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16.4. PROBLEM SET IV 220

Answers: Problem set IV

1. (a) From Ehrenfests theorem, the equation of motion for the spin is given

byidSmdt = [H, Sm]. Making use of the spin commutation relation, wehave (summation on repeated spin indicies assumed)

idSmdt

= J[Sm, Sm](Sm+1+ Sm1) = iJSm(Sm+1+ Sm1) .

We thus obtain the required equation of motion.

(b) Since Sm+1+Sm1 2S|x=m+2S|x=mand, for classical vectors,SS=0, we obtain the required equation of motion.

(c) Substituting the expression forS(x, t), we find that the equation is solvedwith(k) = J k2

S2 c2. The corresponding spin configuration is shown

right.

(d) Substituting for the spin raising and lowering operators, the identity isclear. Expanding to the spin raising and lowering operators to leading

order in aa2S about the ferromagnetic ground state (in which all spins are

aligned along ez

, we obtain

H= J N S2 + J Sm

amam+ a

m+1am+1

amam+1+ h.c.

+ O(S0) ,

where h.c. denotes the Hermitian conjugate. Rearranging, we obtain therequired expression for the Hamiltonian.

(e) With the definitions given in the problem,

[ak, ak ] =

1

N

m,n

eikm+ikn [am, a

n]

mn

= 1

N

m

ei(kk)m =kk.

Then subtituted into the Hamiltonian,

H= J N S2 + Skk

1Nm

ei(kk

)m kk

(eik 1)(eik

1)akak

= J N S2 + Sk

|eik 1|2akak.

From this result we obtain the required dispersion relation.

2. Standard bookwork allows a derivation of the amplitudecn(t). In the presentcase, with V(t) = eE0ze

t/, the matrix element2s|z|1s = 0 since the 1sand 2s wavefunctions both have even parity while z has odd parity. Thereforethe probability of finding the atom in the 2s state is identically zero.

The matrix elements2p1|z|1s = 0 since the part of the integral willvanishes,

2p1|z|1s 20

dei = 0 .

The only non-zero matrix element is:

2p0 |z|1s =

1

32a50

1/2 1

a30

1/2 r2dr r2er/a0er/2a0

2 sin d cos2

= 1

4

2a40

4!

(3/2a0)5

4

3 =

256a0

243

2.

Taking the limit ast , the t

integral is given by,

0

dtet/ei(E2pE1s)t

/ = 1

1/ iE/,

Advanced Quantum Physics

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where E = E2p E1s = 3R/4. Putting all this together we obtain theprobability of being in the 2p0 state after a long time as

|c2p0()|2 = e2E20a

202

15

310

1

E2 +2/2.

3. From the lecture notes, the decay rate for unpolarized light is given by,

A=3|dkj |

2

30c3 ,

and the lifetime is thus = 1/A. Take for example the 2p0 state of Hydrogendecaying to 1s (the other 2p states must have the same lifetime, but this onedepends on the same matrix elements that we computed in in the previousquestion. Only the z-component of d is non-zero for this transition, (the integral yields zero if you compute the matrix elements ofx or y ) giving,

2p0| ez |1s = 256ea0243

2

= 6.31 1030 Cm .

The energy of the emitted photon is

= 3

4R=

3

4

me4

2(40)22 = 1.56 1016 Hz .

Hence, the lifetime of the state is = 1.56 109 s.The only lower lying state to which 3s can decay is 2p according to the selectionrules. We can expect the matrix element3s| ez |2p ea0 on dimensionalgrounds, and thus not very different from2p| ez |1s. The main differencebetween the lifetimes of the 3s and 2p levels will arise from the difference in3. For the 3s2p transition,

= (1

41

9)R=

5

36R .

The ratio of the lifetimes is therefore approximately

(3s)

(2p)

3

4

36

5

3 150 .

The only state lying below 2s is 1s, but the decay 2s1s is not allowed bythe electric dipole selection rules. The 2s state is metastable. The dominantdecay is actually via two-photon emission, a process which can arise throughsecond order perturbation theory, and occurs very slowly. In practice, atomsmay well make transitions from 2s to 2p (for example) before decay takes placeas a result of collision processes. Alternatively, decay of the 2s state may beinduced by the application of an external electric field, which mixes 2s and 2pthrough the Stark effect.

4. From the lecture notes, the Born Approximation gives,

d

d =

m22

2

V(r)eird3r

2 ,where is the difference beweeen incoming and outgoing wave vectors, ofmagnitude 2k sin2(/2). In the case whereV(r) = V(r), i.e. where the potentialis centrally symmetric, it is convenient to take as the axis of polar coordinatesfor the purpose of integration, so that r = ||r cos . The integral thusbecomes

V(r)eird3r= V(r)eir cos

2 sin dr2dr

= 2

V(r)r2dr

eir cos

ir

0

= 4

V(r)rdr sin(r) ,


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and hence

d

d =

2m

2

2

V(r)rdr sin(r)

2 .TakingV(r) =

V0 for r

a, and V(r) = 0 otherwise, the integral becomes

(integrating by parts),

V0 a0

r sin(r)dr= V0

r cos(r)

a0

+

a0

cos(r)

dr

= V02

(sin(a) a cos(a)) ,

and thus

d

d=

2mV023

(sin(a) a cos(a))2

.

In the low energy limit, 0,

sin(a) a cos(a) a 1

3! (a)3

a(1 (a)2

/2) = (a)3

/3 ,

and hence

d

d=

2mV0a

3

32

2.

This is independent of and hence independent of, so isotropic, as required.The total cross-section is obtained by integrating over solid angles, which simplyinvolves multiplying by 4 in this case

tot = 4

2mV0a

3

32

2.

5. (a) WhenkR 1, s-wave scattering dominates. In this case, the problem isequivalent to a one-dimensional scattering problem with an infinite wallat the origin and a -function repulsive potential at r = R.

The wavefunction has the solution,

u(r) =

Csin kr r < Rsin(kr+0) r > R

From the continuity condition on the wavefunction and the derivative, weobtain

A sin(kR) = sin(KR +0)

kA cos(kR) k cos(kR+0) =U0sin(kR+0) .

From the first equation, we obtain A = sin(kR+0)sin(kR) which substituted into

the second equation, leads to the relation

0= tan1

k tan(kR)

k U0tan(kR)

kR .

The structure is similar to that obtained for the spherical square potentialbut with different resonant behaviour.

(b) With U01/R,k, and U0tan(kR)k, we obtain the resonance condi-tion

k tan(kR)k U0tan(kR) kU0tan(kR) 0 ,

i.e. 0 kR, the value that it would have for a hard sphere.


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(c) Now supose that tan(kR) is small. In this case, we have a resonance whenk U0tan(kR) = 0, i.e. tan(kR) = kU0 1, and

0 =

2 kR

2.

The cross-section 0

= 4k2 sin

2 0

4

k2 . The resonance is near tan(kR) =

0, which implies thatkR = (2n+1)/2, the quasi-bound state of the well.

6. Substituting the definition ofS() into the defining condition we obtain1 i

4

1 + i

4

= +

i

4[,]

+

= (g ) .Rearranging the left and right hand sides, we obtain

i

4[,]

= g g,

from which we obtain the required identity. The latter equation is shown to beconsistent with the solution = (i/2)[, ] by making use of the anticom-mutation relation of thematrices.

7. Using the identity

[ p, S p] = pipj

0 [i,j ]

[i, j ] 0

= 2iijk pipj

0 kk 0

.

Therefore, since p p = 0, we find that the Hamiltonian commutes with theHelicity operator.

Turning to the angular momentum, taking each term separately,

[H, Li] =ijk [ p,xj pk] =ijk(lplxj pk xj pklpl)=ijk(illj pk) = i p .

[H, S] = [ p, S] = 1

2(ipij jipi)

= 1

2

0 p p 0

,

00

=

1

2

0 11 0

[ p,]

= i

0 11 0

p = ip .

Putting these terms together we find [ H, J] = 0.

8. Applying the plane wave solution of the Dirac equation(p) =epxu(p) (de-fined in this form for positive and negative energy states) to the two edges ofthe potential step, we obtain the boundary conditions

10p

E+m0

eipa/2 +r

10

pE+m0

eipa/2

= t

10p

E+m0

eipa/2 + r

10

pE+m0

eipa/2

t 10p

E+m0

eipa/2 + r 10 pE+m0

eipa/2 =t 10pE+m

0

eipa/2,


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where the reflection and transmission coefficients are defined in the figure.

From these equations we obtain

2eipa/2 = t(1 +)eipa/2 + r(1 )eipa/2

2reipa/2 = t(1 )eipa/2 + r(1 +)eipa/2

teipa/2 = eip

a/2t + eip

a/2r

teipa/2 =

eipa/2t eipa/2r

.

Rearranging these equations we obtain

r = 2

1 +

1

eipa 1eipa e

i(pp)a/2,

where = (1 )/(1 +). Finally, with this result, we obtain

t=eipa 1

cos(pa) i sin(pa)(1 +2)/2

From this result, we obtain the expression for the transmitted current shownin the question.

For energies E > m, the particles traverse the barrier as a plane wave. Inparticular, when pa = n there is perfect transmission. For m > E >m,

p is imaginary and exchange of particles occurs by resonant tunnelling acrossthe barrier. For energies E

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