2001augqualquant

8/2/2019 2001AugQualQUANT

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Ben SauerwinePractice for Qualifying Exams

Thanks to Yossef and Shiang Yong for their input in this problem.

Problem Source: CMU August 2001 Qualifying Exam

Electron in a spherical well

An electron is in a spherical well of radius a and depth , i.e., the non-relativisticHamiltonian is

0V

V m

p H +=

2

2v

(1)

with m = the mass of an electron and

ar V

ar V V >=

=0

0 (2)

In this problem, we take 12

= h

. The solutions of the Schrodinger equation are of

the form ( ) ( ) slmY r R , , where ( )r R is the radial wave equation, ( ) ,lmY is aspherical harmonic, and s is a non-relativistic spinor. ( )r R is a solution to theequation

( ) ( ) 0121 22

2=

++

R

r ll

V E mdr dR

r dr d

r (3)

where E is the energy of the state.

Note that in spherical coordinates, the operator has the form2

( ,11 22

22 +

=r dr

dRr

dr d

r ) (4)

where is an operator in the angular variables. The differential equation satisfied

by spherical Bessel or Hankel functions is

( ) ( ) ( ) 0111 22

2=

++

xF

xll

dx

xdF x

dxd

x ll (5)


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(a) Explain what properties of the spherical harmonics are used to obtainequation (3) starting with the Hamiltonian given in equation (1). Note theform of the operator given above.2

( )[ ] 02

2

2

22

=+=

+=

V E m

E H V m H

h

Next, separation of variables is assumed so that:

( ) ( ) ( ) 0,22 =+ lmY r RV E m

Now when one expands the operator, one gets2

( ) ( ) ( ) ( )0,2,11 22

2=

++

lmY r RV E m

r dr d

r dr d

r

This is particularly convenient, since the spherical harmonics are orthogonaleigenfunctions of the spherical harmonic differential equation, here written ( ) , , witheigenvalue . At this point, it is clear that by allowing the spherical harmonicdifferential equation to commute with the radial portion and act on the sphericalharmonic, then dividing out the spherical harmonic portion will give (3) exactly.

( 1+ ll )

(b) For our spherical well problem, what conditions must

( )r R satisfy at:

(i) 0=r Since this wave function must be continuous in space, I see that it must without a doubthave a derivative of zero at the origin or otherwise the spherical harmonics approachingfrom either side could force a discontinuity as one approaches from either side:

( )otherwiseconstrano

lif R

int

000' ==

Second, I see that a discontinuity could be imposed by the spherical harmonics if thespherical harmonic is not spherically symmetric:

( )( ) otherwise finite R

lif R0

000 =

(ii) ar =

At this boundary, the wave function must be continuous:


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( ) ( ) ( ) ( )a Ra Rand a Ra R ar ar ar ar == ''

(iii) r

The wave function must be normalizable

( ) 0= R

(c) The solutions for for bound electrons are spherical Bessel functions( )r R( r j l ) inside the well and spherical Hankel functions ( )r hl outside the well.

From the form of the differential equation for the spherical Bessel or Hankelfunctions given above, what are and in terms of the quantities given?The energies depend on the principal quantum number n as well as l. How isthis reflected in the radial equations?

Obeying the Bessel functions inside the well, I have

( )

( ) ( )

( )

( )

( )

( )0

222

2

222

22

2222

22

22

2

22

011

01

11

011111

,

1

,

01

11

V E mmE Identify

Rr

llC

dr dR

r dr d

r

Rr C

lldr dR

r dr d

r C

Rr C

lldr dR

C r C

dr d

C r C

Substitutedr d

C dr d

dxdr

dxd

Cr x Define

x j xhsolutionshas

R xll

dxdR

xdxd

x

ll

+==

=

++

=

++

=

++

==

=

=

++

The principal quantum number as well as the orbital angular momentum will be tied up in

the energy of which bothln E , and are functions, as shown above.

(d) Write down the conditions from which you can find the energy eigenstatesfor bound states of the spherical well.

Valid solutions will be fixed by the boundary condition at the border of the sphere:


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) ( ))( ) ( )( )

( ) =+=

+=

spaceall

lnllnl

lnllnl

r

aV E m BjamE Ah

aV E m BjamE Ah

1

2'2'

22

2

0,,

0,,

v

Many energies E may be possible for a given l, corresponding to the radial portion.

(e) Two electrons are placed in the well. Neglecting interactions and assumingthat the electrons are distinguishable, write down the wave functions of thetwo-particle system. Your solutions should correspond to total orbital andtotal spin quantum numbers.

These will be given by Clebsch-Gordan coefficients. I have:

{ }1,0S

( ) ( ) ( )

( ) (( )

)

( ) ( )

[ ] f M f mnk M k ji f

S

S M

f M f SS

i

Li

Li j

i

ik

L

L M k M jk i L L

n mmnS L

S L

S

S

L

L

x x M S f M f

M Lk M jk i

r Rr Rv

x x

count onwavefunctivalid v

= =

=

+

= = =

=

=

=

=

1

2

1;

2

1

,,;

1,

21,

21

21

0 0,max22,11,

0 02121,

vv

vv

Note that I also have removed the case where the particles had the same wave function.

(f) There is a spin-spin interaction between the two electrons, which has theapproximate form ( )21 SSg S

rv . Show how this interaction arises from the

classical interaction between two magnetic dipoles and express in termsof the properties of the electron. Do not attempt a detailed derivation, but

give a qualitative explanation of the origin of the spin-spin force.

Sg

Suppose two electrons were overlap one another but interact only via their magneticmoments (think of their spins as current loops). There would be a torque on one from thefield on the other, and vice-versa.


5/8

me

Magneton Bohr B 2:

h=

I then have: spinother B B E vv= and similarly,

ee

Sm

e vv =

For a magnetic dipole,

( )[ ] ( )r r r r

Bvvvvv

3

23

40

30 =

Then,

( )[ ] ( )

( )( )[ ] ( ) ( )

( )( ) ( )r SSm

eSSr Sr S

m

e

r

r r r r

r r r r

E

ee

B B B B B B

B

vvvvvvv

vvvvvvv

vvvvv

212

20

21212

2

30

210

212130

03

0

32

34

323

4

32

34

=

=

=

Taking only the portions depending only on 21 SSrv ,

( ) 2122

02

2

30

32

4SSr

me

me

r g

eeS

vvv

=

(g) Treating the interaction of part (f) as a small perturbation, estimate thechange in the energies of the lowest-energy states you listed in part (e).

dV g Sspaceall * will give an estimate. Recall that

( )( )

2

222

21

2

2121

2122

21

2

21

SSSSSS

SSSSSSvvvv

vv

vvvvvv

+=

++=+

so that only the total spin and individual spins take part. There is then shifting but nosplitting in the states above.

( ) ( ) ( )141

043

222

22

21

2

2121

===+= Sif Sif SSSSSS hhvvvv

vv

and the integration is something like


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( ) ( ) dr r r me

me

r r RSS E

een

22

20

2

2

30

0

221 3

24

4

=

vv

which is on the order of

[ ] ( ) ( )

=

13

2

4

14 23

0

22

20

21 dr r r r

r Rm

eSS E n

eabovegiven

vv

(h) The electrons are identical fermions. How are the wave functions of part (e)modified? Prove that if two electrons are placed in the same l-orbit,solutions with total spin S = 1 (triplet states) vanish. Give a qualitativeexplanation in terms of the Pauli Exclusion Principle.

The states in part (e) are anti-symmetrized if the electrons are considered identical. Thismeans that similar states will be combined so that the exchange of particle numbers givesan overall -1 factor.

In the case of identical l-orbit, then, I see from part (e) that all portions of the wavefunctions match under exchange except the spin part:

( ) ( )( )

= =

2

1,

2

1212

1;

21

f

S

S M f M f SS

S

S x x M S f M f vv

In an antisymmetric combination, when I switch the two particles to combine I need an

overall negative factor. In the possibilities { } { }1,01,021

21

;21

21

, the

only qualifying coefficient pair that is nonzero and brings a negative factor in swapping is

is 0021

21

;21

21

0021

21

;21

21 = .

In terms of the Pauli exclusion principle, one can think of a electrons in the same radialand angular momentum orbital. This orbital must be occupied by one spin-up and onespin-down electron.

(i) The system is put in a heat bath at temperature T. If an electron is put in thewell and comes to thermal equilibrium, what is the ratio of probabilities thatthe electron is in the first excited versus the ground state of the system?Assume that the quantum energy levels are not changed by the heat bath.

Using the Canonical Ensemble in a two-state system, I can assume that this ratio isroughly


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0

1

1

1

kT

kT

e

e

I will assume that for the one-electron system, the lowest energy states are both for thelowest radial state and for different total angular momentum states, 0 and 1. The form of the momentum operator then gives

( ) ( ) ( ) ( )

( ) ( )

( ) ( ) ( ) ( )

( ) ( )

( ) ( ) ( ) ( )

( ) ( ) ( )[ ] ( )r RV E mr RV E mr

r RV E mdr d

r dr d

r r RV E m

dr d

r dr d

r

r RV E mdr d

r dr d

r

Y r RV E mr dr

d r

dr d

r

and

r RV E mdr d

r dr d

r

Y r RV E mr dr

d r

dr d

r

10112

102

2112

2

112

2

10122

2

102

2

00122

2

222

21

221

0221

0,2,11

021

0,2,11

=

+

+

=

+

=

+

=

++

=

+

=

++

This integration to collapse the radius, however, is nontrivialat least I get an order of magnitude from the additional term:

( ) ( ) ( )[ ] ( )

( ) ( ) ( )

ma E E

V E mV E ma

dr r RV E mdr r RV E mr

a a

2

01

01

0 010112

ln

22ln2

222

h+=

=+

=

+

where time-averaging over states assumed stationary will remove the seconds from the h-bar units.

akTm

kT

kT ee

e

e ln1

1 2

0

1 h

=

(j) You were not given the value of T in part (i). Describe an experiment thatwould enable you to measure the temperature of the system. Briefly discuss


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the possible errors and the design of your experiment to achieve themeasurement. Assume that a is atomic size and has the scale of electron-volts.

0V

First, notice that the transition from the zero angular-momentum to one angular-

momentum state here could plausibly come from excitation or emission of a circularly-polarized photon.

Assume that the heat-transfer efficiency is perfect between the object and the heat bath.Using this probability, I see that no transition would occur from a photon incident on thewell if the particle were already in an excited state and that absorption and re-emissionwould occur if the particle were in the ground state.

In this case, then, I fire a fine beam of photons of the desired energy into the object, andthen observe the proportion that is allowed to pass through. The temperature can then bededuced from part (i) since the proportion of photons that passes through is equal to the

proportion of measurements that found the electron in the first excited-state.

2001augqualquant

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