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16.1. PROBLEM SET I 197

Answers: Problem set I

1. (a) In one dimension, the current operator is specified by j = 12m(p+

(p)). Applied to the left hand side of the system (outside the regionof the potential), where (x) =Aeikx + Beikx, we havejleft =

km(|A|

2 |B|2); similarly jright =

km(|C|

2

|D|2). Current conservation demands

thatjleft = jright, i.e. |A|2 + |D|2 =|C|2 + |B|2 all of the incoming beamis transferred to the outgoing beam, a statement of particle conservation.This conservation of current can be written asin|in =out|out.Therefore, using the expression for |out, we have

in|in=out|out=inSS!=I

in ,

leading to the unitarity condition.

(b) From the unitarity condition, it follows that

SS= I =

|t|2 +|r|2 rt+ rt

rt+rt |t|2 +|r|2

Comparing the matrix elements, we obtained the required result.

(c) For the -function scattering problem, the wavefunction is given by

(x) =

eikx +reikx x < 0teikx x > 0

For the-function potential, the wavefunction must remain continuous asx= 0, and

x|+x| =2maV02

(0) .

This translates to the conditions 1 + r= t, and ik(t 1 + r) =2maV0

2

t.As a result, we obtain

t= ik

ik+ maV02

,

as required. Using the result from (b), since the potential is symmetric,we have

r2 = t

t(|t|2 1) = ik+ ik+

k2

k2 + 2 1

=

ik+ ik+

2

k2 + 2 =

2

(ik+ )2

As a result, we obtain the required expression for r .

2. (a) We are told that H|1 = E1|1 and H|2 = E2|2, where E1= E2.Therefore1|H|2 =

1 H2 dx =

1E22 dx = E21|2. Since

His Hermitian, we can write

1|H|2=

(H1)2 dx=

(E11)

2 dx=E11|2=E11|2 ,

sinceE1and E2are real. Therefore, (E1E2)1|2= 0 and, ifE1=E2then1|2= 0, i.e. |1 and |2 are orthogonal.

(b) IfA|1 = |2 and A|2 = |1, then adding them, A(|1+|2) =|1+ |2and subtracting, A(|1 |2) = |2 |1=(|1 |2).Hence we have an eigenvector ofa = +1 corresponding to a normalized

eigenvector 12(|1+|2) and an eigenvalue a =1 corresponding to

eigenvector 12

(|1 |2).

Advanced Quantum Physics

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(c) The time-dependent Schrodinger equation is H = E = it, hence(t) = (t= 0)eiEt/. Since|1and |2are eigenstates of the Hamil-tonian Hthen we can write, |(t)= 1

2[|1eiE1t/ |2eiE2t/].

P =|(t= 0)|(t)|2 =14

[1| 2|][|1eiE1t/ |2eiE2t/]

2

=12

[1 + cos((E1 E2)t/)] = cos2((E1 E2)t/2)

3. (a) Differentiating the left hand side of the given expression with respect to ,one obtains

ea

[a, a] =1

ea

= 1 .

Integrating, we therefore have that ea

aea

=+integration constant.By setting = 0 we can deduce that the constant must be a yieldingthe required result. Using this result, we have that

ea

a|= eaaea =+a

|0= (+a)|0= |0 .

Lastly, to obtain the normalization, we have that

|= N20|ea|=N20|n=0

(a)n

n! |

=N2n=0

()n

n! 0|=N2e||2 != 1 .

i.e. N=e||2/2 as required.

(b) For the harmonic oscillator, the creation and annihilation operators are

related to the phase space operators by x = 2m (a+ a), and p =i

m2 (a a). Therefore, we have

x=

2m(+ ), p=i

m

2 () .

Then, using the identity (x)2 =(x x)2=x2 x2, we have

x2= 2m

|(a2 + aa + aa + (a)2|= 2m

(1 + (+ )2),

p2=m2

|(a2 aa aa + (a)2|= 2m

(1 + ()2) .

As a result, we find that x= 2m and p= m

2 leading to the required

expression.

(c) The equation follows simply from the definion of the operator a and thesolution may be checked by substitution.

(d) Using the time-evolution of the stationary states,|n(t)= eiEnt/|n(0),whereEn = (n+ 1/2), it follows that

|(t)= eit/2e||2/2 n

n!

eint|n= eit/2|eit .

Therefore, during the time-evolution, the coherent state form is preservedbut the centre of mass and momentum follow that of the classical oscilla-tor,

x0(t) =A cos( + t), p0(t) = mA sin( + t) .

The width of the wavepacket remains constant.


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4. (a) Starting with the electron Hamiltonian H= 12m(p qA)2, substitution ofthe expression for the vector potential leads to the required result.

(b) Setting = m = = 1, we immediately obtained the required dimension-less form of the Hamiltonian.

(c) Straightforward substitution of the differential operators leads to the re-

quired identities. As a result, we can confirm that

[a, a] = 2

4([z,z] [z, z ]) = 1 ,

and similarly [b, b] = 1. In the complex coordinate representation,

H= 1

2

i(z+ z) i

4(zz)

2+

1

2

(z z) +1

4(z+ z)

2

=2zz+ 18

zz +1

2(zzzz) = 2

z+ z

4

z+

z

4

+

1

2 =aa +

1

2

(d) The angular momentum operator is given by (= 1)

Lz = (x p)z =i(xy yx) = i2

((z+ z)(z z) (zz)(z+ z))

=2z+ z

4

z+

z

4

+ 2

z+ z

4

z+

z

4

=(aa bb) .

(e) In the coordinate representation, the conditiona|0, 0= 0 translates to theequation

2z+

z

4

r|0, 0= 0

We thus obtain the Gaussian expression for the wavefunction. Then, using

the relation |0, m= (b)mm!

|0, 0, we have

r|0, m= 1m!

2m/2z+ z

4m 1

2ezz/4 = 1

22mm!zmezz/4 ,

as required.

(f ) If we populate the states of the lowest Landau with electrons, startingfrom states of the lowest angular momentum m, the wavefunction is aSlater determinant involving entries m(rj) = z

mj e

|zj |2/4. Taking thedeterminant, and making use of the Vandemonde determinant identity,one obtains the required many-electron wavefunction.

5. The spin operator in the (,) direction, S, can be found by forming the

scalar product of the spin operator S with a unit vector in the (,) direction,(sin cos, sin sin, cos ). Therefore

S =

2

cos ei sin ei sin cos

.

We need the eigenvalues of the matrix, i.e.

2

cos sin ei

sin ei cos

uv

=

uv

.

Eliminatinguand v, we find 2 = 1 and hence the eigenvalues ofS are /2,as expected. Substituting the values = 1 back into the equations relatinguand v , we can infer the ratios, uv =e

i cot(/2) andei tan(/2). So, inmatrix notation, the eigenstates are

cos(/2)ei sin(/2)

and

sin(/2)ei cos(/2)

,


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for eigenvalues +/2 and/2 respectively. The spin states in thex-directionare obtained by setting = /2, = 0, and the spin states in the y-directionare obtained by setting = /4 in these general formulae.

6. The four states are given by 1 = +(1)+(2), 2 = (1)(2), 3 =12

[+(1)(2)+(1)+(2)], and4 = 1

2[+(1)(2)(1)+(2)], where

1, 2 and3 are symmetric under particle interchange and 4 is antisymmet-ric.

Since S+S = S2x+ S2y + i(SySxSxSy) = S2 S2z + Sz,

S2 = S+S+ S2z Sz. (16.1)

Noting that Sz = S(1)z +S

(2)z , and likewise forS, and using the basic results

for the operation of the spin operators on the single particle states,

S(1)z (1) =1

2(1), S

(1) (1) = (1), S

(1) (1) = 0 ,

we can now calculate the effect of applyingS2 to each state. For example

S21= 2[2+(1)+(2) + +(1)+(2) +(1)+(2)] = 221 ,

where the three terms in square brackets correspond to the three operators onthe right hand side of Eq. (16.1). Likewise2. Clearly Sz3 = 0 = Sz4, soonly the S+S term need be considered. We find

S+S+(1)(2) = S+(1)(2) = 2(+(1)(2) + (1)+(2))

223 .

Similarly S+S(1)+(2) =

223 so that

S23 = 223, S

24= 0 .

Thus, 1, 2, 3 all have eigenvalue 22 and hence S= 1, while 4 hasS= 0.

The state is clearly an eigenstate ofSz with eigenvalue 0, and must thereforebe a linear combination of3 and 4. The S= 1 component is thus the 3term in the wavefunction with amplitude

c3 =3|=

2

3+(1)(2) +

1

3(1)+(2)|

1

2+(1)(2) +

1

2(1)+(2)

=

1/3 +

1/6

The probability ofS= 1 is therefore |c3|2 = 3+2

2

6 = 0.971.

7. Taking as a basis the states of the Sz operator, we can deduce the form of theSx operator most easily from the action of the spin raising and lower operators.Noting that, for spin S= 1, S|S= 1, m= [2 m(m+ 1)]1/2|1, m1, wecan construct the matrix elements ofS. The latter are given by

S+ =

2

0 1 00 0 1

0 0 0

, S = S+ .

Then, using the relation, Sx = 12 (S+S), we obtained the matrix elements

of the operator and corresponding eigenstates,

Sx = 2

0 1 01 0 10 1 0

,mx = 1 mx = 0 mx =

1

1

2

121

12

101

12

121


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When placed in a magnetic field, B, the molecules will acquire an energyBmz, where is the magnetic moment of the molecule, which in this caseequals twice the magnetic moment of the proton, and mz is the eigenvalue ofSz . Att = 0, the molecules enter the magnetic field in the mx = 1 state, afterwhich their wavefunction evolves with time in the usual way, i.e.

(t) = eiBSzt/(0) = 1

2

eiBt

0 00 0 00 0 eiBt

121

= 12

eiBt

2

eiBt

.Thus, ifBt = (2n+ 1), with n an integer, the molecules will be in a puremx =1 state, and none will pass the second filter. The time is given byt=L/v = L

m/2E, whereL= 20 mm and m and Eare the mass and energy

of the molecules respectively. We thus have

= (2n+ 1)

BL

2E

m

1/2= 2.84 1026 JT1

and hence the proton magnetic moment is 1.42 1026 JT1.

Note that the result can also be obtained by treating the problem as one of

classical precession. The couple = B = L, where L = is the angularmomentum and the angular frequency of precession. Ift= (2n+ 1), themolecules have precessed into themx =1 state, and the result readily follows.

8. Substituting for the definition of the spin raising and lowering operators usingthe Holstein-Primakoff transformation, the commutator is obtained as

1

2S2[S+, S] =

1 a

a

2S

1/2 aa + 1aa

1 a

a

2S

1/2 a

1 a

a

2S

a

=

1 a

a

2S

+aa

1 a

a

2S

aa+ a

aaa

2S = 1 a

a

S .

With Sz = (Saa), we obtain the required commutation relation [S+, S] =2Sz.

M (m1, m2)3 (1, 2)2 (1, 1) (0, 2)1 (1, 0) (0, 1) (1, 2)0 (1, 1) (0, 0) (1, 1)

1 (1, 2) (0, 1) (1, 0)2 (0, 2) (1, 1)3 (1, 2)

9. For the case 1 = 1, 2 = 2, a table of possible ways of forming each valueof M = m1+ m2 is shown right. The largest value of M is 3, so the largestvalue ofL must be 3. There must also be a state withM= 2 correspondingto L = 3, but we have two states with M = 2. Therefore there must be astate with L = 2 as well. We need two states with M= 1, one for each of theL = 3, 2 multiplets, but we actually have three states with M = 1, so theremust be an L = 1 state as well. All of the M states are now accounted for.

M (m1, m2)4 (3, 1)3 (3, 0) (2, 1)2 (3, 1) (2, 0) (1, 1)1 (2, 1) (1, 0) (0, 1)0 (1, 1) (0, 0) (1, 1)

1 (0, 1) (1, 0) (2, 1)2 (1, 1) (2, 0) (3, 1)3 (2, 1) (3, 0)4 (3, 1)

For the case 1 = 3, 2 = 1, we can again for a table (see right). Following the

same logic as before, we see that states with L = 4, 3, 2 just account for all thestates.

To construct the states explicitly, we start by writing the L = 3 M= 3 state,since there is only one way of forming M= 3, viz. |3, 3 = |1, 1 |2, 2. Wethen operate with the lowering operator L, which is simply the sum of thelowering operators for the two separate particles. Recalling that:

L|, m=( + 1) m(m 1)|, m 1 ,

we obtain

6|3, 2 = 2|1, 0 |2, 2 + 4|1, 1|2, 1, where the firstterm on the right hand side comes from lowering the = 1 state and thesecond from lowering the = 2 state. Hence |3, 2 =

1/3|1, 0 |2, 2+

2/3|1, 1|2, 1. The state|2, 2must be the orthogonal linear combination,

i.e. |2, 2= 2/3|1, 0 |2, 2 1/3|1, 1 |2, 1. Further states could becomputed in the same way if required.


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10. The commutation relations are given by [Ji, Jj ] = iijkJk.

(a) First define the eigenstates m: Jzm = mm. To see if Jm is aneigenstate ofJz, we need to look atJzJm, which is equal to JJzm[J, Jz]m. The required commutator is [J, Jz ] = [Jx, Jz] i[Jy, Jz ],from the definition ofJ. From the basic commutators given at the start,

this is [J, Jz] = ) iJy Jx) = J (if treating like a numberis confusing, do this separately for J+ and J). Going back to JzJm,we can now write this as JJzm + Jm. The first term is justJmm, so this is (m 1)(Jm). Thus, Jm is an eigenstate ofJz, with eigenvalue (m1). This establishes the raising and loweringproperty ofJ.

(b) Two electrons would have a total spin ofS= 1 or 0. Adding a third spin1/2 particle creates total spinS= 3/2 or 1/2 from theS= 1 two-particlestate. The S= 0 two-particle state becomes S= 1/2 only on adding thethird particle, so total S= 3/2 or 1/2 are the only possibilities.

(c) The states with well-defined values of m1, m2, and m3 for the z spincomponents of all particles are the uncoupled basis. Where all particles

are spin up, this state may be written as |. This state is also themS= 3/2 state of total S= 3/2 (there is no other way to get m1+ m2+m3 = 3/2 in the uncoupled basis). We can therefore write |S= 3/2, mS=3/2= |. To get from here to |S= 3/2, mS= 1/2, we need to applyJ = S

(1) + S

(2) + S

(3) . In other words, the total lowering operator is the

sum of the lowering operator for each separate spin (reasonably enough)

this follows from the definition of J and Jx = S(1)x +S

(2)x +S

(3)x , etc.

Now, we need to use the given normalization result. This says that

J|S= 3/2, mS= 3/2=

15/4 3/4|S= 3/2, mS= 1/2=

3|S= 3/2, mS= 1/2 .

Notice that the total quantum number, J, is the same as the overall spinquantum number, S in this case. Therefore |S = 3/2, mS = 1/2 =(1/

3)J|S = 3/2, mS = 3/2. Using the given normalization result

again for a single state, S|1/2, 1/2 =

3/4 + 1/4|1/2, 1/2. Thisestablishes the required result.


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