2s = a2po = 0. a2p−1and a2p+1

are undetermined. The state that corresponds to thisstate is any linear combination of the 3rd and 4th states. We choose∣∣∣φE1

⟩=∣∣∣φo2p−1

⟩(II.120)∣∣∣φE2

⟩=∣∣φ2p−1

⟩(II.121)

(b) Next we consider E(1)3 = +w12: −w12 w12 0 0

w12 −w12 0 00 0 −w12 00 0 0 −w12

a2s

a2p0

a2p−1

a2p+1

=

0000

(II.122)

This gives a2p−1 = a2p+1 = 0. The other two equations are

− w12a2s + w12a2po = 0 (II.123)

w12a2s − w12a2po = 0 (II.124)

This requires that a2s = a2po .

(c) The final eigenvectors for E4 = −w12 are given by a2s = −a2p0 , a2p−1= a2p+1

= 0.

We have fixed the components of the eigenvectors (the energy corrections). We should fix the normalization ofthe last two states (we include the previous states for concreteness):∣∣∣φE(1)

3

⟩=

1√2

(|φo2s〉+

∣∣φo2po⟩) (II.125)∣∣∣φE(1)4

⟩=

1√2

(|φo2s〉 −

∣∣φo2po⟩) (II.126)∣∣∣φE(1)1

⟩=∣∣∣φo2p−1

⟩(II.127)∣∣∣φE(1)

2

⟩=∣∣∣φo2p+1

⟩(II.128)

3. Summary and interpretation

(a) Splitting of energy levels is shown in figure

(b) Note that⟨φo2`,m

∣∣∣ z |φo2`m〉 = 0 hence the original states (φo2s, φ2p0 , φo2p±1

) have no static dipole moment.

However the Stark states∣∣∣φE(1)

3

⟩,∣∣∣φE(1)

4

⟩do have non zero static dipole moment

p0z,3 = −

⟨φE

(1)3

∣∣∣ z ∣∣∣φE(1)3

⟩) (II.129)

= 3a.u. (II.130)

pz,4 = −3a.u. (II.131)

(c) The new states are shown in figure 6

17

F

E m=+/- 1

m=0

m=0E4

E3

FIG. 5. The linear Stark splitting of energy levels

z

2s

x=y=0

2p0

z

2s

x=y=0

2p0

z

E

x=y=0

3

z

E

x=y=0

4

FIG. 6. The Stark states

(d) Note that we still have azimuthal symmetry.

[`z, w] = 0 (II.132)

Hence m is still a good quantum number

(e) The last question to consider is the effect on the n = 3 shell states. We would need to consider the

(3s, 3po, ..., 3d±2) (II.133)

18

This involves solving the secular equation of a 9× 9 matrix.

Lecture 10, January 30th, 2012

(f) To solve the Stark effect on the n = 3 states one can decompose the matrix into blocks. To show thismethod consider the L-shell problem one more time

det

−E(1) w12 0 0w12 −E(1) 0 00 0 −E(1) 00 0 0 −E(1)

(II.134)

The matrix is block diagonal and we have 3 subspaces since the m states don’t mix (this can also by seenby [`z,W ] = 0. For block diagonal matrices of the form:

det

A1 0 ...

0 A2

...... ...

. . .

= detA1 detA2... (II.135)

detA = 0 ⇐⇒ detAi = 0; ∀i. For this case:

m = ±1 det(−E(1)

)= 0 ⇐⇒ E

(1)1 = E

(1)1 = 0 (II.136)

m = 0 det

(−E(1) w12

w12 −E(1)

)= 0 ⇐⇒

(E(1)

)2

− w212 = 0 (II.137)

In the n = 3 shell we can subdivide the states by

m =

0 3s, 3po, 3do−1 3p−1, 3d−1

1 3p1, 3d1

−2 3d−2

2 3d2

(II.138)

III. INTERACTION OF ATOMS WITH RADIATION

Consider the scheme of Quantum Theory shown in figure 7.

A. The semiclassical Hamiltonian

1. Consider a classical particle with charge q in an electromagnetic field. The force acting on the particle is simplythe Lorentzian force.

FL = q (E + v ×B) (III.1)

Introduce the electromagnetic potentials, A is the vector potential and φ is the scalar potential

E = −∇φ− ∂

∂tA (III.2)

B = ∇×A (III.3)

Inserting these equations gives

FL = q

(−∇φ− ∂A

∂t+ v × (∇×A)

)(III.4)

19

FIG. 7. Scheme of Quantum Theory

Define a generalized potential energy,

U = q (φ−A · v) (III.5)

The Lagrangian is given by

L = T − U (III.6)

=m

2v2 − qφ+ qv ·A (III.7)

The Lagrangian equations of motion

∂

∂t

∂L

∂vi− ∂L

∂xi= 0 ⇐⇒ ma = FL (III.8)

Hence the generalize potential energy yields the correct force. The Hamiltonian can now easily be extracted

H = p · v − L (III.9)

Note that

p =∂L

∂v= mv + qA (III.10)

is the canonical momentum. We can rearrange this equation for velocity:

v =1

m(p− qA) (III.11)

We can now rewrite our Hamiltonian without v (we need to eliminate this variable to have a proper Hamiltonian):

H =p

m· (p− qA)− 1

2m(p− qA)

2+ qφ− q

m(p− qA) ·A (III.12)

=1

2m(p− qA)

2+ qφ (III.13)

2. Next consider the quantum mechanical Hamiltonian for an electron.

• To consider bound states we add another scalar potential, V due to a nucleus.

• The charge for the electron is q = −e

20

• Next we perform quantization (p→ p = ~i∇ ). Hence

H =1

2m(p + eA)

2 − eφ+ V (III.14)

Hψ(r, t) =1

2m

(~i∇+ eA(r, t)

)(~i∇+ eA(r, t)

)φ(r, t)− eφ(r)ψ (r, t) + V (r)φ (r, t) (III.15)

=1

2m

(−~2∇2φ+

~ei∇ (Aψ) +

~eiA∇ψ + e2A2ψ

)− eφψ + V ψ (III.16)

=

Ho︷ ︸︸ ︷(− ~2

2m∇2 + V

)ψ +

e~mi

A · ∇ψ +

(e~

2mi∇ ·A

)ψ +

(e2

2mA2 − eφ

)ψ (III.17)

= Ho +

W (t)︷ ︸︸ ︷e

mA · p +

e

2mp ·A +

e2

2mA2 − eφ (III.18)

where Ho is the Hydrogen Hamiltonian without electromagnetism and W (t) is the time dependent pertur-bation.

• Assume for the following the electromagnetic is a free electromagnetic field which is characterized by nocharges and no currents, ρ = 0; J = 0. Here its convenient to choose the Coloumb gauge, ∇ ·A = 0. Thisis useful because then we arrive at the equations

∇2A− 1

c2∂2A

∂t2= 0 (III.19)

and

φ = 0 (III.20)

Note this doesn’t mean that there is no electric field (since E = −∂A∂t ). Consider the monochromaticsolution to the wave equation, a plane wave.

A(r, t) = Π |Ac| cos (k · r− ωt+ α) (III.21)

One can perform two checks. Firstly

∇ ·A = −Π · k |Ac| sin (k · r− ωt+ α) = 0 (III.22)

Hence Π ⊥ k. Thus the vector potential is a transverse wave. The second check is the wave equation

∇2A =1

c2∂2A

∂t2(III.23)

−k2A = −ω2

c2A (III.24)

Thus we see that this is a solution given that k = ωc . Going back to the perturbation we have (due to the

Coloumb gauge p ·A one term disappears and another disappears because we don’t have a scalar potential)

W (t) =e

mA · p +

e2

2mA2 (III.25)

Assume that A is weak. If this is the case we can neglect the second term:

W (t) ≈ e

mA · p (III.26)

21

Lecture 11 - February 1st, 2012Test 1 is up to (and including) last lecture. In other words up to this point.

We can write the Hamiltonian due to an external field by

H = Ho +W (t) (III.27)

where

Ho =p2

2m+ V (r); W (t) =

e

mA(r, t) · p +

e2

2mA2(r, t) (III.28)

Note that since we have a time dependent Hamiltonian we cannot use the stationary Schrodinger equation. Thus weuse time dependent perturbation theory.

B. Time-Dependent Perturbation Theory

1. General Formulation

Consider a perturbation W (t) = λw(t) on a Hamiltonian which is stationary.

H(t) = Ho + λw(t) (III.29)

The goal is to solve the time-dependent Schrodinger equation:

i~∂

∂t|ψ(t)〉 = H(t) |ψ(t)〉 (III.30)

The SE is a initial value problem. We assume we know the state at some time. Assume for t ≥ to that

W (t) = 0; |φ(to)〉 = |φo〉 (III.31)

where Ho |ψj〉 = εj |ψj〉 (j = 0, 1, ...). We assume that W is switched on at to and we look into how the systemdevelops in time.

Define |ψj(t)〉 = e−iεjt/~ |φj〉 then inserting into the SE (before perturbation) by

i~∂

∂t|ψj(t)〉 = i~

(∂

∂te−iεjt/~ |φj〉

)(III.32)

= εje−iεjt/~ |φj〉 (III.33)

= Ho |ψj〉 (III.34)

Hence if |φ〉 solves the SE then so does |ψ〉. We denote the |ψ(t)〉 by the solution after the perturbation has beenturned on. We expand these states as

|ψ(t)〉 =∑j

cj |ψj(t)〉 (III.35)

and insert into the time dependent SE

i~∂

∂t

∑j

cj(t) |ψj(t)〉

= (Ho +W (t))∑j

cj(t) |ψj(t)〉 (III.36)

∑j

(i~cj(t) + εj) e−iεjt/~ |φj〉 = (Ho +W (t))

∑j

cj(t)e−iεjt/~ |φj(t)〉 (III.37)

(III.38)

22

We project these states onto the state 〈ψk(t)| = 〈φk| eiεkt/~:

∑j

ei~ (εk−εj)t

δj,k︷ ︸︸ ︷〈φk|φj〉 (i~cj(t) + cjεj) =

∑j

ei~ (εk−εj)tcj(t) 〈φk|Ho + λw(t) |φj〉 (III.39)

i~ck(t) +ck(t)εk =

ck(t)εk +∑j

ei~ (εk−εj)tcj(t) 〈φk|λw(t) |φj〉 (III.40)

i~ck =∑j

ei~ (εk−εj)tcj(t)λwkj (III.41)

where wkj ≡ 〈φk|w |φj〉. Note that thus far we have not made any approximations. These are sometimes calledcoupled-channel equations. If we assume that W (t > tf ) = 0 then for t > tf the matrix elements on the right sideare 0 and hence

ck(t > tf ) = 0 (III.42)

and the states are constant. The probabilities for transitions |φ0〉 → 〈φk〉 (equivalently |ψ0〉 → |ψk〉 ). The probabilityfor transition to state k is

pk = |ck|2t>tf = |〈ψk|ψ〉|2t>tf = |〈φk|ψ〉|2t>tf (III.43)

To check the answer one can check that∑k pk = 1. This turns out to be true. We finally introduce perturbation

theory. Notice that the only thing we can expand is the expansion coefficients. We use a power series expansion

ck(t) = c(0)k (t) + λc

(1)k (t) + λ2c

(2)k + ... (III.44)

We insert this into the coupled channel equations

i~c(1)k + λc

(1)k + ...

= λ

∑j

ei~wkj(εk−εj)t

c(0)j + λc

(1)j + ...

(III.45)

consider zeroth order, λ0:

i~c(0)k = 0 (III.46)

Next consider first order

i~c(1)k =

∑j

c(0)j e

i~ (εk−εj)twkj (III.47)

Lastly consider second order

i~c(2)k =

∑j

c(1)j e

i~ (εk−εj)twkj (III.48)

Notice that you can in principle get a solution for all orders using this method. Once you have first order you can getsecond order and once you have second order you can get third order etc. In other words we solve these successively.For first order

c(0)k (t) = const = δk,0 (III.49)

Our initial condition was the ground state. Hence the coefficients are zero except the ground state coefficient. Inzeroth order we just stay in the ground (initial) state. Input this into higher orders:

i~c(1)k =

∑j

δj,0ei~ (εk−εj)twk,j (III.50)

= ei~ (εk−ε0)t 〈φ|w(t) |φ0〉 (III.51)

c(1)k (t)−

*0

c(1)k (t0) = − i

~

∫ t

to

ei~ (εk−ε0)t′ 〈φk|w(t′) |φj〉 dt′ (III.52)

c(1)k (t) = − i

~

∫ t

to

ei~ (εk−ε0)t′ 〈φk|w(t′) |φj〉 dt′ (III.53)

23

where c(1)k (t0) = 0 since our initial conditions say that the system is in the ground state. In second order we have

i~c(2)k =

∑j

c(1)j e

i~ (εk−εj)twk,j (III.54)

i~c(2)k = − i

~∑j

∫ t′

to

ei~ (εj−ε0)t′′ 〈φj |wj(t′′) |φ0〉 dt′′e

i~ (εk−εj)t 〈φk|w(t) |φj〉 (III.55)

c(2)k (t) = − 1

~2

∑j

∫ t

t0

dt′∫ t′

t0

dt′′ei~ (εk−εj)t′e

i~ (εj−ε0)t′′wk,j(t

′)wj,0(t′′) (III.56)

Lecture 12 - February 6th, 2012

2. Comments

1. ‘Exact’ calculations beyond 1st order are in general impossible (due to infinite sums)

2. Practical calculations of second order often rely on the ‘closure approximation ’. Notice that the second ordercalculation is not an infinite sum if the εj are constant (you can use the completeness relation formed fromwk,j(t

′)wj,0(t′′) = 〈φk|w(t′) |φj〉 〈φj |w(t′′) |φ0〉). Thus one can approximate that

εj → ε (III.57)

where ε is an average energy value. This produces a second order correction of

c(2)k (t) = − 1

~2

∫ t

t0

dt′∫ t′

t0

dt′′ei~ (εk−ε)t′e

i~ (ε−ε0)t′′ 〈φk|w(t′)w(t′′) |φ0〉 (III.58)

3. The interpretation of the results are as follows. The first order result is called a direct transition since

|φ0(t0)〉 w−→ |φk(t)〉 (III.59)

One the contrary for second order we have

|φ0〉w−→ |φj〉

w−→ |φk〉 (III.60)

We call this a transition through a ‘virtual’ state (two steps). The |φj〉 state serves as an intermediate step. Wecan see this pictorially as shown in figure 8.

3. Discussion of 1st order result

ck(t) ≈ c(0)k + λc

(1)k (III.61)

= δk0 −i

~

∫ t

t0

eiωk0t′Wk0(t′)dt′ (III.62)

where ωk0 ≡ εk−ε0~ and Wk0(t) = 〈φk|W (t) |φ0〉 = λ 〈φk|w(t) |φ0〉 is called a transition matrix element. The proba-

bility to transition to state k 6= 0 (from the initial, ground state) is

P 1st−order0→k =

1

~2

∣∣∣∣∫ t

t0

eiwk0t′Wk0(t′)dt′

∣∣∣∣2 (III.63)

24

Time

t

t

w(t')

0

FIG. 8. Diagram of first order perturbation theory

For k = 0 we have an elastic collision (the particle stays in the initial state)

P 1st−order0→0 =

∣∣∣c(0)0 + λc

(1)0 + λ2c

(2)0 + ...

∣∣∣ (III.64)

≈ 1 + λ(c(1)0 + c

(1)∗0

)+ λ2

(∣∣∣c(1)0

∣∣∣2 + c(2)0 + c

(2)∗0

)(III.65)

= 1−∑k 6=0

P 1st−order0→k (III.66)

This shows norm conservation even in first order.

4. Example: Slowly varying perturbation

1. Slowly varying perturbation. An example of such a slowly varying perturbation is shown in figure 9. for k 6= 0

t0t

W(t)

FIG. 9. Slowly varying perturbation

25

we have

ck(t) = − i~

∫ t

t0

eiωk0t′Wk0(t′)dt′ (III.67)

= − i~

1

iωk0eiωk0t

′Wk0(t′)

∣∣∣∣tt0

− 1

iωk0

∫ t

t0

eiωk0t′

≈0︷ ︸︸ ︷Wk0(t′) dt′

(III.68)

≈ − 1

~ωk0Wk0(t)eiωk0t (III.69)

= −〈φk|W (t) |φ0〉εk − ε0

eiωk0t (III.70)

p0→k = |ck|2 (III.71)

=|〈φk|W (t) |φ0〉|2

(εk − ε0)2 (III.72)

= 0(if for t > tf , W (tf ) = const) (III.73)

This only works for a non-degenerate initial state (due to the energies in the denominator).

5. Solution of TDSE up to 1st order

Consider the solution to the TDSE and insert in the 1st order coefficient

|ψ(t)〉 =∑k

ck(t)e−i~ εkt |φk〉 (III.74)

= c0(t)e−i~ ε0t |φ0〉+

∑k 6=0

〈φk|W (t) |φ0〉ε0 − εk

e−i~ εkt |φk〉 (III.75)

Since we are using perturbation theory we require that c0 ≈ 1 in this case we have

|ψ(t)〉 ≈

|φ0(t)〉︷ ︸︸ ︷

|φ0〉+∑k 6=0

〈φk|W |φ0〉ε0εk

|φk〉

e−i~ ε0t (III.76)

∣∣∣φ0(t)⟩

describes state of the system at time t up to 1st order corresponding to perturbed energy eigenvalue

ε(1)0 (t) = ε0 + 〈φ0|W (t) |φ0〉 (III.77)

The system remains in the ground state of the total (instantaneous) Hamiltonian, H(t), at all times. This is oftencalled an adiabatic situation. The approximation of neglecting the time derivative of W is called the adiabaticapproximation.

6. Comments

1. This argument can be generalized to strong perturbations (all orders). If perturbation varies slowly with timethe system is found in an eigenstate of the total Hamiltonian H(t) = H0 + W (t) at all times (“adiabaticapproximation”). Ref: D.Bohm, Quantum Theory, Chapter 20

2. Realizations of this formulism can found in

• Slow atomic collisions

26

• Stern-Gerlach experiment

Lecture 13, Feb 13th, 2012

7. Example: Sudden Perturbation

Consider the following perturbation:

W (t) =

0 t ≤ t0W t > t0

(III.78)

The first order amplitude is

ck(t) = − i~

∫ t

0

ei(ω−ωk0)tWk0(t′)dt′ (III.79)

where ωk0 = εk−ε0~ and Wk0 = 〈k|W (t) |0〉. Thus

ck(t) = − i~Wk0

∫eiωk0dt′ (III.80)

= −Wk0

~ωk0

(eiωk0t − 1

)(III.81)

P0→k(t) = |ck(t)|2 =|Wk0|2

~2ωk0 2− 2 cosωk0t (III.82)

=4 |Wk0|2

~2f (t, ωk0) (III.83)

where f (t, ωk0) =

(sin2(ωk0t2 )

ω2k0

). The function f is independent of the particular perturbation. The perturbation is

shown in figure 10. Note that noticeable transitions only occur within ∆Ω = 2πt . This corresponds to an energy range

DΩ= 2 Π

t

-15-10 -5 5 10 15Ωk0

0.05

0.10

0.15

0.20

0.25f

FIG. 10. The function f which determines the probability of transition

∆E = 2π~t .

8. Example:Periodic perturbation

Consider the potential below:

W (t) =

0 t ≤ t0 = 0

Beiωt +B†e−iωt t > t0 = 0(III.84)

27

Note that W = W †. We consider the first order amplitude for a transition from state |i〉 to |f〉. The amplitude is

cfi(t) = − i~

∫ t

0

Wfi(t′)eiωfit

′dt′ (III.85)

where ωfi =εf−εi

~ .

cfi(t) = − i~

〈f |B |i〉

∫ t

0

ei(ωfi+ωt′dt′) + 〈f |B† |i〉

∫ t

0

ei(ωfi−ω)t′dt′

(III.86)

we now define Bfi ≡ 〈f |B |i〉 and note that 〈f |B† |i〉 = 〈i|B |f〉 = B∗if . We can now write

cfi(t) = −

Bfi~ (ωfi + ω)

(ei(ωfi+ω)t − 1

)+

B∗if~ (ωfi − ω)

(ei(ωfi−ω)t − 1

)(III.87)

Pi→f (t) = |cfi(t)|2 (III.88)

=|Bfi|2

~2 (ωfi + ω)2

∣∣∣ei(ωfi+ω)t − 1∣∣∣2 +

|Bif |2

~2 (ωfi − ω)

∣∣∣ei(ωfi−ω)t − 1∣∣∣2 + cross terms (III.89)

The function is plotted in figure 11.

fi = - fi = +

FIG. 11. The probability of a transition with oscillatory perturbation

We summarize the observations below

• Consider the case of ωfi = −ω ⇐⇒ εf = εi−~ω. Hence we can only have a noticeable transition if the appliedfrequency corresponds to the difference between energy of the two energy levels is ~ω. This is called stimulatedemission since this required a perturbation to emit energy. This is also called resonance de-excitation.

• For the second peak we have ωfi = ω ⇐⇒ εf = εi − ~ω. Hence we can only have a noticeable transition if wehave absorption of ~ω. This is also called resonant excitation.

Now we consider the connection to atom-radiation interaction. We found earlier that

W (t) =e

mA(r, t) · p (III.90)

with

A(r, t) = Π |A0| cos (k · r− ωt+ α) (III.91)

=Π

2

A0e

i(k·r−ω)t +A∗0e−i(k·r−ωt)

(III.92)

where A0 ≡ |A0| eiα. Hence we have

W (t) =e

2m

A0e

i(k·r−ωt)Π · p +A∗0e−i(k·r−ωt)

(III.93)

Recall that we had W (t) = B†e−ωt +Beiωt thus we an make the identification

B =e

2mA∗0e

−ik·rΠ · p

28

We need to check the criterion to avoid the overlap of resonances (neglect cross terms) this requires

∆ω =2π

t ω ⇐⇒ t 2π

ω(III.94)

Now we check the validity of first order time dependent perturbation theory.

Pi→f =4 |Bfi|2

~2

t2/4︷ ︸︸ ︷f (t, ωji ± ω = 0) =

|Bfi|2

~2t2 (III.95)

To be valid this must be much less then 1. I.e.|Bfi|2~2 t2 1. We combine these results to say that

2π

ω=

2π

|ωfi| ~|Bfi|

(III.96)

This is true only if

|εf − εi| |Bfi| (III.97)

Hence the matrix element which causes the transition must be small when compared to the energy difference betweenthe states.

Lecture 14th, February 15th, 2012

C. Photoionization

1. Transitions into the continuum: Fermi’s golden rule (FGR)

Suppose that we don’t transition to a single state but into a band of states as shown in figure ?? We need to replace

i

f-f+

FIG. 12. Jumping into an energy band

pi→f = |〈φf |ψ(t(f))〉|2 (III.98)

by

Pi→f =

∫ εf+∆

εf−∆ε

pi→f (εf ′) ρ (εf ′) dεf (III.99)

with ρ(εf ′) is the density of states. Free particle continuum states are

φf (r) = 〈r|p〉 =1

(2π~)3/2

ei~p·r (III.100)

29

〈p|p′〉 =

∫r

〈p|r〉 〈r|p′〉 d3r (III.101)

=1

(2π~)3

∫ei~ (p′−p)d3r (III.102)

= δ (p− p′) (III.103)

Consider

1 = 〈ψ|ψ〉 (III.104)

=

∫〈ψ|p〉 〈p|ψ〉 d3p (III.105)

=

∫|φ(p)|2 d3p (III.106)

We can transfer this integral to an energy integral using εf = p2

2m . We can use

d3p = p2dpdΩp = p2 dp

dεfdεfdΩp (III.107)

We use

p = (2mεf )1/2 ⇒ dp

dεf=

(m

2εf

)1/2

(III.108)

This gives

d3p = 2mεf

√m

2εfdεfdΩp (III.109)

=√

2m3εfdεfdΩp (III.110)

We now go back to our normalization equation

1 =

∫|ψ(p)|2

√2m3εfdεfdΩp (III.111)

=

∫dεf

√2m3εf

ρ(εf )

pi→f (εf )︷ ︸︸ ︷∫dΩp |ψ(p)|2 (III.112)

Note that this was only for free particles. Hence for a free particle he density of states is√

2m3εf . If we want todiscuss photoionization we need to discuss

Pi→f =

∫ εf+∆ε

εf−∆ε

pi→f (ε′f )ρ(ε′f )dεf (III.113)

Recall our previous result for first order pi→f

Pi→f =4

~2

∫ εf+∆ε

εf−∆ε

|Bfi|2f(t, ω′fi + ω

)+ f(t, ω′fi − ω)

ρ(ε′f )dε′f (III.114)

If we assume that the interval range is small then ρ(ε′f ) and Bfi don’t vary much accross that interval. In this case.

Furthermore the meximan hat (f) functions are only large at particular ω. If we are looking at absorption then only

f(t, ω′fi − ω

)is large across this band. In this case

P absi→f ≈4

~2|Bfi|2 ρ(εf )

∫ εf+∆ε

εf−∆ε

f(t, ω′fi − ω)dε′f (III.115)

30

To carry out the integral we define ω ≡ ω′fi − ω and hence dεf = ~dω

P absi→f ≈4

~|Bfi|2 ρ(εf )

∫sin2 ωt

2

ω2dω (III.116)

Technically the integral is from εf−∆ε to εf+∆ε. However if we are centered on an absorption peak then contributionsfrom other parts of the function are small. Thus we may as well just extend the limits of the integral to ±∞, whichis a well known, analytic integal. With this we have

P absi→f =2πt

~|Bfi|2 ρ (εf ) t (III.117)

with εf = εi + ~ω One can easily show that we get a similar equation for stimulated emission:

PSEi→f =2πt

~|Bfi|2 ρ (εf ) t (III.118)

where εf = εi − ~ω. We define the transition rate by

Wi→f =d

dtPi→f (III.119)

This is given by

Wi→f =2π

~|Bfi|2 ρ(εf ) (III.120)

where εf = εi ± ~ω. This is called Fermi’s Golden Rule (FGR).

2. Dipole Approximation

Typical situation is that the wavelength of the light used is large compared to the characteristic distance of atoms.i.e. λ = 2π

k a0. In this case we use the dipole approximation that says

eik·r = 1 + k · r + ... (III.121)

≈ 1 (III.122)

Hence the field doesn’t change spatially across the atom. On Monday we said that

Bfi =e

2mA∗0Π 〈φf | e−ik·rp |φi〉 (III.123)

Applying the dipole approximation we have

Bfi ≈e

2mA∗0Π · 〈φf |p |φi〉 (III.124)

We now use a commutator relation:

p =im

~[H0, r] (III.125)

where Ho = p2

2m + V . With this relation

〈φf |p |φi〉 ≈im

~〈φf |Hor− rH0 |φi〉 (III.126)

If we have φf and φi as eigenstates then we have

〈φf |p |φi〉 =im

~(εf − εi)

Dipole Matrix Elements︷ ︸︸ ︷〈φf | r |φi〉 (III.127)

31

Thus we have (inserting back into previous equation)

Bdipfi =ie

2~A∗0 (εf − εi) Π · 〈φf | r |φi〉 (III.128)

For Π = z we have the standard selection rules, ∆m = 0,∆` = ±1. We can now figure out the transition rates usingFGR. Using this one will find that

W dip,zi→f ∝ cos2 θ (III.129)

for φi = s-states. This is plotted in figure 13

z

FIG. 13. The transition probability as a function of θ for the 1s state

Lecture 15th - February 27th, 2012

D. Outlook on Field Quantization

We need to find a Hamiltonian for atom and electromagnetic field

H = HA +HF +W (III.130)

where HA is the Hamiltonian due to the atom, HF is the Hamiltonian due to the field, and W represents the interactionof the atom with the field. We can write

H = Ho +W (III.131)

where Ho = HA +HF . The steps toward a 1st-order PT treatment are

1. Determine H0 = HA + HF . We already know HA = p2

2m + e2

r . However we don’t know HF . One requirement

we will use is that HF will be Hermitian, i.e. HF = H†F .

2. Solve the eigenvalue problem of H0. We already know the original eigenstates and energies:

HA |φj〉 = εj |φj〉 (III.132)

but we don’t know the states and energies below

HF |ρk〉 = εk |ρk〉 (III.133)

32

If we denote |ψ`〉 as the full unperturbed states then we have

H0 |ψ`〉 = (HA +HF ) |φ`〉 |ρ`〉 (III.134)

= (HA |φ`〉) |ρ`〉+ |φ`〉HF |ρ`〉 (III.135)

= ε` |ψ`〉+ ε` |ψ`〉 = (ε` + ε`) |ψ〉 (III.136)

3. Determine W . We use the ansatz

W =e

mA · p (III.137)

4. Obtain 1st - order transition rates (apply Fermi’s Golden Rule)

• Need

〈ψf |W |ψi〉 (III.138)

1. Construction of HF

The energy of a classical electromagnetic field in a vacuum of volume L3 is (denote this energy WEM )

WEM =εo2

∫ (E2 + c2B2

)d3r (III.139)

Recall for free electromagnetic waves (electric potential is zero) we have with the Coloumb gauge that

∇2A− 1

c2∂2A

∂t2= 0 (III.140)

Assume periodic boundary conditions for each side of cube

A (x, y, z = 0) = A (x, y, z = L)

A (x, y = 0, z) = A (x, y = L, z)

A (x = 0, y, z) = A (x = L, y, z)

This gives (kj = kx, ky, kz for j = 1, 2, 3)

1 = eikjL (III.141)

and hence

kz =2π

Lnz (III.142)

The total vector potential is given by

A(r, t) =∑λ

Aλ(r, t) (III.143)

where

Aλ(r, t) =Π

L3/2

qλe

i(kλ·r−ωλt) + q∗λe−i(kλ·r−ωλt)

λ is the mode index. Each mode is characterized by

kλ, Πλ, ωλ

. Πλ is the unit polarization vector, ωλ = ckλ,

and kλ = 2πL

(nλx, n

λy , n

λz

)where nλx, n

λy , n

λz ∈ Z. We can now calculate E = −∂A∂t and B = ∇ ×A. By taking these

derivatives and inserting into the equation for WEM above we get (after a lot of manipulations)

WEM = 2ε0∑λ

w2λq∗λqλ (III.144)

33

We now perform a substitution amplitudes given by

Qλ ≡√ε0 (qλ + q∗λ)

Pλ ≡ i√ε0ωL (q∗λ − qλ)

This gives

qλ =1

2√ε0

(Qλ + i

Pλωλ

)(III.145)

This gives

WEM = 2ε0∑λ

x2λ

(1

4ε0

)(Q2λ +

P 2λ

ω2λ

)(III.146)

=1

2

∑λ

(P 2λ + ωλQ

2λ

)(III.147)

We can now quantize this energy by promoting Pλ and Qλ to operators. We also demand them to be Hermitian and

follow the commutation relations for position and momentum. i.e.Pλ = P †λ and Qλ = Q†λ and

[Qλ, Pλ′ ] = i~δλ,λ′ (III.148)

Since we have a quantized WEM we have HF :

HF =1

2

∑λ

(P 2λ + ω2

λQ2λ

)= H†F (III.149)

We know the eigenvalues of this Hamiltonian if it has the commutation relations of position and momentum. Theenergies are

En1,n2,... =∑λ

~ωλ(nλ +

1

2

)(III.150)

where nλ = 0, 1, 2, .... Lecture 16th - February 29th, 2012Consider the following summary and discussion of our results.

1. For our expressions we have assumed that we have periodic boundary conditions. This forced us to have discretewavelengths inside our system. This enables us to change our integrals to sums∫

d3k →∑λ

2. Note that

WEM =1

2

∑λ

(P 2λ + ω2

λQ2λ

)(III.151)

is independent on time (since the numbers Pλ and Qλ don’t change with time). i.e.

dWEM

dt(III.152)

This means that the surface integral of the Poynting vector must be zero (from classical electrodynamics).

3. We quantized Pλ and Qλ by elevating them to Hermitian operators that obey the canonical commutationrelations

[Qλ, Pλ′ ] = i~δλ,λ′ (III.153)

34

4. Energy spectrum of our Hamiltonian is given by

E =∑λ

~ωλ(nλ +

1

2

)(III.154)

with nλ = 0, 1, 2, ....

5. The “conventional” but wrong interpretation is to associate each mode with a particle in a parabolic potentialand then the eigenenergies in this mode Enλ = ~ω

(nλ

+ 12

)are the ground and excited state energy levels

6. The alternate interpretation is to associate each mode with nλ particles (or quanta) in the same state. All thesequanta carry the same energy, ~ωλ (forgetting about the 1

2~ωλ term). Note that this only works because theenergy is linear in nλ otherwise. For example if you had 3 quanta (i.e. nλ = 3) we would have and energy of~ω + ~ω + ~ω = 3~ω. This is the photon interpretation.

7. However we have not yet described the 12~ωλ factor. This is the zero-point energy. The full zero-point energy is

E0 =∑λ

~ωλ2→∞ (III.155)

since this is an infinite sum. To make this finite we require a technique called renormalization. The zero-pointenergy is the energy without any photons there. This relates to the idea that we have energy in a vacuum (whichcauses effects such as spontaneous emission).

2. Creation and Annihilation Operators

We introduce creation and annihilation operators given by

bλ =1√

2~ωλ(ωλQλ + iPλ)

b†λ =1√

2~ωλ(ωλQλ − iPλ)

These operators obey [bλ, b

†λ′

]= δλ,λ′ (III.156)

and

[bλ, bλ′ ] =[b†λ, b

†λ′

]= 0 (III.157)

This gives

HF =∑λ

~ωλ(b†λbλ +

1

2

)(III.158)

=∑λ

~ωλ(nλ +

1

2

)(III.159)

where nλ ≡ b†λbλ is called the occupation number operator. The eigenvalue equation with HF =∑λH

(λ)F is

H(λ)F = |ψnλ〉 = ~ω

(nλ +

1

2

)|ψnλ〉 (III.160)

where nλ = 0, 1, 2, ... and we have ⟨ψnλ |ψn′λ

⟩= δnλn′λ (III.161)

35

The occupation number obeys

nλ |ψnλ〉 = nλ |ψnλ〉 (III.162)

We use shorthand notation of |ψnλ〉 = |nλ〉. One must be careful with this notation. Since we may write the followingequation

nλ |nλ + 1〉 = (nλ + 1) |nλ + 1〉 (III.163)

Aˆwas used to differentiate the operator nλ from the number. We now consider the state

|nλ = 0〉 ≡ |0〉 (III.164)

which we call the vacuum state. This gives

nλ |0〉 = |∅〉 (III.165)

This is not the same as |0〉! This is really the zero state. However since we have a ket on the left side we don’t wantto write just 0. One can show that

b†λ |nλ〉 =√nλ + 1 |nλ + 1〉

bλ |nλ〉 =√nλ |nλ − 1〉

In particular we have

bλ |0〉 = |∅〉 (III.166)

Lecture 17th - March 5th, 2012Here we generalize these results. We use the rule that if the eigenvalue of your system is the sum of a set of eigenvaluesthen the resultant eigenvalues are the product of the eigenstates:

HF |n1, n2, ...〉 = E |n1, n2, ...〉 (III.167)

where E =∑λ ~ωλ

(nλ + 1

2

)=∑λ ~ωλnλ+E0 and |n1, n2, ...〉 = |n1〉⊗|n2〉⊗ ... are the product states. The operator

nλ counts the number of modes in mode λ:

nλ |n1, n2, ..., nλ, ...〉 = nλ |n1, n2, ..., nλ, ...〉

b†λ |n1, n2, ..., nλ, ...〉 =√nλ + 1 |n1, n2, ..., nλ + 1, ...〉

bλ |n1, n2, ..., nλ, ...〉 =√nλ |n1, n2, ..., nλ, ...〉

3. Interaction Between Photon Field and Electrons

In the beginning we discussed that we are using the semiclassical interaction with a perturbing Hamiltonian:

W =e

mA · p (III.168)

with

A (r, t) =∑λ

Πλ

L3

qλe

i(kλ·r−ωλt) + q∗λe−i(kλ·r−ωλt)

(III.169)

To quantize this operator we made the transformation given earlier:

qλ =1

2√ε0

(Qλ + i

Pλωλ

)=

√~

2ε0ωλbλ (III.170)

36

Hence we have the vector potential in its new form given by

A (r, t) =∑λ

Πλ

γ︷ ︸︸ ︷1

L3

√~

2ε0ωλ

bλe

i(kλ·r−ωλt) + b†λe−i(kλ·r−ωλt)

(III.171)

This operator is time dependent. However we want an operator thats time independent. To achieve this we define

bHλ = bλe−iωλt (III.172)

For this to make sense we need to show that

i~d

dtbHλ =

[bHλ , H

Hλ

](III.173)

where we denote operators in the Heisenberg picture by superscript H.Proof of statement above is shown below. The Hamiltonian in the Schrodinger interpretation is the same as in the

Heisenberg interpretation. Thus we have

RHS =[bHλ , H

(λ),HF

]= e−iωλt

[bλ, ~ω

(nλ +

1

2

)](III.174)

= ~ωλe−iωλt [bλ, nλ] (III.175)

= ~ωλe−iωλtbλ (III.176)

LHS = i~ω(−iωλ)bλe−iωλt = ~ωλe−iωλtbλ (III.177)

= RHS (III.178)

Hence our construction gave us an operator in the Heisenberg picture. Thus removing the time dependence is easyand gives (where γ was defined above)

A(r) = γ∑λ

Πbλe

ikλr + b†λe−ikλ·r

(III.179)

This is our quantized vector potential in the Schrodinger picture where the interaction is

W =e

mA · p (III.180)

= −i

β︷ ︸︸ ︷√e2~3

2ωλε0m2L3Πeikλr∇bλ + e−ikλ·r∇b†λ

(III.181)

4. The Transition Matrix Elements

〈φf |W |i〉 =∑λ

〈φf |Wλ |φi〉 (III.182)

=∑λ

〈ψf | ⊗⟨nf1n

f2 ...∣∣∣ (Wλ)

∣∣ni1ni2...⟩⊗ |ψi〉 (III.183)

where |ψ〉 are the electron states while |n1n2, ...〉 are the photon states. We have two parts to the equation the productstates are made up of a photon part and an electron part.

〈φf |W |φi〉 = −iβ∑λ

〈ψf | eikλ·rΠλ · ∇ |φi〉 ⊗

⟨nf1n

f2 ...∣∣∣ bλ ∣∣ni1ni2...⟩+ 〈φf | e−ikλ·rΠλ∇ |φi〉 ⊗

⟨nf1n

f2 ...∣∣∣ b†λ ∣∣ni1ni2...⟩

(III.184)

37

Note that we already know the electron matrix elements since we dealt with them earlier (we will get back to themin more detail shortly). We now consider the photon matrix elements.⟨

nf1nf2 ...n

fλ...∣∣∣ bλ ∣∣ni1ni2...niλ...⟩ = (δnf1 ,ni1

δnf2 ,ni2...δnfλ,niλ−1...)

√niλ (III.185)

This matrix element is highly selective. Furthermore we have⟨nf1n

f2 ...n

fλ...∣∣∣ b†λ ∣∣ni1ni2...niλ...⟩ = (δnf1 ,ni1

δnf2 ,ni2...δnfλ,niλ+1...)

√niλ + 1 (III.186)

Hence the condition to have a non zero transition elements are that the photon numbers don’t change by more thenone. Furthermore there can only be a single photon interaction at a time. The annihilation of a photon correspondsto absorption of a photon, while the creation of a photon corresponds to an emission of a photon.Summary:The matrix elements

〈φf |W |φi〉 6= 0 (III.187)

are non-zero if and only if

1. ∣∣ni1ni2...⟩ and∣∣∣nf1nf2 ...⟩ (III.188)

differ in exactly one mode (by one photon).

2. If we apply the dipole approximation then we have the dipole selection rules for the electronic part in thenon-zero mode (eikλ·r ≈ 1). In this case we can use the typical selection rules of ∆` = ±1,∆m = 0

3. Recall that we found in the constant perturbation or sudden approximation that the transition is small unlessenergy is conserved. i.e.

Ei = Ef (III.189)

H0 |φi〉 = (HA +HF ) |ψi〉∣∣ni1ni2...⟩ (III.190)

= (HA |φi〉) + |φi〉HF

∣∣ni1ni2...⟩ (III.191)

=

(εi +

∑λ′

~ωλ′(niλ′ +

1

2

))|φi〉 (III.192)

This gives

Ef = εf +∑λ′

~ωλ′(nfλ′ +

1

2

)(III.193)

Energy conservation Ei = Ef implies that

εf = εi +∑λ′

~ωλ′(niλ′ − n

fλ′

)(III.194)

εi ± ~ωλ (III.195)

where the plus sign corresponds to absorption while the minus sign corresponds to emission.

Lecture 18 - March 7th, 2012Note: Look carefully on last question of new assignment, it may be a test question

Recall we can write

〈φf |W |φi〉 = −iβ ×

〈ψf | e−ikλ·rΠλ · ∇ |ψi〉

√niλ + 1

〈ψf | eikλ·rΠλ · ∇ |φi〉√niλ

0

(III.196)

Note we can only have one of above.

38

5. Spontaneous Emission

Thus is a special case of the term

− iβ 〈ψf | e−ikλ·rΠλ · ∇ |ψi〉√niλ + 1 (III.197)

with niλ = nfλ − 1 = 0. Energy conservation needs to be fulfilled. In other words

εf = εi − i~ωλ (III.198)

Fermi’s Golden rule says that

W spon.emissioni→f =

2π

~|〈φf |W |φi〉|2 ρ(Ef ) (III.199)

where ρ is the density of states. We first need to find the density of states.

Initial state =

|ψi〉 : (discrete) Excited atomic state

|0〉 : No photon(III.200)

Final state =

|ψf 〉 : (discrete) atomic ground state

|1〉 : One photon with ω =εi−εf

~(III.201)

We want to find the density of (photon) states. We look at the density of states with respect to k-space. Recall that

kλ =2π

L

(nλx, n

λy , n

λz

)(III.202)

where nλi ∈ Z.

ρ(k) =∆N

∆Vk=

∆nx∆ny∆nz∆kx∆ky∆kz

(III.203)

We use the relation between k and n shown in equation III.202. It’s easy to see that

ρ(k) =

(2π

L

)3

(III.204)

We can rewrite the differential as

d3k = k2dkdΩ = k2 dk

dEdEdΩ (III.205)

For photons we have

E = ~ω = ~ck (III.206)

Hence we have

dE

dk= ~c ⇐⇒ dk

dE=

1

~c(III.207)

With this relation we have

d3k =1

~c

(E2

~2ω2

)dEdΩ (III.208)

39

We can now find the number differential

dN = ρd3k =

(L

2π

)3ω2

~c3dEdΩ (III.209)

ρ(E)dEdΩ (III.210)

Fermi’s Golden Rule says that

dWSpon.Emissioni→f

dΩ=

2π

~

(L

2π

)3ω2

~c3

(e2~3

2ε0ωm2L3

) ∣∣∣〈ψf | e−ik·rΠ · ∇ |φi〉∣∣∣2 (III.211)

=ω

8π2c3

(e2~ε0m2

) ∣∣∣〈ψf | e−ik·rΠ · ∇ |φi〉∣∣∣2 (III.212)

If we apply the dipole approximation then e−k·r ≈ 1 (this says that the size of the atoms is much smaller then thewavelength of the light). This leaves us to consider

〈ψf | Π · ∇ |ψi〉 =i

~〈ψf | Π · p |ψi〉 (III.213)

However one can show that

im

~[HA, r] = p (III.214)

if HA = p2

2m + V (r). Using this relation we have

〈ψf | Π · ∇ |ψi〉 = −m~2

Π · 〈ψf |HAr− rHA |ψi〉 (III.215)

= −m~2

~ω︷ ︸︸ ︷(εf − εi) Π · 〈ψf | r |ψi〉 (III.216)

Putting this result together we have

dWSpon.Emissioni→f

dΩ=

e2ω3

8πε0~c3×∣∣∣Π · 〈ψf | r |ψi〉∣∣∣2 (III.217)

=

α︷ ︸︸ ︷e2ω3

8π2ε0~c3∣∣∣Π · rif ∣∣∣2 (III.218)

We now sum over all polarizations thus we integrate. We choose our wave propagation direction, k, such that rifis along the z axis and define the angle between these two axes as θ. The two linearly independent polarizationdirections of the light must be perpendicular to the motion of the wave (⊥ k). If we choose one polarization to beperpendicular to rif then this contribution is zero. This sets the other polarization direction to be

Π2 · rif = |r| cos(π

2− θ)

= rif sin θ (III.219)

Hence we have

WS.Ei→f =

∫ (|Π1 · rif |2 + |Π2 · rif |2

)dΩ (III.220)

= α

∫r2if sin2 θdΩ (III.221)

= α8π

3r2if (III.222)

Thus the total transition rate in the dipolar approximation is

WS.E.i→f =

e2ω3

3πε0~c3|rif |2 (III.223)

where rif = 〈ψi| r |ψf 〉.Discussion:

40

1. As an example consider a Hydrogen 2p→ 1s transition

H(2p)

H(1s)

The lifetime is given by

T dip2p→1s =1

WS.E2p→1s

≈ 1.6× 10−9s (III.224)

In a classical picture it takes the electron 10−16s to circle around the nucleus. Thus this is a very long lifetimewith respect to this value.

2. We can consider a different decay of Hydrogen 2s→ 1s It turns out that

T 1storder2s→1s =∞ (III.225)

However experimentally we have

T experiment2s→1s = 0.1s (III.226)

This state is metastable. We need second order perturbation theory (more then FGR) to do this. This corre-sponds to a two-photon process.

3. For an N -photon process one needs to consider N th order perturbation theory. The W operator is linear in band b† so in order to contribute a N photon process we need to combine more of these operators.

Lecture 19th - March 12th, 2012Recall that the total transition rate in the dipole approximation is

W s.e.i→f =

e2ω3

3πε0~c3(|xif |2 + |yif |2 + |zif |2

)(III.227)

The lifetime is simplify given by

Ti→f =1

W s.e.i→f

(III.228)

Consider if we have an electron in the 3p state: φi = H(3p). This is shown in figure 14 The total decay rate is the

2s

1s

3p

FIG. 14. A decay from a Hydrogen 3p state

sum of the two rates:

W s.e. =e2

3πε0~c3∑

f(εf<εi)

ω3if |rif |

2(III.229)

41

Here the rates are uncoupled (the rate of 3p→ 2s doesn’t effect the rate of 3p→ 1s). In practice of course this is notthe case.Concluding remarks on photons:

Here we defined photons as the quanta of an electromagnetic field. The properties of the photon are

• Can be created or annihilated (hence they are not stable)

• Carry energy ~ωλ

• One can show that they carry momentum ~kλ. The reasoning is as follows. One could start from a classicalexpression for momentum of the electromagnetic field:

pEM = ε0

∫V=L3

(E×B) d3r (III.230)

For our cube we know the vector potential (Aλ(r, t) =∑λ ...) and hence we can find the electromagnetic fields

E = −∂A∂t

B = ∇×A

we can then insert this result into the momentum and find the momentum of the fields:

pEM = 2ε0∑λ

kλωλq∗λqλ (III.231)

Quantizing this operator gives

pF =∑λ

~kλb†λbλ =∑

~kλnλ (III.232)

Hence we have

pF |n1, n2, ...〉 =∑λ

~kλnλ |n1, n2, ...〉 (III.233)

From this it is clear that momentum of each mode is ~kλ.

• Similarly one can show that photons also carry angular momentum (often called photon spin) given by ±~. Thisis reasoned as follows. The angular momentum of an electromagnetic field is

LEM = ε0

∫L3

r× (E×B) d3r (III.234)

This gives a spin of ±~.

• Since one can show that the angular momentum is an integer the photons are bosons. Note that we didn’t findany restriction for the number of photons that can be in a given mode. This is another way of showing thatphotons are bosons.

• The photons move with the velocity of light v = c since that the velocity of electromagnetic waves in a vacuum.By Einstein’s theory of special relativity we know that the photons have zero mass

• The photon treatment can be found in the following resources:

– Friedrich, Theoretical Atomic Physics

– Sakurai and Napolitano, Modern Quantum Mechanics

– Schiff, Quantum Mechanics

• The semiclassical atom-radiation interaction can be found in

– Liboff (Chap 13.5 - 13.9)

– Cohen-Tannoudiji (Chap 13)

42

As an aside we now go over the solution to the practice problem put online. Consider the harmonic oscillatorHamiltonian:

H0 =p2

2m+m

2ω2

0x2 (III.235)

with the perturbation

W =1

2mω2x2 cosωt (III.236)

c1storderk (t) = δk0 −i

~

∫ t

t0

eiωk0t′〈k|W (t′) |0〉 dt (III.237)

We first need to find the matrix elements.

〈k| W |0〉 ≡ 〈k|x2 |0〉 (III.238)

Here we use the following

x =

√~

2mω0

(a+ a†

)(III.239)

where a and a† are the annihilation and creation operators respectively.

x2 =~

2mω0

(a2 + a† 2 + aa†a†a

)(III.240)

〈k|W |0〉 =~

2mω0

(

:0〈k| a2 |0〉+ 〈k| aa† |0〉+

:0〈k| a†a |0〉+ 〈k| a† 2 |0〉

)(III.241)

but a† |0〉 = |1〉 , a |1〉 = |0〉 , and a† |1〉 =√

2 |2〉. Thus

〈k|W |0〉 =~

2mω0

(〈k|0〉+

√2 〈k|2〉

)(III.242)

=~

2mω0

(δk0 +

√2δk2

)(III.243)

We now summarize our results as

〈k|W (t) |0〉∣∣∣∣k 6=0

=

√2~

2mω0

mω2

4

(eiωt + e−iωt

)(III.244)

Thus

c2(t) = −i√

2

8

ω2

ω0

∫ t

0

ei2ω0t′(eiωt

′+ e−iωt

)dt′ (III.245)

where we have ω20 = 2ω0.

c2(t) = −i√

2

8

ω2

ω0

(1

i (2ω0 + ω)

(ei(2ω0+ω)t − 1

)+

1

i (2ω0 − ω)

(ei(2ω0−ω)t − 1

))(III.246)

For the case of ω = 2ω0 (the resonance condition) we have (must go back to the integral to show this)

c2(t) = −i√

2

2ω0t (III.247)

The probability is given by

P2(t) =ω2

0

2t2 (III.248)

Lecture 20th, March 19th, 2012

43

IV. BRIEF INTRODUCTION TO RELATIVISTIC QUANTUM MECHANICS

A. Klien-Gordon Equation

1. Setting up a relativistic wave equation

Recall the non-relativistic case. We start from a classical expression for the energy of a free particle

E − p2

2m= 0 (IV.1)

We then quantize this expression by promoting the energy and momentum parts to operators:

E → i~∂

∂t(IV.2)

p2 → −i~2∇2 (IV.3)

This gives (i~∂

∂t+ ~2∇2

)ψ = 0 (IV.4)

The relativistic case is as follows. We start with the relativistic energy momentum relation:

E2 = p2 +m2 (IV.5)

We then quantize this equation using the prescription above(−~2 ∂

2

∂t2+ ~2c2∇2 −m2c4

)ψ = 0 (IV.6)

or in natural units (KG equation) (− ∂2

∂t2+∇2 −m2

)ψ = 0 (IV.7)

B. Discussion of KG equation

(i) KG equation is a second order PDE in both space and time

(ii) KG equation is Lorentz covariant (this means that this equation behaves under LT the way that it should)

(iii) Time development is determined from two initial conditions (since it’s second order in time) for the wavefunctionand for the time deriviative. i.e.,

ψ(t0) =∂ψ(t0)

∂t(IV.8)

However this is not compatible with one of the postulates of quantum mechanics which says that to get the fullwavefunction we just need to act on the wavefunction at a certain time with the time evolution operator.

ψ(t0)U−→ ψ(t) (IV.9)

(iv) Another problem arises with the continuity equation. One can derive the probability current in the KG equation:

∂ρ

∂t= −∇ · J = 0 , (IV.10)

where ρ = i~2mc2

(ψ∗ ∂ψ∂t − ψ

∂ψ∗

∂t

)is the probability density and J = i~

2m (ψ∇ψ∗ − ψ∗∇ψ) is the probability

current. However ρ is not positive definite and can be negative!

44

(v) Consider the following trial solution of the KG equation:

ψ(r, t) = A sin (k · r− ωt) (IV.11)

∂2ψ

∂t2= −Aω2 sin (k · r− ωt) = −ω2ψ (IV.12)

∇2ψ = −k2ψ (IV.13)

substitution into the KG equation gives

~2ω2ψ =(~2c2k2 +m2c4

)ψ (IV.14)

This gives

E2 = c2p2 +m2c4 (IV.15)

with E = ~ω and p = ~k. However this solution doesn’t solve the Schrodinger equation! The time dependentSchrodinger equation cannot be solved by a real solution. However this is not true for the KG equation.Subsitution into the current and probability densities given above for this solution gives

J = 0 (IV.16)

ρ = 0 (IV.17)

This solution is clearly problematic. On the other hand we can try another solution for the KG equation:

ψ(r, t) = Aei(k·r−ωt) (IV.18)

with the same energy momentum relation:

E2 = p2c2 +m2c4 (IV.19)

The Klien Gordon equation does not seem to have any restriction on energy thus we can have

E = ±√p2c2 +m2c4 (IV.20)

(vi) Adding the Coloumb potential to the free KG equation and solving gives solutions that don’t agree with exper-iments. This is what bothered Schrodinger. This is because the KG equation does not take spin into account.

(vii) In 1934 the KG equation was recognized as correct wave equation for spin 0 particles.

C. Dirac Equation

1. Free Particles

Dirac wanted an equation first order in space and time. He hoped this would remove some of the problems withthe KG equation. Dirac tried the an satz was

i~∂ψ

∂t= Hψ (IV.21)

He had a set of requirements that his equation was to fulfill.

(i) Compatibility with the relativistic energy relation: E2 = p2c2 +m2m4

(ii) Covariant with respect to Lorentz transformations.

(iii) An equation that is consistent with the continuity equation and probability interpretation

(iv) He didn’t want to invent any new quantization rules

45

We have our energy momentum relation

E2 = p2c2 +m2c4 (IV.22)

If we had the equation (which is not true!)

E = pc+mc2 (IV.23)

then we are first order and space and time. Dirac’s idea was to write

H = α · pc+ βmc2 . (IV.24)

Dirac realized that this an satz can fulfill the energy momentum relation if α and β are matrices. Trying to fulfillthe requirements listed above gives the restrictions on α and β. All we know so far is that we have three α matricesand 1 β matrix. We call these N ×N matrices. Since these are matrices we require wavefunctions with a number ofcomponents equal to the dimensionality of the matrices. These wavefunctions are called ‘spinor wavefunctions’. Wecan denote this spinor wavefunction as follows

Ψ =

ψ1(r, t)ψ2(r, t)

...ψN (r, t)

(IV.25)

where N is the dimensionality of the matrices. The first requirement says that each component ψi(r, t) must fulfillthe KG equation.

Lecture 21 - March 21st, 2012We now go through Dirac’s wishlist in order to find conditions for α and β: First each component of ψi(r, t) must

fulfill the Klien Gordon equation. Consider the Dirac equation:(i~∂

∂t−H

)Ψ = 0 (IV.26)

iterating the Dirac equation gives(i~d

dt−H

)(i~d

dt−H

)Ψ = 0 (IV.27)

−~2 d2

dt2+H2 − 2i~

(d

dtH

)Ψ = 0 (IV.28)

−~2 d2

dt2+H2 − 2H2Ψ = 0 (IV.29)

− ~2 d2

dt2= H2Ψ 0 (IV.30)

−~2 ∂2

∂t2Ψ =

(cα · p + βmc2

) (cα · p + βmc2

)Ψ

=

c~i

∑j

αjd

dxj+ βmc2

(c~i

∑k

αkd

dxk+ βmc2

)Ψ

=

−~2c2∑j,k

αjαk∂2

∂xj∂xk+

~mc3

i

∑j

(αjβ − βαj)d

dxj+ β2m2c4

Ψ

=

−~2c2∑j,k

αjαk + αkαj2

∂2

∂xj∂xk+

~mc3

i

∑j

(αjβ − βαj)d

dxj+ β2m2c4

Ψ (IV.31)

46

Recall the Klien Gordan equation says that

− ~2 ∂2

∂t2ψ =

(−~2c2∇2 +m2c4

)ψ (IV.32)

The left hand sides are trivially equal to one another. Thus we require certain conditions. Clearly there are no firstorder derivatives in the KG equation. Thus we require

αjβ + βαj ≡ αj , β = 0 . (IV.33)

The KG equation only has second derivatives in the form of a Laplacian. Thus all the cross terms such as ∂2

∂x1x2must

be zero. This can be done by introducing a Kronecker delta. However the term must be 1 for each second derivative

of a given variable (e.g. ∂2

∂x21):

αjαk + αkαj = 2δjk (IV.34)

Lastly we require

β2 = 1 (IV.35)

This conditions must be fulfilled for all j, k = 1, 2, 3. Further requirements(one that wasn’t mentioned in Dirac’swishlist but important) was that the Dirac Hamiltonian should be Hermitian:

H = H† (IV.36)

This requires

αj = α† ; β = β† . (IV.37)

Hence they have real eigenvalues. Notice our requirements tell us that

α2j = 1 and β2 = 1

Thus the eigenvalues must be ±1 for all four of these matrices. Consider the equation

αjβ + βαj = 0 .

If we multiply this equation on the right with β we we have

αjβ2 = −βαjβαj = −βαjβ

Tr(αj) = −Tr(βαjβ)

Tr(αj) = −Tr(αj) (IV.38)

where in the last step we used β2 = 1. Hence the trace of αj must be zero. The same can be shown for β bymultiplying by αj instead of β in the beginning. Thus we have another condition on αj and β, αj and β are traceless.Since the matrices are tracelss the sum of the eigenvalues is zero (easy to prove). Since the eigenvalues are ±1 andthe only way to have a sum of 0 we need to have an even dimension! We need 4 anticommuting matrices.

The simplest choice for the dimension would be N = 2. However we already know that the Pauli matrices makeup three independent anticommuting matrices and there cannot be a fourth. Hence we need to go up to a largerdimension. We choose the next lowest dimension, namely N = 4. Matrices are obey all the conditions we have aboveare

αx =

0 0 0 10 0 1 00 1 0 01 0 0 0

; αy =

0 0 0 −i0 0 i 00 −i 0 0i 0 0 0

; αz =

0 0 1 00 0 0 −11 0 0 00 −1 0 0

; β =

1 0 0 00 1 0 00 0 −1 00 0 0 −1

(IV.39)

47

Note that we can rewrite these as

αx =

(0 σxσx 0

); αy =

(0 σyσy 0

); αz =

(σz 00 σz

); β =

(1 00 −1

)(IV.40)

We now take a look at the Dirac equation more explicitly:

i~∂

∂t

ψ1

ψ2

ψ3

ψ4

=c (αxpx + αypy + αzpz) + βmc2

ψ1

ψ2

ψ3

ψ4

(IV.41)

These are four coupled equations because α have all off-diagonal terms. Working out these equations gives (when youmultiply by α and β matrices)

i~∂

∂tψ1 =

~ci

((∂

∂x− i ∂

∂y

)ψ4 +

∂

∂zψ3

)+mc2ψ1 (IV.42)

i~∂

∂tψ2 =

~ci

((∂

∂x+ i

∂

∂y

)ψ3 −

∂

∂zψ4

)+mc2ψ2 (IV.43)

i~∂

∂tψ3 =

~ci

((∂

∂x− i ∂

∂y

)ψ2 +

∂

∂zψ1

)−mc2ψ3 (IV.44)

i~∂

∂tψ4 =

~ci

((∂

∂x+ i

∂

∂y

)ψ1 −

∂

∂zψ2

)−mc2ψ4 (IV.45)

2. Solutions of the free Dirac equation

To solve them we use the ansatz

ψj(r, t) = ujei(k·r−ωt) (IV.46)

with E = ~ω and p = ~k. This gives four linearly independent solutions. The two solutions

u(1) =

10χ1

χ2

; u(2) =

01χ′1χ′2

(IV.47)

correspond to an energy of

E = +√p2c2 +m2c4 (IV.48)

The factors χ1, χ2, χ′1, χ′2 are kinematic factors. In the limit of v c these factors are zero. The two other solutions,

u(3) =

φ1

φ2

10

; u(4) =

φ′1φ′201

(IV.49)

correspond to negative energy, E = −√p2c2 +m2c4 with φ1, φ2, φ

′1, φ′2vc−−−→ 0. These negative solutions are difficult

to interpret. Dirac’s interpretation is shown in figure 15. Dirac says that the negative energy levels are all occupiedby negative energy electrons. Since electrons are fermions they cannot all pile in these negative energy states. Theseparticles make up the vacuum. Thus we use a photon with high enough energy we can excited these particles andcreate a hole in the Dirac sea. Dirac inferred that these holes are observable as antiparticles called the positrons.These are holes in the negative energy spectrum and hence have positive energy.

This interpretation says that the vacuum is a a fully occupied Dirac sea with no electrons with energy E > mc2.

48

0

E

-mc2

2mc

Enegy Gap

Filled States ("Dirac Sea")

Filled States

Hole

(2 x 511 keV)

FIG. 15. The Dirac Interpretation of the solutions to the Dirac equation. He says that all the negative energy levels are filled.However we cannot observe these particles

Lecture 22 - March 26th, 2012As an aside recall that Dirac’s problem with the KG equation was that it didn’t give physical answers for thecontinuiuty equation. For the Dirac equation one can show that

ρ = Ψ†Ψ = |ψ1|2 + |ψ2|2 + |ψ3|2 + |ψ4|2 > 0 (IV.50)

Jk = cΨ†αkΨ (IV.51)

for k = 1, 2, 3. The Dirac equations gives strictly positive probabilities as expected and parallels the probabilities inthe Schrodinger equation.

We now come back to the Dirac interpretation. The process of pair creation is given by

γ → e+ + e− (IV.52)

To conserve energy and momentum this process must occur in the presence of a nucleus. Annihilation can producephotons:

e+ + e− → 2γ (IV.53)

The consequence of the Dirac interpretation is that particle number is not conserved in this theory. Not only photonscan be created and annihilated but also photons can. This suggests that this is a many-particle theory.

Recall that we have 4 energy solutions. This is called a doublet structure. This represents the spin degree offreedom:

u(1), u(3) “spin up”

u(2), u(4) “spin down”

At this point from our analysis its not clear that this spin has anything to do with angular momentum.Now consider the components in the u vectors labeled χ. These are non-vanishing ‘small’ components which

implies that e+ and e− are intrinsically connected. For small velocity (v c) the Schrodinger (Pauli) equation canbe recovered from the Dirac equation.

49

3. Add Electromagnetic Potentials

We now add EM potentials to the Dirac theory. This is often called the “minimal coupling prescription”. Recallthat the get from the free Schrodinger equation to one with potentials we add a potential:

i~∂

∂tψ =

p2

2mψ (IV.54)

In EM field we have

i~∂

∂tψ =

1

2m(p + eA)

2 − eφ (IV.55)

We can extract a recipe from this.

p→ p + eA

i~∂

∂t→ i~

∂

∂t+ eφ

Recall the free Dirac equation is

i~∂ψ

∂t=(α · pc+ βmc2

)ψ (IV.56)

Using our recipe we have

i~∂ψ

∂t=(α · (p + eA) + βmc2 − eφ

)ψ (IV.57)

One can show that this equation is also Lorentz covariant which says that it has the same form in all inertial referenceframes.

4. The relativistic hydrogen problem

This corresponds to a given choice for the potential in the Dirac equation

A = 0; φ =Ze

4πε0r. (IV.58)

However this is not Lorentz covariant since we chose a nonconvariant form for the potential (if you have a boost thenthe vector potential would be nonzero). This introduces a small error. To solve the Dirac equation we use the usualan satz

Ψ(r, t) = Φ(r)e−iEt/~ (IV.59)

this gives cα · p + βmc2 − Ze2

4πε0r

Φ(r) = EΦ(r) (IV.60)

Lecture 23 - March 28th, 2012The bound-state eigenenergies for the bound states can be obtained and written as

En,j = mc2

[−

√1 +

(Zα)2

(n− δj)2

](IV.61)

where δj ≡ j + 12 −

√(j + 1

2

)2 − (Zα)2

and α = ~ma0c

≈ 1137 .

n = 1, 2, 3, ...

is still the principle quantum number and

j =1

2,

3

2,

5

2, ..., n− 1

2

is the total angular momentum quantum number.Discussion

50

(i) δj ≥ 0 for Zα ≤ 1

(ii) En,j 6= En,j′ this accounts for fine-structure.

(iii) One can expand En,j in powers of the small parameters (Zα)2. This gives

En,j = mc2

(1− (Zα)

2

2n2− (Zα)

4

2n3

(1

j + 12

− 3

4n

))(IV.62)

The first term (0th order) is the rest energy mc2. The second term (1st order) we have the non relativisticbinding energy since

mc2α2 =~

ma20

(IV.63)

and

E(2)n = − ~

2ma20

Z2

n2(IV.64)

The third term (2nd order) gives a relavistic correction which is j dependent (fine structure). The fine structureis shown in figure ??:

Continuum

E=mc2

n=1 1s1.8 x 10^-4 eV1s1/2

n=2

DiracSchrodinger

2s,2p

2s1/2

2p3/2, 2p1/2

10^-4 eV

n=33s,3p,3d

3s1/2

3p3/2, 2p1/2 , 3d1/2

, 3d 3/2

3d5/2

Notice that in Schrodinger’s theory our reference frame was zero. On the other hand with Dirac we have relativityand we use E = mc2 as our new reference frame.Further corrections of bound state hydrogen problem:

(i) Hyperfine structure coupling of magnetic moments of the electron to the nuclear magnetic moment. e.g.

Continuum

E=mc2

n=1 1s1/2

'trilet'

'singlet'

6 x 10 eV-6

(ii) QED effects (the Lamb shift) e.g.

51

Continuum

E=mc2

n=2 2s1/24.4 x 10 eV-6

,2p1/2 1/22p2s 1/2

5. Nonrelativistic limit of the Dirac equations

The starting point is the stationary Dirac equation:(cα · p + βmc2 + V (r)

)Φ = EΦ(r) (IV.65)

we write the 4-spinor as

Φ =

(φχ

)=

φ1

φ2

χ1

χ2

(IV.66)

We can write αj =

(0 σjσj 0

)with σj as the Pauli matrices. Further we have β =

(1 00 −1

). Inserting into the

Dirac equation gives

c

(0 σσ 0

)· p(φχ

)=

(E − V (r)−mc2

(1 00 −1

))Φ(r) (IV.67)

This gives two coupled matrices for φ and χ:

cσ · pχ =(E − V (r)−mc2

)φ

cσ · pφ =(E − V (r) +mc2

)χ

Isolating the second equation for χ gives

χ =c

E − V (r) +mc2σ · pφ (IV.68)

Inserting this equation into the top equation gives

σ · p

c2

E − V (r) +mc2σ · p

φ =

(E − V (r)−mc2

)φ (IV.69)

Lecture 24th, 2012Isolating for φ we have

φ =1

(E − V (r)−mc2)σ · p

c2

E − V (r) +mc2σ · p

φ (IV.70)

We now make a weak relativistic approximation. We define

ε = E −mc2 mc2

Further we assume

V (r) mc2

52

We now expand consider the term in the left hand side of equation IV.69:c2

ε+ 2mc2 − V (r)

=

1

2m(1 + ε−V

2mc2

) ≈ 1

2m

(1− ε− V

2mc2

)(IV.71)

This gives

1

2m

[σ · p

(1− ε− V

2mc2

)σ · p

]φ = (ε− V )φ (IV.72)

1

2m

[(1− ε− V

2mc2

)(σ · p)

2+

~iσ · ∇V

2mc2σ · p

]φ = (ε− V )φ (IV.73)

Now for Pauli matrices we can write

(σ ·A) (σ ·B) = ∇ (A ·B) + iσ · (A×B) (IV.74)

In our case we have A×B = p×p = 0. Further we have that the gradient of a function is (given no radial dependence

∇V =1

r

dV

drdr (IV.75)

Inserting in these relations we have

1

2m

[(1− ε− V

2mc2

)p2 +

~i

1

r

dV

dr

(σ · rσ · p2mc2

)]φ = (ε− V )φ (IV.76)

p

2m

[(1− ε− V

2mc2

)+

~i

1

4m2c2r

dV

dr(r · p) +

~4m2c2r

dV

drσ · L

]φ = (ε− V )φ (IV.77)

where we have used equation IV.74 We define these terms T1, T2 and T3 respectively.Interpretation of TermsAs preparation consider a term acting on a non-relativistic wavefunction

(ε−mc2 − V (r)

)ψ =

p2

2mψ

This is not the term we have. In our case we have

T1φ =

(1− ε− V

2mc2

)p2

2m≈ p2

2m− 1

2mc2

(p2

2m

)2

(IV.78)

=p2

2m− p4

8m3c2(IV.79)

Here we see the relativistic correction of kinetic energy. The second term T2 is not Hermitian. However it can easilybe made Hermitian by considering Hermitian average

T2 ≡1

2

(T2 + T †2

)(IV.80)

=1

2

~i

1

r

dV

drr · p− ~

i

1

r

dV

drr︸ ︷︷ ︸

F(r)

·p

† (IV.81)

Now

(F · p)†

= p†F†(r) (IV.82)

= p† · F† (IV.83)

= p · F (IV.84)

53

With this we have

T2 =1

8m2c2

(~i

dV

drr · p− ~

ip · r1

r

dV

dr

)(IV.85)

By inserting p = ~i∇ and doing some algebra (exercise) we get

T2 =~2

8m2c2∇2V ≡ HD (IV.86)

This term is called the Darwin hamiltonian. Note that

∇2

(1

r

)= −4πδ(r) (IV.87)

Thus we finally have

HD =Ze2~2

8m2c2ε0δ(r) (IV.88)

This represents “Zitterbewgung” or trembling motion. Lastly consider the third term

T3 =~

4m2c21

r

dV

drσ · L (IV.89)

=1

2m2c21

r

dV

drS · L (IV.90)

with S = ~2σ. This is a spin orbit coupling term. This is the reason we call spin-orbit coupling a relativistic effect.

The weak relativistic limit of the Dirac equation takes the form

H |φ〉 = ε |φ〉 (IV.91)

with

H = fracp22m− Ze2

4πε0r+

(− p4

8m3c2

)+HD + T3 (IV.92)

= H0 +W (IV.93)

You can account for W in first order perturbation theory. We obtain

∆E(1) = 〈W 〉 =mc2

2

(Zα)4

n3

(1

j + 12

− 3

4n

)(IV.94)