phys 481 homework 1 solutions

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Phys 481 Homework 1 Solutions

January 16, 2016

4.25) In this problem we are treating the electron to behave as a classical sphere (this is a rather naivemodel, for electrons are point particles so we do not in general treat them like rigid bodies as we

are in this problem), so we will use the fact that the moment of inertia of a solid sphere is given byI = 2/5mr2c , where

rc = e2

4π0mc2,

as given in the problem statement. From classical mechanics, we know that L = I ω = 2/5mr2c (v/rc),so using the fact that L = /2, we see that

2 =

2

5mrcv,

thus

v = 5 4mrc

= 5π 0c2

e2 = 5π(1.055 × 10

−34

J · s)(8.85 × 10−12 C 2

N ·m2 )(3 × 108

m/s)2

1.6 × 10−19C 2 ≈ 5.16 × 1010m/s.

Clearly this model cannot make sense, for we obtained a velocity that is over 100 times greater thanthe speed of light!

4.27) (a) We know that our spin state |χ is to be normalized, so we require that χ| χ = 1, so:

χ| χ = |A|2 −3i 43i4

=

|A

|2(9 + 16) = 25

|A

|2 = 1

⇒ A = 15

.

(b) We’re looking for S i ≡ χ|S i |χ = 2 χ|σi |χ, where i = x, y, z and the σi’s denote the 2 × 2Pauli matrices:

S x = 2χ|σx |χ

=

2

1

5−

3i 40 11 0

1

5

3i4

=

50

4 −3i3i

4

=

50(12i − 12i)

= 0

S y = 2χ|σy |χ

=

50 −3i 40 −ii 0

3i4

=

50

4i −33i

4

=

50(−12 − 12)

= −24

50


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S z = 2χ|σz |χ

=

50

−3i 41 00 −1

3i4

=

50−3i −4

3i

4

=

50(9 − 16)

= −7

50 .

(c) To find uncertainties σSi ≡ S 2i − S i2, i = x, y, z, we first note that the square of any of

the Pauli matrices gives the identity matrix: i.e.

σ2x = σ2y = σ

2x =

1 00 1

(you should check this on your own!).

From here, we see that for any normalized spinor

|χ =

ab

, where |a|2 + |b|2 = 1,

it follows that

S 2i =

2

4

χ σ2i χ

=

2

4 χ| I |χ

=

2

4

a∗ b∗1

1a

b

=

2

4

a∗ b∗

ab

=

2

4 (|a|2 + |b|2)

=

2

4 ,

where i = x, y, z. Now that we’ve established S 2i , we find that

σSx = S 2x − S x2

=

2

4 − 0

=

2.

σSy = S 2y − S y2

=

2

4 −

24

50

2

= 7

50.

σSz = S 2z − S z2

=

2

4 − 49

2

2500

= 12

25 .

(d) The uncertainty principles that we need to check are listed as follows:

σSxσSy

≥

2|

S z

|, σSxσSz

≥

2|

S y

|, and σSyσSz

≥

2|

S x

|.


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Multiplying the products of the uncertainties found in part (c) above, we find:

σSxσSy = 7 2

100 =

2|S z|

σSxσSz = 12 2

50 =

2|S y|

σSyσSz =

96 2

1250 > 0 =

2 |S y|,so our results are consistent with all three of the uncertainty principles.

4.28) Given the most general normalized spinor

|χ =

ab

,

we see

S x = 2χ|σx |χ

=

2

a∗ b∗

0 11 0

a4b

=

2

b∗ a∗

ab

=

2(b∗a + a∗b)

S y = 2χ|σy |χ

=

2

a∗ b∗

0 −ii 0

ab

=

2

b∗i −a∗ia

b

= i

2 (ab∗ − a∗b)

S z = 2χ|σz |χ

=

2

a∗ b∗

1 00 −1

ab

=

2

a∗ −b∗a

b

=

2(|a|2 − |b|2).

We already showed in part (c) of the previous problem that S 2i = 2

4 . From this result, we see that

S 2x

+S 2y

+S 2z

= 3 2

4 . Explicitly computing

S 2

:

S 2 = 3 2

4

a∗ b∗

1 00 1

ab

= 3 2

4

a∗ b∗

ab

= 3 2

4 (|a|2 + |b|2)

= 3 2

4 .

We know this must be true because

|χ

is normalized, so it follows that

|a

|2 +

|b

|2 = 1.

4.29) (a) We know that for any spin-1/2 state |χ, a measurement of any spin operator will yield aneigenstate with eigenvalues ±

2 (you can check this by explicitly determining the eigenvalues of

the 2×2 matrix S y, as you would any 2×2 matrix). Now it amounts to finding the eigenspinorsof S y:

2

0 −ii 0

a1b1

= +

2

a1b1

so a1 = 1, and b1 = i

2

0 −ii 0

a2b2

= −

2

a2b2

so a2 = 1, and b2 =−

i


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Substituting our values in and normalizing, we find our eigenspinors to be

χy+ = 1√ 2

1i

for spin up, and

χy− = 1√ 2 1

−ifor spin down.

(b) As was indicated in part (a), a measurement of S y on the general state |χ will yield values of ±

2. We can determine the relative probabilities of measuring either

2 or −

2, by looking at

| χy+ χ |2 or | χy− χ |2, respectively:

| χy+ χ |2 = 1√ 2

1 i

ab

2

= 1√ 2 (a + bi)2

= 1

2|a + bi|2

| χy− χ |2 =

1√ 2

1 −ia

b

2

= 1√ 2 (a − bi)

2

= 1

2|a − bi|2

so the probability of measuring + 2

is 12|a+bi|2, and the probability of measuring −

2 is 1

2|a−bi|2.

As a sanity check, let’s make sure that these two probabilities add to unity:

| χy+ χ |2 + | χy− χ |2 = 12 |a + bi|2 + 12 |a − bi|2=

1

2(a + bi)(a∗ − b∗i) + 1

2(a − bi)(a∗ + b∗i)

=

1

2|a|2 + ab∗i − ba∗i + |b|2 +

1

2|a|2 − ab∗i + ba∗i + |b|2

= |a|2 + |b|2= 1.

(c) As we have shown earlier, S 2i = 2

4 , so we would expect to measure a value of +

2

4 , with a

probability of 1.

4.31) When constructing the spin-1 operators, note that there are now three allowed projections of angularmomentum, namely ms = 1, 0,−1. Given this, we can construct our three spin-1 states |s, ms as:

|s = 1, ms = 1 = 1

00

, |1, 0 =

0

10

, |1,−1 =

0

01

.

Using the fact that in the basis where S z is diagonalized, S z |s, m = m |s, m, it follows that thematrix operator S z in the spin-1 case is given by:

S z =

1 0 00 0 0

0 0 −1

.

Aside : We can rigorously test this by construcing the individual matrix elements of S z, this canbe done by computing

s, m

|S z

|s, m

= m

s, m

|s, m

= m δ ssδ mm , where δ ss and δ mm are


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the kronecker delta symbols. We know that the matrix elements evaulate to this because of theorthonormality of the |s, m states. Analyzing the expression s, m|S z |s, m = m s, m| s, m =m δ ssδ mm carefully, we see that the spin operator S z forms a diagonal 3 × 3 matrix with diagonalentries given by ms (and of course a multipicative constant of ). We could use this expression toshow that for spin-3/2 particles S z is given by times a 4× 4 diagonal matrix with diagonal entriesgiven by 3/2, 1/2,−1/2,−3/2, respectively. This also generalizes to the diagonal spin operator forhigher spins.

Back to the main task at hand: with the construction of our S z operator, we can act with ladderoperators S ± on our three spin-1 states to determine the spin-1 operators S x and S y. To do this, wemust keep the following in mind (it is a good exercise to check these):

S ± |s, m =

s(s + 1) − m(m ± 1) |s, m ± 1 (1)

and

S x = S + + S −

2 S y =

S + − S −2i

. (2)

Applying (1) to our spin-1 kets, we see

S + |1, 1 =00

0

S + |1, 0 =√

2 |1, 1

=√

2

10

0

S + |0, 1 =√

2 |1, 0

=√

2

01

0

Piecing together the three column vectors, will give us our S + operator for spin 1 (we know thatS + |1, 1 must be 0, for |1, 1 is the highest ms state for a spin-1 particle):

S + =0

√ 2 0

0 0 √ 2 0 0 0

We can now apply the same procedure to construct S −:

S − |1, 1 =√

2 |1, 0

=√

2

01

0

S − |1, 0 =√

2 |1,−1

=√

2

00

1

S − |1,−1 =

000

Like before, we may now piece together the three column vectors to get

S − =

0 0 0√ 2 0 0

0√

2 0

.

Now that we have matrix representations of S + and S −, we simply substitute them into (2) to obtain

S x = S + + S −

2 =

√ 2

0 1 01 0 10 1 0

and S y =

S + − S −2i

= i √

2

0 −1 01 0 −10 1 0

.


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4.32) Let’s start by writing out |χ+x ,χ+y , and |χ+z ; i.e., the spin up states in the x, y, and z bases. The

reason why this is important is because the probability of measuring + 2

for the component of spinangular momentum along the n̂ direction at time t is given by | χ+

n̂

χ(t) |2, where |χ(t) denotesthe spin state of interest given in Griffiths eqn (4.163). Fortunately, |χ+x , and |χ+z are given inGriffiths, and

χ+y was computed in problem 4.29, so I won’t belabor the calculations to obtainspinors for these states and will just list them below:

χ+x = 1√ 2

11

χ+y = 1√ 2

1i

χ+z = 10

.

(a)

| χ+x χ(t) |2 = 121 1

cos α

2eiγB0t/2

sin α2

e−iγB0t/2

2

= 1

2

cos α2

eiγB0t/2 + sin α

2e−iγB0t/2

2=

1

2 cos

α

2eiγB0t/2 + sin

α

2e−iγB0t/2

cos

α

2e−iγB0t/2 + sin

α

2eiγB0t/2

= 1

2

cos2 α

2 + cos α

2 sin α

2

eiγB0t/2 + e−iγB0t/2

+ sin2 α

2

=

1

2

1 + cos

α

2 sin

α

2 ∗ 2 cos(γB0t)

=

1

2 (1 + sin α cos(γB0t))

(b)

| χ+y

χ(t) |2 = 1

2

1 −i cos α2 eiγB0t/2sin α

2e−iγB0t/2

2

= 1

2cos α2 eiγB0t/2 − i sin α2 e−iγB0t/2

2

= 1

2

cos

α

2eiγB0t/2 − i sin α

2e−iγB0t/2

cos

α

2e−iγB0t/2 + i sin

α

2eiγB0t/2

=

1

2

cos2

α

2 + i cos

α

2 sin

α

2

eiγB0t/2 − e−iγB0t/2

+ sin2

α

2

=

1

2

1 + i cos

α

2 sin

α

2 ∗ 2i sin(γB0t)

=

1

2 (1 − sin α sin(γB0t))

(c)

| χ+z χ(t) |2 =1 0

cos α

2eiγB0t/2

sin α2

e−iγB0t/2

2

= cos2α

2

.

phys 481 homework 1 solutions

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