Assignment 2 Solutions

Problem 1

We will need the change of basis matrices between the x-, y-, and z- bases. These change of basis

matrices are given by the eigenvectors of the spin operators.

Sz =1

2

DiagonalMatrix@83, 1, −1, −3<D;

Sx =1

2

0 3 0 0

3 0 2 0

0 2 0 3

0 0 3 0

;

Sy =1

2

0 −I 3 0 0

I 3 0 −I 2 0

0 I 2 0 −I 3

0 0 I 3 0

;

Mxz = Transpose@Table@Normalize@Permute@Eigenvectors@SxD, 84, 1, 3, 2<D@@iDDD, 8i, 1, 4<DD;

Myz = Transpose@Table@Normalize@Permute@Eigenvectors@SyD, 84, 1, 3, 2<D@@iDDD,

8i, 1, 4<DD;

Mzx = Inverse@MxzD;

Mzy = Inverse@MyzD;

Mxy = Mzy.Mxz;

Myx = Mzx.Myz;

Case 1: express the original state in the x basis

8 Mzx.81, 0, 0, 0< êê MatrixForm

1

− 3

3

−1

So 3

2]z

=1

8II 3

2^x

- 31

2^x

N - I -3

2^x

- 3-1

2^x

NN,

where the first two terms are the A state and the last two terms are the B state. The SGz devices don’t

change these states, but simply split them into several beams. The original state is therefore just given

by A\ + B\, up to normalization. The probabilities of getting an A or a B outcome is therefore 50%.

Recombining the states without measurement into an SGx device is the same as simply passing the

original state through an SGx and will result in 4 separate beams, with probabilities 3

8 for the spin

≤1

2

outcomes and 1

8 for the spin

≤3

2 outcomes.

So 3

2]z

=1

8II 3

2^x

- 31

2^x

N - I -3

2^x

- 3-1

2^x

NN,

where the first two terms are the A state and the last two terms are the B state. The SGz devices don’t

change these states, but simply split them into several beams. The original state is therefore just given

by A\ + B\, up to normalization. The probabilities of getting an A or a B outcome is therefore 50%.

Recombining the states without measurement into an SGx device is the same as simply passing the

original state through an SGx and will result in 4 separate beams, with probabilities 3

8 for the spin

≤1

2

outcomes and 1

8 for the spin

≤3

2 outcomes.

Case 2:

8 Mzx.80, 0, 0, 1< êê MatrixForm

1

3

3

1

So -1

2]z

=1

8II 3

2^x

+ 3-1

2^x

N - I -3

2^x

- 31

2^x

NN. (note the order of terms, for later conve-

nience)

All four components are sent through individual SGz devices, but these have no effect since we send

the outcomes for each SGz the same way. The top SGx receives the first two terms in the decomposi-

tion above and the bottom one receives the last two terms. This is similar to the first case. Each elec-

tron will pass through the top or the bottom SGx with 50% probability, if we were to measure it.

The top SGx will give x-spin outcomes of 3

2 or

-1

2 with overall probability

3

8 and

1

8 respectively, while the

bottom SGx will give outcomes of -3

2 or

1

2 with probability

3

8 and

1

8 respectively.

Case 3:

We will assume the original two states are actually components of a superposition state (i.e. the two

beams were prepared by a single process and then split and directed into the apparatus).

Mzy.81, 0, 0, 0<Mzy.80, 0, 0, 1<

:−ä

2 2

,

1

2

ä3

2

, −1

2

ä3

2

,

ä

2 2

>

:1

2 2

,

3

2

2

,

3

2

2

,

1

2 2

>

The bottom two components of the first state and the top two components of the second then get fed

into an SGx, so we need to convert to the x-basis

Myx.:1

2 2

,

3

2

2

, −1

2

ä3

2

,ä

2 2

> êê FullSimplify

:1

2− ä

2

, 0,

3

2

2

, 0>

and the top two components and bottom two components go through their respective SGy , so we

change bases again

2 QM_A2Sol.nb

Mxy.:ä

2 2

, 0, 0, 0>

Mxy.:0, 0,

3

2

2

, 0>

:−

1

8+

ä

8

2

, −1

8

+ä

8

3

2

,

1

8

+ä

8

3

2

,

1

8−

ä

8

2

>

:3

8−

3 ä

8

2

,

1

8

+ä

8

3

2

,

1

8

−ä

8

3

2

,

3

8+

3 ä

8

2

>

Finally we recombine the appropriate components with the ones from the original SGy devices to get

the states entering each SGz

:−

1

8+

ä

8

2

, −1

8

+ä

8

3

2

, 0, 0> + :−ä

2 2

,1

2

ä3

2

, 0, 0>

:0, 0,1

8

+ä

8

3

2

,

1

8−

ä

8

2

> + :3

8−

3 ä

8

2

,1

8

+ä

8

3

2

, 0, 0>

:0, 0,1

8

−ä

8

3

2

,

3

8+

3 ä

8

2

> + :0, 0,

3

2

2

,1

2 2

>

:−

1

8+

5 ä

8

2

, −1

8

+5 ä

8

3

2

, 0, 0>

:3

8−

3 ä

8

2

,

1

8

+ä

8

3

2

,

1

8

+ä

8

3

2

,

1

8−

ä

8

2

>

:0, 0,

5

8

−ä

8

3

2

,

7

8+

3 ä

8

2

>

Converting them to the z-basis we get.

QM_A2Sol.nb 3

s1 = Myz.:−

1

8+

5 ä

8

2

, −1

8

+5 ä

8

3

2

, 0, 0>

s2 = Myz.:3

8−

3 ä

8

2

,1

8

+ä

8

3

2

,1

8

+ä

8

3

2

,

1

8−

ä

8

2

>

s3 = Myz.:0, 0,5

8

−ä

8

3

2

,

7

8+

3 ä

8

2

>

:5

8

+ä

16

,

5 ä 3

16

,

ä 3

16

, −1

8

+5 ä

16

>

:1

16

+ä

16

, −1

16

+3 ä

16

3 , −1

16

−ä

16

3 ,

5

16

+ä

16

>

:3

16

+ä

4

, −1

16

−ä

8

3 , −1

16

+3 ä

8

3 ,

11

16

>

From which we can read off the relative probabilities of the various outcomes for each SGz by squaring

the components.

s1 Conjugate@s1D êê FullSimplify

s2 Conjugate@s2D êê FullSimplify

s3 Conjugate@s3D êê FullSimplify

:101

256

,

75

256

,

3

256

,

29

256

>

:1

128

,

15

128

,

3

128

,

13

128

>

:25

256

,

15

256

,

111

256

,

121

256

>

In all three cases, recombining all the final outputs will yield the original state (or sum of input states, in

case 3) and the outcome of feeding it into an SGz device is 3

2 for case 1,

-3

2 for case two, and

3

2 or

-3

2

with equal probability in case 3.

Problem 2

In 1+1 dimension, the electric fields made by point charges are constant, proportional to the charge

(compare to infinite charged planes in 3+1 dimensions). So if the stationary charges are Q1 and Q2,

located at ≤d

2 respectively the electric potential energy is

Ue = -1

e0

qJQ1Jx +d

2N + Q2J d

2- xN N

where we assume -d

2< x <

d

2

The springs also contribute to the potential energy as

Us = k1Jx +d

2N2

+ k2Jx -d

2N2

where we can set k1 = 0 for the single-spring case.

4 QM_A2Sol.nb

The springs also contribute to the potential energy as

Us = k1Jx +d

2N2

+ k2Jx -d

2N2

where we can set k1 = 0 for the single-spring case.

U = k1Jx +d

2N2

+ k2Jx -d

2N2

-1

e0

qJQ1Jx +d

2N + Q2J d

2- xN N

This is a shifted quadratic potential potential with a minimum at xmin =d

2

Hk2-k1L+q

e0

HQ1-Q2Lk1+k2

SolveBDBk1 x +d

2

2

+ k2 x −d

2

2

−1

e

q Q1 x +d

2

+ Q2d

2

− x , xF � 0, xF êê FullSimplify

::x →d e H−k1 + k2L + q HQ1 − Q2L

2 e Hk1 + k2L>>

The linear terms shift the equilibrium point, but not the frequency of the oscillator, which will remain

w =k1+k2

m

and the energy levels are ÑwIn +1

2M + U0 as usual with U0 = UHxminL and the wavefunctions given by the

usual harmonic oscillator wavefunctions centered around x =d

2

Hk2-k1L+q

e0

HQ1-Q2Lk1+k2

.

Throughout the problem we assume that the charges are far enough apart (or the difference of charges

is small) and the springs are sufficiently stiff that the lowest lying states are localized between the

massive charges. Outside the region between the charges, the electric fields change signs and the

potential becomes a different (but still quadratic) function, giving corrections to the energies and

wavefunctions.

Problem 3

The time dimension is basically irrelevant. We’re looking for eigenstates on a torus with dimensions

l1, l2 along x and y respectively. The torus topology imposes periodic boundary conditions in x and y.

The allowed wavefunctions are of the form

y = expJi m p

l1

xN expJi n p

l2

yNwith energy, up to factors of mass and Ñ,

Em,n = J m p

l1

N2

+ J n p

l2

N2

If we populate these energy levels with 2N spin 1

2 particles, by Pauli exclusion, we will populate the

bottom N energy levels with two particles per level, so the minimum energy of the system will be a sum

of the bottom N energy levels times two.

For large N, the number m, n can be treated as continuous and the sum as an integral over an ellipse

with volume N

E = 2 ‡ dm dn J m p

l1

N2

+ J n p

l2

N2

= 2l1 l2

p2 Ÿ dm dn Im

2+ n

2M = 4l1 l2

pŸ dr r

3

where the first change of variables changes the ellipse to a circle and the second expresses it in polar

coordinates and does the angular integral. The volume of the ellipse was N, so the volume of the circle

is Np

2

l1 l2

and the limits on the radial integration is then from zero to Np

l1 l2

. Performing the final inte-

gral, we get

QM_A2Sol.nb 5

E = N2 p

l1 l2

6 QM_A2Sol.nb