Physics 3 for Electrical EngineeringPhysics 3 for Electrical Engineering

Ben Gurion University of the Negevwww.bgu.ac.il/atomchip, www.bgu.ac.il/nanocenter

Lecturers: Daniel Rohrlich, Ron Folman Teaching Assistants: Daniel Ariad, Barukh Dolgin

Week 10. Quantum mechanics – Schrödinger’s equation for the hydrogen atom • eigenvalues and eigenstates • atomic quantum numbers • Stern-Gerlach and spin • Pauli matrices • spin-orbit couplingSources: Feynman Lectures III, Chap. 19 Sects. 1-5; Merzbacher (2nd edition) Chap. 9;Merzbacher (3rd edition) Chap. 12;

1-2, פרקים 8פרקים בפיסיקה מודרנית, יחידה Tipler and Llewellyn, Chap. 7 Sects. 2-5.

Schrödinger’s equation for the hydrogen atom

The most common isotope of hydrogen contains just one proton and one electron. Their potential energy is

so Schrödinger’s equation depends on rp and re , i.e. on six coordinates:

But a clever change of variables makes the equation simpler:

, 4

)(0

2

epep

eV

rrrr

. ),,( 422 0

22

22

2

te

mmti ep

ep

ee

pp

rrrr

Schrödinger’s equation for the hydrogen atom

Let r = re – rp and then Schrödinger’s

equation becomes

where is called the reduced mass. For a proton

and an electron, the reduced mass is essentially the electron mass, since mp ≈ 2000 me. But for a positronium atom (which has a positron in the place of the proton), μ = me /2.

;pe

ppee

mm

mm

rrR

pe

pe

mm

mm

, 42)(2 0

22

22

2

r

e

mmti

pe rR

Schrödinger’s equation for the hydrogen atom

Let r = re – rp and then Schrödinger’s

equation becomes

where is called the reduced mass. There are

solutions of the form Ψ = T(t)χ(R)ψ(r), where T(t) is the usual time dependence and χ(R) is a solution to the free Schrödinger equation in R (the center-of-mass coordinate), which has no potential term.

;pe

ppee

mm

mm

rrR

, 42)(2 0

22

22

2

r

e

mmti

pe rR

pe

pe

mm

mm

Schrödinger’s equation for the hydrogen atom The equation for ψ(r) is then

with the Coulomb energy as a central potential V(r).

We have seen that, for a central potential, Schrödinger’s equation reduces to

, ψ 42

)ψ(0

22

2

r

eE

rr

. ψ)(ψ ˆ

2),,(ψ

2

22

2

2rV

rr

rrmrE

L

We replace m by μ and V(r) by

and obtain the eigenvalues of ; they are . We solved this equation by expressing ψ(r,θ,φ) as a product of two functions:

ψ(r,θ,φ) = R(r)Ylm(θ,φ) ,

where

, ψ4

ψ ˆ

2),,(ψ

0

2

2

22

2

2

r

e

rr

rrrE

L

2)1( ll2L

. ˆ

,)1(ˆ 22

ml

mlz

ml

ml

YmYL

YllY

L

:4 0

2

r

e

Eigenvalues and eigenstates

Hence the equation for R(r) is

The solutions Rnl(r) of this equation have the form

with Fnl (r/a0) a polynomial and n ≥ l + 1. The constant a0 is called the Bohr radius and equals

. )(4

)( )1(

2)(

0

22

2

2rR

r

erRll

rr

rrrER

, )/()( 0/ 0 arFrR nlnar

nl e

. m10 29.5 529.0 4 11

2

20

0

ea

Å

Eigenvalues and eigenstates

The energies depend only on n:

where 1 eV = 1.602176 × 10-19 J.

Question: What is the normalization condition for ψnlm(r,θ,φ)?

,

eV 6.13

24 22220

4

nn

enE

Eigenvalues and eigenstates

The energies depend only on n:

where 1 eV = 1.602176 × 10-19 J.

Question: What is the normalization condition for ψnlm(r,θ,φ)?

Answer:

,

eV 6.13

24 22220

4

nn

enE

. sin ),,(ψ1 22 drddrrnlm

Eigenvalues and eigenstates

Question: What is the normalization condition for Rn0(r)?

Eigenvalues and eigenstates

Question: What is the normalization condition for Rn0(r)?

Answer:

. )(

)4( 4

)(

sin ),,(ψ1

220

22

0

2200

drrrR

drrrR

drddrr

n

n

n

Eigenvalues and eigenstates

Here are the lowest normalized solutions ψnlm = Rnl(r)Ylm(θ,φ)

of the hydrogen atom Schrödinger equation (see also here): n l m ψnlm

Radial eigenfunctions Rnl(r) and probability distributions Pnl(r) for the lowest eigenstates of the hydrogen atom.From here.

Question: How can Pn0(r) vanish at r = 0 if Rn0(r) does not?

Radius (a0)

Rnl(r) Pnl(r)

Radial eigenfunctions Rnl(r) and probability distributions Pnl(r) for the lowest eigenstates of the hydrogen atom.From here.

Question: How can Pn0(r) vanish at r = 0 if Rn0(r) does not?

Answer: On a sphere of radius r, we have Pn0(r) = r2 |Rn0(r)|2.

Radius (a0)

Rnl(r) Pnl(r)

For a single electron bound by a nucleus containing Z protons, the solutions of Schrödinger’s equation are almost unchanged; the reduced mass μ is even closer to me, while the potential term is

Thus by replacing e2 with Ze2 in the eigenstates and eigenvalues we obtained for Z = 1, we obtain the Z > 1 eigenstates and eigenvalues.

For example, we obtain the energy eigenvalues

. 4

)(0

2

epep

ZeV

rrrr

.

eV 6.13

24

)( 2

2

2220

22

nZ

n

Zen

E

Exercise: Show that the energy of the ground state is half the expectation value of the potential energy in the ground state.

Exercise: Show that the energy of the ground state is half the expectation value of the potential energy in the ground state.

Solution: The ground state energy is

The expectation value of the potential energy is

.

4

4

4 4

1

4 4

)(ψ 4

20

4

00

2

2

0

2/23

0

2

0

22

1000

2

0

erdre

a

e

drrr

ee

a

drrr

er

r

e

r

ar

.

24

1 22

0

4

e

E

Atomic quantum numbers

According to what we have seen so far, every eigenstate of the hydrogen atom can be associated with three quantum numbers n, l and m, where

n = 1, 2, 3, … “principle” quantum numberl = 0, …, n – 1;m = –l, –l +1, …, l – 1, l .

The degeneracy of the energy eigenvalue En is therefore

1 + 3 + 5 + … + [2(n – 1) + 1] = n2 .

In atomic physics, the l quantum numbers have special names: s for l = 0, p for l = 1, d for l = 2, f for l = 3, etc. Then 2s means n = 2 and l = 0; 3p means n = 3 and l = 1; and so on.

Since the energy of an electron in a stationary hydrogen atom can only be one of the En , where

a stationary atom can absorb and emit photons only if the photon energy equals Ephoton = En – En’ . And since the energy of a photon is related to its frequency ν by Ephoton = hν, the frequencies of electromagnetic radiation emitted or absorbed by a hydrogen atom must obey the rule

for some n and n´. This formula, derived by Bohr 13 years before Schrödinger, was soon verified via spectroscopy.

,

eV 6.13

24 22220

4

nn

enE

,

)'(

11s10 3.3

)'(

11

4 22115

22320

3

4

nnnn

e

Energy levels for hydrogen, and transitions among the levels

Energy levels for hydrogen, and transitions among the levels

Corresponding spectral lines

Setup for emission spectroscopy:

Light from source

The “magnetic” quantum number m:

When an atom is immersed in a uniform magnetic field, the energies En split! Let B be the strength of the field and let point up the z-axis. A state with quantum numbers n, l, m has energy

where μB = eħ/2me is the Bohr magneton. How can we explain this effect? An electron moving in a circular orbit of radius r, at speed v, produces a current I = ev/2πr and a magnetic moment μz = I(πr2) = evr/2 = eLz/2me. Since Lz = mħ, we have μz = μBm.

The corresponding extra potential term for the hydrogen atom is

, μeV 6.13

B2mB

nEnlm

. ˆˆ2

ˆˆ BLBμ e

B m

eV

The “magnetic” quantum number m:

But this is not the only magnetic effect. In a uniform magnetic field, there is a torque on the atom (to make it anti-parallel to the magnetic field). But in a non-uniform magnetic field, there is also a force on the atom. Assume again that the field points up the z-axis, so that VB = –μz B. Then if dB/dz ≠ 0, the force is

So, what happens if a beam of neutral atoms with –lħ ≤ Lz ≤ lħ crosses a non-uniform magnetic field?

. dz

dBV

dz

dF zBz

The “magnetic” quantum number m:

But this is not the only magnetic effect. In a uniform magnetic field, there is a torque on the atom (to make it anti-parallel to the magnetic field). But in a non-uniform magnetic field, there is also a force on the atom. Assume again that the field points up the z-axis, so that VB = –μz B. Then if dB/dz ≠ 0, the force is

So, what happens if a beam of neutral atoms with –lħ ≤ Lz ≤ lħ crosses a non-uniform magnetic field?

We expect the beam to split into 2l+1 beams, one for each value of μz. In some cases, the beam indeed splits into 2l+1 beams.

. dz

dBV

dz

dF zBz

Stern-Gerlach and spin

But O. Stern and W. Gerlach saw a beam of silver atoms split into two beams!

Stern-Gerlach and spin

But O. Stern and W. Gerlach saw a beam of silver atoms split into two beams!

How can have an even number of eigenvalues?

G. Uhlenbeck and S. Goudsmit suggested that each electron has its own intrinsic angular momentum – “spin” – with only two eigenvalues.

But electron spin has odd features. For example, its magnitude never changes, just its direction – and it has only two directions.

ˆzL

Stern-Gerlach and spin

Let’s try to understand spin better by reviewing the algebra of

Consider l = 1 and m = –1, 0, 1. The

matrix representation of in a basis of eigenstates of is

since the eigenvalues are 0 and ±ħ.

.ˆ and ,ˆ ,ˆ ,ˆ 2Lzyx LLL

ˆzL ˆ

zL

,

100

000

001

ˆ

zL

Stern-Gerlach and spin

What is ? We know it must equal but what is a?

We have

.21ˆˆˆ 1

1ˆˆˆ 1

1ˆˆ10*0

222

22

zz

zyx

LL

LLL

LLaa

L

1ˆ L

.

000

200

020

ˆ

L

,0a

0 2 1ˆ L

1 2 0ˆ L

Hence , up to an overall phase. Similarly, we

can show that , hence

Stern-Gerlach and spin

Similarly, Now, since ,

we can write Since

we can write

.

020

002

000

ˆ

L

,ˆ2ˆˆyLiLL .

2010

101

010

ˆ

xL

xLLL ˆ2ˆˆ

.2

00

0

00

ˆ

i

ii

i

Ly

Pauli matrices

It is straightforward to check that these matrix representations

have the correct commutation relations:

But Pauli discovered 2 × 2 matrices with the same commutation relations:

(The “Pauli matrices” are these matrices without the ħ/2 factors.)These are the operators for the components of electron spin!

,ˆˆ , ˆ zyx LiLL

.ˆˆ , ˆ ,ˆˆ , ˆ yxzxzy LiLL LiLL

. 210

01ˆ , 20

0ˆ , 2

01

10ˆ

zyx S

i

iSS

Pauli matrices

We can write the eigenstates of

as for Sz = ħ/2

and as for Sz = –ħ/2.

Since for s = ½,

we refer to electron spin as “spin-½”.

210

01ˆ

zS

0

1or

1

0or

22

222 )1( 410

013ˆˆˆ

ssSSS zyx

Stern-Gerlach and spin

More odd features of electron spin:

The eigenvalues of are ±ħ/2.

We can write an eigenstate of with eigenvalue mħ as

but an analogous eigenstate of , namely

, would not be single-valued. Yet experiments show

that these electron spin eigenstates are not invariant under rotation by 2π, but they are invariant under rotation by 4π!

This is reminiscent of a trick with a twisted ribbon….

zyx SSS ˆ and ˆ ,ˆ

zS

ime2

1

zL

2/

2

1

ie

Stern-Gerlach and spin

This is reminiscent of a trick with a twisted ribbon…one twist cannot be undone, but two twists are equivalent to no twist.

Stern-Gerlach and spin

One more odd feature of electron spin:

For orbital angular momentum, we found that μz = eLz/2me.

For spin angular momentum, experiment shows that μz = eSz/me. That is, electronic spin produces an anomalous “double” magnetic moment.

Therefore, the total magnetic moment of an electron with orbital angular momentum mħ and spin angular momentum ±ħ/2 is

. 2

mm

e

ez

Atomic quantum numbers (again)

We associated every eigenstate of the hydrogen atom with three quantum numbers n, l and m. But now we have to introduce a fourth quantum number, the spin: ms = ±½ .

The degeneracy of the energy eigenvalue En is therefore not n2 but 2n2, since there are two spin states for every set of quantum numbers n, l and m.

The nucleus, too, has spin angular momentum. But its magnetic moment is relatively tiny because the mass of a proton is about 2000 times the electron mass. In this course we neglect the spin and magnetic moment of the nucleus.

Exercise: Show that the superposition of wave functions

is normalized if each wave function is, and calculate and ΔLz.

2

1,1,1,2

2

1,1,1,2

2

1,1,1,2

ψ5

3ψ

5

1ψ

5

1

ˆ ,ˆzz SL

smmln ,,,ψ

Exercise: Show that the superposition of wave functions

is normalized if each wave function is, and calculate and ΔLz.

Solution: Since the components have different eigenvalues, they are orthonormal, and the normalization is obtained from the absolute value of the squares of the coefficients:

2

1,1,1,2

2

1,1,1,2

2

1,1,1,2

ψ5

3ψ

5

1ψ

5

1

ˆ ,ˆzz SL

. 15

3

5

1

5

1ψ

2223

2

,,,

xd

smmln

smmln ,,,ψ

Exercise: Show that the superposition of wave functions

is normalized if each wave function is, and calculate and ΔLz.

Solution:

2

1,1,1,2

2

1,1,1,2

2

1,1,1,2

ψ5

3ψ

5

1ψ

5

1

ˆ ,ˆzz SL

smmln ,,,ψ

. 105

3)

2(

5

1)

2(

5

1)

2(ˆ

5

3

5

3)(

5

1)(

5

1)(ˆ

222

222

z

z

S

L

Exercise: Show that the superposition of wave functions

is normalized if each wave function is, and calculate and ΔLz.

Solution:

2

1,1,1,2

2

1,1,1,2

2

1,1,1,2

ψ5

3ψ

5

1ψ

5

1

ˆ ,ˆzz SL

smmln ,,,ψ

. 5

4

5

3ˆˆ

5

3)(

5

1)(

5

1)(ˆ

2222

22

22

22

22

zzz

z

LLL

L

Exercise: What happens in a Stern-Gerlach experiment, if each electron in an incident beam of hydrogen atoms has l = 1?

Exercise: What happens in a Stern-Gerlach experiment, if each electron in an incident beam of hydrogen atoms has l = 1?

Solution: The magnetic moment of the electron depends on Lz and Sz according to

Since m = –1, 0, 1 and, independently, ms = ±½, we get five possible values of m +2ms : 2, 1, 0, –1, –2. We therefore expect to see 5 separate spots on the screen.

. 22

22 s

ezz

ez mm

m

eSL

m

e

Spin-orbit coupling

We discussed atomic magnetic moments in a magnetic field that is uniform or non-uniform. But even without any external magnetic field, an electron feels an effective field. Why?

Spin-orbit coupling

We discussed atomic magnetic moments in a magnetic field that is uniform or non-uniform. But even without any external magnetic field, an electron feels an effective field. Why?

The electron moves relative to the nucleus. Transforming the Coulomb field to the electron’s rest frame yields a magnetic field B' = –v × E/c2. Since E is radial, –v × E/c2 is –dV(r)/dr times r × p /emec2r = L/emec2r. Since the electron’s magnetic

moment e /me interacts with B', the spin-orbit interaction

contains also . It enters the Hamiltonian as

where V(r) is the Coulomb potential.

, ˆˆ)(

2

1ˆ22SO LS

dr

rdV

rcmH

e

LS ˆˆ

S

Spin-orbit coupling

To compute the eigenvalues of , we must know how to add

angular momenta. Defining , we find that

and so on, i.e. the components of follow exactly the same algebra as the components of and . We immediately infer that the eigenvalues of are j(j+1)ħ2 and that the eigenvalues of are –jħ, (–j+1)ħ,…, (j–1)ħ, jħ.

LS ˆˆ

SLJ ˆˆˆ

, ˆˆˆˆˆ , ˆˆ ˆ , ˆ zzzyyxxyx JiSiLiSLSLJJ

L S

zJ

2J

J

Spin-orbit coupling

To compute the eigenvalues of , we must know how to add

angular momenta. Defining , we find that

and so on, i.e. the components of follow exactly the same algebra as the components of and . We immediately infer that the eigenvalues of are j(j+1)ħ2 and that the eigenvalues of are –jħ, (–j+1)ħ,…, (j–1)ħ, jħ.

Now from we derive

LS ˆˆ

SLJ ˆˆˆ

, ˆˆˆˆˆ , ˆˆ ˆ , ˆ zzzyyxxyx JiSiLiSLSLJJ

L S

zJ

2J

LSSLJ ˆˆ2ˆˆˆ 222

J

. 2

)1()1()1(ˆˆˆ2

1ˆˆ2

222 sslljjSLJLS

Exercise: The spin-orbit coupling splits the degeneracy between

the hydrogen states ψ2,1,1, –½ and ψ2,1,1,½ by ΔE = 4.5 × 10-5 eV.

Estimate the magnetic field B' felt by the electron.

Exercise: The spin-orbit coupling splits the degeneracy between

the hydrogen states ψ2,1,1, –½ and ψ2,1,1,½ by ΔE = 4.5 × 10-5 eV.

Estimate the magnetic field B' felt by the electron. Solution: The energy splitting is due to the interaction of the electron’s magnetic moment with the effective magnetic field B'. In the rest frame of the electron only the spin magnetic moment contributes: μzB' = (eSz/me)B' = ±eB'ħ/2me, hence ΔE = eB'ħ/me and B' = meΔE/eħ

= ΔE/2μB

= (4.5 × 10-5 eV) / 2 × (5.79 × 10-5 eV/T)

= 0.39 T .