Today’s Outline - October 30, 2013 - IITphys.iit.edu/~segre/phys405/13F/lecture_18.pdf · 2013. 10. 31. · C. Segre (IIT) PHYS 405 - Fall 2013 October 30, 2013 1 / 17. Today’s

Today’s Outline - October 30, 2013

• Problems 4.29 and 4.52

• Electrons in a magnetic field

• Adding angular momentum

• Total spin of hydrogen


• Problem 4.55

Midterm Exam 2: Monday, November 11, 2013 covers through Chapter4.4.2 (at least!)

Homework Assignment #09:Chapter 4: 20,23,27,31,43,55due Wednesday, November 6, 2013

C. Segre (IIT) PHYS 405 - Fall 2013 October 30, 2013 1 / 17







• Problem 4.55










• Problem 4.55










• Problem 4.55










• Problem 4.55










• Problem 4.55










• Problem 4.55










• Problem 4.55




Problem 4.29

(a) Find the eigenvalues and eigenspinors of Sy

(b) If you measured Sy on a particle in the general state

χ = aχ+ + bχ−

what values might you get, and what is theprobability of each? Check that the probabilitiesadd up to 1.

(c) If you measured S2y , what values might you get,

and with what probabilities?


Problem 4.29a

Starting with the eigenvalue equation

Syχ =~2

(0 −ii 0

)= λχ

the eigenvectors are obtained by as-suming

χ =

(αβ

)0 = det

∣∣∣∣ −λ −i~/2i~/2 −λ

∣∣∣∣0 = λ2 −

(~2

)2

λ = ±~2

~2

(0 −ii 0

)(αβ

)= ±~

2

(αβ

)→

(−iβiα

)= ±

(αβ

)

β = ±iα → χ(y)+ =

1√2

(1i

), χ

(y)− =

1√2

(1−i

)


Problem 4.29a


Syχ =~2

(0 −ii 0

)= λχ


χ =

(αβ

)0 = det

∣∣∣∣ −λ −i~/2i~/2 −λ

∣∣∣∣0 = λ2 −

(~2

)2

λ = ±~2

~2

(0 −ii 0

)(αβ

)= ±~

2

(αβ

)→

(−iβiα

)= ±

(αβ

)

β = ±iα → χ(y)+ =

1√2

(1i

), χ

(y)− =

1√2

(1−i

)


Problem 4.29a


Syχ =~2

(0 −ii 0

)= λχ


χ =

(αβ

)0 = det

∣∣∣∣ −λ −i~/2i~/2 −λ

∣∣∣∣0 = λ2 −

(~2

)2

λ = ±~2

~2

(0 −ii 0

)(αβ

)= ±~

2

(αβ

)→

(−iβiα

)= ±

(αβ

)

β = ±iα → χ(y)+ =

1√2

(1i

), χ

(y)− =

1√2

(1−i

)


Problem 4.29a


Syχ =~2

(0 −ii 0

)= λχ


χ =

(αβ

)

0 = det

∣∣∣∣ −λ −i~/2i~/2 −λ

∣∣∣∣

0 = λ2 −(~2

)2

λ = ±~2

~2

(0 −ii 0

)(αβ

)= ±~

2

(αβ

)→

(−iβiα

)= ±

(αβ

)

β = ±iα → χ(y)+ =

1√2

(1i

), χ

(y)− =

1√2

(1−i

)


Problem 4.29a


Syχ =~2

(0 −ii 0

)= λχ


χ =

(αβ

)

0 = det

∣∣∣∣ −λ −i~/2i~/2 −λ

∣∣∣∣0 = λ2 −

(~2

)2

λ = ±~2

~2

(0 −ii 0

)(αβ

)= ±~

2

(αβ

)→

(−iβiα

)= ±

(αβ

)

β = ±iα → χ(y)+ =

1√2

(1i

), χ

(y)− =

1√2

(1−i

)


Problem 4.29a


Syχ =~2

(0 −ii 0

)= λχ


χ =

(αβ

)

0 = det

∣∣∣∣ −λ −i~/2i~/2 −λ

∣∣∣∣0 = λ2 −

(~2

)2

λ = ±~2

~2

(0 −ii 0

)(αβ

)= ±~

2

(αβ

)→

(−iβiα

)= ±

(αβ

)

β = ±iα → χ(y)+ =

1√2

(1i

), χ

(y)− =

1√2

(1−i

)


Problem 4.29a


Syχ =~2

(0 −ii 0

)= λχ


χ =

(αβ

)0 = det

∣∣∣∣ −λ −i~/2i~/2 −λ

∣∣∣∣0 = λ2 −

(~2

)2

λ = ±~2

~2

(0 −ii 0

)(αβ

)= ±~

2

(αβ

)→

(−iβiα

)= ±

(αβ

)

β = ±iα → χ(y)+ =

1√2

(1i

), χ

(y)− =

1√2

(1−i

)


Problem 4.29a


Syχ =~2

(0 −ii 0

)= λχ


χ =

(αβ

)0 = det

∣∣∣∣ −λ −i~/2i~/2 −λ

∣∣∣∣0 = λ2 −

(~2

)2

λ = ±~2

~2

(0 −ii 0

)(αβ

)= ±~

2

(αβ

)

→(−iβiα

)= ±

(αβ

)

β = ±iα → χ(y)+ =

1√2

(1i

), χ

(y)− =

1√2

(1−i

)


Problem 4.29a


Syχ =~2

(0 −ii 0

)= λχ


χ =

(αβ

)0 = det

∣∣∣∣ −λ −i~/2i~/2 −λ

∣∣∣∣0 = λ2 −

(~2

)2

λ = ±~2

~2

(0 −ii 0

)(αβ

)= ±~

2

(αβ

)→

(−iβiα

)= ±

(αβ

)

β = ±iα → χ(y)+ =

1√2

(1i

), χ

(y)− =

1√2

(1−i

)


Problem 4.29a


Syχ =~2

(0 −ii 0

)= λχ


χ =

(αβ

)0 = det

∣∣∣∣ −λ −i~/2i~/2 −λ

∣∣∣∣0 = λ2 −

(~2

)2

λ = ±~2

~2

(0 −ii 0

)(αβ

)= ±~

2

(αβ

)→

(−iβiα

)= ±

(αβ

)

β = ±iα

→ χ(y)+ =

1√2

(1i

), χ

(y)− =

1√2

(1−i

)


Problem 4.29a


Syχ =~2

(0 −ii 0

)= λχ


χ =

(αβ

)0 = det

∣∣∣∣ −λ −i~/2i~/2 −λ

∣∣∣∣0 = λ2 −

(~2

)2

λ = ±~2

~2

(0 −ii 0

)(αβ

)= ±~

2

(αβ

)→

(−iβiα

)= ±

(αβ

)

β = ±iα → χ(y)+ =

1√2

(1i

)

, χ(y)− =

1√2

(1−i

)


Problem 4.29a


Syχ =~2

(0 −ii 0

)= λχ


χ =

(αβ

)0 = det

∣∣∣∣ −λ −i~/2i~/2 −λ

∣∣∣∣0 = λ2 −

(~2

)2

λ = ±~2

~2

(0 −ii 0

)(αβ

)= ±~

2

(αβ

)→

(−iβiα

)= ±

(αβ

)

β = ±iα → χ(y)+ =

1√2

(1i

), χ

(y)− =

1√2

(1−i

)


Problem 4.29b

The general spinor can be written in terms of the Sy eigenspinors as wellas the usual ones

χ = aχ+ + bχ− = c+χ(y)+ + c−χ

(y)−

By measuring Sy , the possible answers are the magnitudes of thesecoefficients squared. The coefficients may be extracted using the Fouriertrick

c+ = χ(y)†+ χ

=1√2

( 1 −i )

(ab

)=

1√2

(a− ib)

c− = χ(y)†− χ

=1√2

( 1 i )

(ab

)=

1√2

(a + ib)

Thus we will measure the eigenvalues with probabilites as follows, whichsum to 1 ir χ is normalized.

+~2, P+ =

1

2|a− ib|2 − ~

2, P− =

1

2|a + ib|2


Problem 4.29b


χ = aχ+ + bχ− = c+χ(y)+ + c−χ

(y)−


c+ = χ(y)†+ χ

=1√2

( 1 −i )

(ab

)=

1√2

(a− ib)

c− = χ(y)†− χ

=1√2

( 1 i )

(ab

)=

1√2

(a + ib)


+~2, P+ =

1

2|a− ib|2 − ~

2, P− =

1

2|a + ib|2


Problem 4.29b


χ = aχ+ + bχ− = c+χ(y)+ + c−χ

(y)−


c+ = χ(y)†+ χ

=1√2

( 1 −i )

(ab

)=

1√2

(a− ib)

c− = χ(y)†− χ

=1√2

( 1 i )

(ab

)=

1√2

(a + ib)


+~2, P+ =

1

2|a− ib|2 − ~

2, P− =

1

2|a + ib|2


Problem 4.29b


χ = aχ+ + bχ− = c+χ(y)+ + c−χ

(y)−


c+ = χ(y)†+ χ

=1√2

( 1 −i )

(ab

)=

1√2

(a− ib)

c− = χ(y)†− χ

=1√2

( 1 i )

(ab

)=

1√2

(a + ib)


+~2, P+ =

1

2|a− ib|2 − ~

2, P− =

1

2|a + ib|2


Problem 4.29b


χ = aχ+ + bχ− = c+χ(y)+ + c−χ

(y)−


c+ = χ(y)†+ χ =

1√2

( 1 −i )

(ab

)

=1√2

(a− ib)

c− = χ(y)†− χ

=1√2

( 1 i )

(ab

)=

1√2

(a + ib)


+~2, P+ =

1

2|a− ib|2 − ~

2, P− =

1

2|a + ib|2


Problem 4.29b


χ = aχ+ + bχ− = c+χ(y)+ + c−χ

(y)−


c+ = χ(y)†+ χ =

1√2

( 1 −i )

(ab

)=

1√2

(a− ib)

c− = χ(y)†− χ

=1√2

( 1 i )

(ab

)=

1√2

(a + ib)


+~2, P+ =

1

2|a− ib|2 − ~

2, P− =

1

2|a + ib|2


Problem 4.29b


χ = aχ+ + bχ− = c+χ(y)+ + c−χ

(y)−


c+ = χ(y)†+ χ =

1√2

( 1 −i )

(ab

)=

1√2

(a− ib)

c− = χ(y)†− χ

=1√2

( 1 i )

(ab

)=

1√2

(a + ib)


+~2, P+ =

1

2|a− ib|2 − ~

2, P− =

1

2|a + ib|2


Problem 4.29b


χ = aχ+ + bχ− = c+χ(y)+ + c−χ

(y)−


c+ = χ(y)†+ χ =

1√2

( 1 −i )

(ab

)=

1√2

(a− ib)

c− = χ(y)†− χ =

1√2

( 1 i )

(ab

)

=1√2

(a + ib)


+~2, P+ =

1

2|a− ib|2 − ~

2, P− =

1

2|a + ib|2


Problem 4.29b


χ = aχ+ + bχ− = c+χ(y)+ + c−χ

(y)−


c+ = χ(y)†+ χ =

1√2

( 1 −i )

(ab

)=

1√2

(a− ib)

c− = χ(y)†− χ =

1√2

( 1 i )

(ab

)=

1√2

(a + ib)


+~2, P+ =

1

2|a− ib|2 − ~

2, P− =

1

2|a + ib|2


Problem 4.29b


χ = aχ+ + bχ− = c+χ(y)+ + c−χ

(y)−


c+ = χ(y)†+ χ =

1√2

( 1 −i )

(ab

)=

1√2

(a− ib)

c− = χ(y)†− χ =

1√2

( 1 i )

(ab

)=

1√2

(a + ib)


+~2, P+ =

1

2|a− ib|2 − ~

2, P− =

1

2|a + ib|2


Problem 4.29b


χ = aχ+ + bχ− = c+χ(y)+ + c−χ

(y)−


c+ = χ(y)†+ χ =

1√2

( 1 −i )

(ab

)=

1√2

(a− ib)

c− = χ(y)†− χ =

1√2

( 1 i )

(ab

)=

1√2

(a + ib)


+~2, P+ =

1

2|a− ib|2

− ~2, P− =

1

2|a + ib|2


Problem 4.29b


χ = aχ+ + bχ− = c+χ(y)+ + c−χ

(y)−


c+ = χ(y)†+ χ =

1√2

( 1 −i )

(ab

)=

1√2

(a− ib)

c− = χ(y)†− χ =

1√2

( 1 i )

(ab

)=

1√2

(a + ib)


+~2, P+ =

1

2|a− ib|2 − ~

2, P− =

1

2|a + ib|2


Problem 4.29c

Measuring S2y is simply a question of applying the

same operator twice, thus pulling out the same eigen-value twice.

χ†SySyχ = χ†Sy (+~2c+χ

(y)+ −

~2c−χ

(y)− )

= χ†(~2

4c+χ

(y)+ +

~2

4c−χ

(y)− )

= χ†~2

4χ =

~2

4

with a probability of 1


Problem 4.29c




(y)+ −

~2c−χ

(y)− )

= χ†(~2

4c+χ

(y)+ +

~2

4c−χ

(y)− )

= χ†~2

4χ =

~2

4



Problem 4.29c




(y)+ −

~2c−χ

(y)− )

= χ†(~2

4c+χ

(y)+ +

~2

4c−χ

(y)− )

= χ†~2

4χ =

~2

4



Problem 4.29c




(y)+ −

~2c−χ

(y)− )

= χ†(~2

4c+χ

(y)+ +

~2

4c−χ

(y)− )

= χ†~2

4χ

=~2

4



Problem 4.29c




(y)+ −

~2c−χ

(y)− )

= χ†(~2

4c+χ

(y)+ +

~2

4c−χ

(y)− )

= χ†~2

4χ =

~2

4



Problem 4.29c




(y)+ −

~2c−χ

(y)− )

= χ†(~2

4c+χ

(y)+ +

~2

4c−χ

(y)− )

= χ†~2

4χ =

~2

4



Problem 4.52

Find the matrix representing Sx for a particle of spin 3/2 (using, as always,the basis of eigenstates of Sz . Solve the characteristic equation todetermine the eigenvalues of Sx .

First we write down the eigenstates of Sz in the S = 3/2 system.

| 32

32〉 =

1000

| 32

12〉 =

0100

| 32− 1

2〉 =

0010

| 32− 3

2〉 =

0001


Problem 4.52



| 32

32〉 =

1000

| 32

12〉 =

0100

| 32− 1

2〉 =

0010

| 32− 3

2〉 =

0001


Problem 4.52



| 32

32〉 =

1000

| 32

12〉 =

0100

| 32− 1

2〉 =

0010

| 32− 3

2〉 =

0001


Problem 4.52



| 32

32〉 =

1000

| 32

12〉 =

0100

| 32− 1

2〉 =

0010

| 32− 3

2〉 =

0001


Problem 4.52



| 32

32〉 =

1000

| 32

12〉 =

0100

| 32− 1

2〉 =

0010

| 32− 3

2〉 =

0001


Problem 4.52



| 32

32〉 =

1000

| 32

12〉 =

0100

| 32− 1

2〉 =

0010

| 32− 3

2〉 =

0001


Problem 4.52 (cont.)

The raising and lowering operators are

S±|sm〉 = ~√

s(s + 1)−m(m ± 1) |s(m ± 1)〉

we build the matrix for S+ by applying the raising operator to each of theeigenstates

S+ = ~

0√

3 0 00 0 2 0

0 0 0√

30 0 0 0

S+| 32 3

2〉 = 0

S+| 32 12〉 =

√32( 5

2)− 12( 3

2)~ | 32 32〉 =√

3~ | 32

32〉

S+| 32 − 12〉 =

√32( 5

2)+ 12( 1

2)~ | 32 12〉 = 2~ | 3

212〉

S+| 32 − 32〉 =

√32( 5

2)+ 32(− 1

2)~ | 32 − 12〉 =√

3~ | 32− 1

2〉




S±|sm〉 = ~√

s(s + 1)−m(m ± 1) |s(m ± 1)〉


S+ = ~

0√

3 0 00 0 2 0

0 0 0√

30 0 0 0

S+| 32 3

2〉 = 0

S+| 32 12〉 =

√32( 5

2)− 12( 3

2)~ | 32 32〉 =√

3~ | 32

32〉

S+| 32 − 12〉 =

√32( 5

2)+ 12( 1

2)~ | 32 12〉 = 2~ | 3

212〉

S+| 32 − 32〉 =

√32( 5

2)+ 32(− 1

2)~ | 32 − 12〉 =√

3~ | 32− 1

2〉




S±|sm〉 = ~√

s(s + 1)−m(m ± 1) |s(m ± 1)〉


S+ = ~

0√

3 0 00 0 2 0

0 0 0√

30 0 0 0

S+| 32 32〉 = 0

S+| 32 12〉 =

√32( 5

2)− 12( 3

2)~ | 32 32〉 =√

3~ | 32

32〉

S+| 32 − 12〉 =

√32( 5

2)+ 12( 1

2)~ | 32 12〉 = 2~ | 3

212〉

S+| 32 − 32〉 =

√32( 5

2)+ 32(− 1

2)~ | 32 − 12〉 =√

3~ | 32− 1

2〉




S±|sm〉 = ~√

s(s + 1)−m(m ± 1) |s(m ± 1)〉


S+ = ~

0√

3 0 00 0 2 0

0 0 0√

30 0 0 0

S+| 32 3

2〉 = 0

S+| 32 12〉 =

√32( 5

2)− 12( 3

2)~ | 32 32〉 =√

3~ | 32

32〉

S+| 32 − 12〉 =

√32( 5

2)+ 12( 1

2)~ | 32 12〉 = 2~ | 3

212〉

S+| 32 − 32〉 =

√32( 5

2)+ 32(− 1

2)~ | 32 − 12〉 =√

3~ | 32− 1

2〉




S±|sm〉 = ~√

s(s + 1)−m(m ± 1) |s(m ± 1)〉


S+ = ~

0

√3 0 0

0

0 2 0

0

0 0√

3

0

0 0 0

S+| 32 3

2〉 = 0

S+| 32 12〉 =

√32( 5

2)− 12( 3

2)~ | 32 32〉 =√

3~ | 32

32〉

S+| 32 − 12〉 =

√32( 5

2)+ 12( 1

2)~ | 32 12〉 = 2~ | 3

212〉

S+| 32 − 32〉 =

√32( 5

2)+ 32(− 1

2)~ | 32 − 12〉 =√

3~ | 32− 1

2〉




S±|sm〉 = ~√

s(s + 1)−m(m ± 1) |s(m ± 1)〉


S+ = ~

0

√3 0 0

0

0 2 0

0

0 0√

3

0

0 0 0

S+| 32 3

2〉 = 0

S+| 32 12〉 =

√32( 5

2)− 12( 3

2)~ | 32 32〉

=√

3~ | 32

32〉

S+| 32 − 12〉 =

√32( 5

2)+ 12( 1

2)~ | 32 12〉 = 2~ | 3

212〉

S+| 32 − 32〉 =

√32( 5

2)+ 32(− 1

2)~ | 32 − 12〉 =√

3~ | 32− 1

2〉




S±|sm〉 = ~√

s(s + 1)−m(m ± 1) |s(m ± 1)〉


S+ = ~

0

√3 0 0

0

0 2 0

0

0 0√

3

0

0 0 0

S+| 32 3

2〉 = 0

S+| 32 12〉 =

√32( 5

2)− 12( 3

2)~ | 32 32〉 =√

3~ | 32

32〉

S+| 32 − 12〉 =

√32( 5

2)+ 12( 1

2)~ | 32 12〉 = 2~ | 3

212〉

S+| 32 − 32〉 =

√32( 5

2)+ 32(− 1

2)~ | 32 − 12〉 =√

3~ | 32− 1

2〉




S±|sm〉 = ~√

s(s + 1)−m(m ± 1) |s(m ± 1)〉


S+ = ~

0√

3

0 0

0 0

2 0

0 0

0√

3

0 0

0 0

S+| 32 3

2〉 = 0

S+| 32 12〉 =

√32( 5

2)− 12( 3

2)~ | 32 32〉 =√

3~ | 32

32〉

S+| 32 − 12〉 =

√32( 5

2)+ 12( 1

2)~ | 32 12〉 = 2~ | 3

212〉

S+| 32 − 32〉 =

√32( 5

2)+ 32(− 1

2)~ | 32 − 12〉 =√

3~ | 32− 1

2〉




S±|sm〉 = ~√

s(s + 1)−m(m ± 1) |s(m ± 1)〉


S+ = ~

0√

3

0 0

0 0

2 0

0 0

0√

3

0 0

0 0

S+| 32 3

2〉 = 0

S+| 32 12〉 =

√32( 5

2)− 12( 3

2)~ | 32 32〉 =√

3~ | 32

32〉

S+| 32 − 12〉 =

√32( 5

2)+ 12( 1

2)~ | 32 12〉

= 2~ | 32

12〉

S+| 32 − 32〉 =

√32( 5

2)+ 32(− 1

2)~ | 32 − 12〉 =√

3~ | 32− 1

2〉




S±|sm〉 = ~√

s(s + 1)−m(m ± 1) |s(m ± 1)〉


S+ = ~

0√

3

0 0

0 0

2 0

0 0

0√

3

0 0

0 0

S+| 32 3

2〉 = 0

S+| 32 12〉 =

√32( 5

2)− 12( 3

2)~ | 32 32〉 =√

3~ | 32

32〉

S+| 32 − 12〉 =

√32( 5

2)+ 12( 1

2)~ | 32 12〉 = 2~ | 3

212〉

S+| 32 − 32〉 =

√32( 5

2)+ 32(− 1

2)~ | 32 − 12〉 =√

3~ | 32− 1

2〉




S±|sm〉 = ~√

s(s + 1)−m(m ± 1) |s(m ± 1)〉


S+ = ~

0√

3 0

0

0 0 2

0

0 0 0

√3

0 0 0

0

S+| 32 3

2〉 = 0

S+| 32 12〉 =

√32( 5

2)− 12( 3

2)~ | 32 32〉 =√

3~ | 32

32〉

S+| 32 − 12〉 =

√32( 5

2)+ 12( 1

2)~ | 32 12〉 = 2~ | 3

212〉

S+| 32 − 32〉 =

√32( 5

2)+ 32(− 1

2)~ | 32 − 12〉 =√

3~ | 32− 1

2〉




S±|sm〉 = ~√

s(s + 1)−m(m ± 1) |s(m ± 1)〉


S+ = ~

0√

3 0

0

0 0 2

0

0 0 0

√3

0 0 0

0

S+| 32 3

2〉 = 0

S+| 32 12〉 =

√32( 5

2)− 12( 3

2)~ | 32 32〉 =√

3~ | 32

32〉

S+| 32 − 12〉 =

√32( 5

2)+ 12( 1

2)~ | 32 12〉 = 2~ | 3

212〉

S+| 32 − 32〉 =

√32( 5

2)+ 32(− 1

2)~ | 32 − 12〉

=√

3~ | 32− 1

2〉




S±|sm〉 = ~√

s(s + 1)−m(m ± 1) |s(m ± 1)〉


S+ = ~

0√

3 0

0

0 0 2

0

0 0 0

√3

0 0 0

0

S+| 32 3

2〉 = 0

S+| 32 12〉 =

√32( 5

2)− 12( 3

2)~ | 32 32〉 =√

3~ | 32

32〉

S+| 32 − 12〉 =

√32( 5

2)+ 12( 1

2)~ | 32 12〉 = 2~ | 3

212〉

S+| 32 − 32〉 =

√32( 5

2)+ 32(− 1

2)~ | 32 − 12〉 =√

3~ | 32− 1

2〉




S±|sm〉 = ~√

s(s + 1)−m(m ± 1) |s(m ± 1)〉


S+ = ~

0√

3 0 00 0 2 0

0 0 0√

30 0 0 0

S+| 32 3

2〉 = 0

S+| 32 12〉 =

√32( 5

2)− 12( 3

2)~ | 32 32〉 =√

3~ | 32

32〉

S+| 32 − 12〉 =

√32( 5

2)+ 12( 1

2)~ | 32 12〉 = 2~ | 3

212〉

S+| 32 − 32〉 =

√32( 5

2)+ 32(− 1

2)~ | 32 − 12〉 =√

3~ | 32− 1

2〉



Similarly, we build the matrix for S− by applying the lowering operator toeach of the eigenstates

S− = ~

0 0 0 0√3 0 0 0

0 2 0 0

0 0√

3 0

S−| 32 32〉 =

√32( 5

2)− 32( 1

2)~ | 32 12〉 =√

3~ | 32

12〉

S−| 32 12〉 =

√32( 5

2)− 12(− 1

2)~ | 32 −12〉 = 2~ | 3

2− 1

2〉

S−| 32 − 12〉 =

√32( 5

2)+ 12(− 3

2)~ | 32 −32〉 =√

3~ | 32− 3

2〉

S−| 32 − 32〉 = 0




S− = ~

0 0 0 0√3 0 0 0

0 2 0 0

0 0√

3 0

S−| 32 32〉 =

√32( 5

2)− 32( 1

2)~ | 32 12〉 =√

3~ | 32

12〉

S−| 32 12〉 =

√32( 5

2)− 12(− 1

2)~ | 32 −12〉 = 2~ | 3

2− 1

2〉

S−| 32 − 12〉 =

√32( 5

2)+ 12(− 3

2)~ | 32 −32〉 =√

3~ | 32− 3

2〉

S−| 32 − 32〉 = 0




S− = ~

0 0 0 0√3 0 0 0

0 2 0 0

0 0√

3 0

S−| 32 32〉 =

√32( 5

2)− 32( 1

2)~ | 32 12〉

=√

3~ | 32

12〉

S−| 32 12〉 =

√32( 5

2)− 12(− 1

2)~ | 32 −12〉 = 2~ | 3

2− 1

2〉

S−| 32 − 12〉 =

√32( 5

2)+ 12(− 3

2)~ | 32 −32〉 =√

3~ | 32− 3

2〉

S−| 32 − 32〉 = 0




S− = ~

0 0 0 0√3 0 0 0

0 2 0 0

0 0√

3 0

S−| 32 32〉 =

√32( 5

2)− 32( 1

2)~ | 32 12〉 =√

3~ | 32

12〉

S−| 32 12〉 =

√32( 5

2)− 12(− 1

2)~ | 32 −12〉 = 2~ | 3

2− 1

2〉

S−| 32 − 12〉 =

√32( 5

2)+ 12(− 3

2)~ | 32 −32〉 =√

3~ | 32− 3

2〉

S−| 32 − 32〉 = 0




S− = ~

0

0 0 0

√3

0 0 0

0

2 0 0

0

0√

3 0

S−| 32 32〉 =

√32( 5

2)− 32( 1

2)~ | 32 12〉 =√

3~ | 32

12〉

S−| 32 12〉 =

√32( 5

2)− 12(− 1

2)~ | 32 −12〉 = 2~ | 3

2− 1

2〉

S−| 32 − 12〉 =

√32( 5

2)+ 12(− 3

2)~ | 32 −32〉 =√

3~ | 32− 3

2〉

S−| 32 − 32〉 = 0




S− = ~

0

0 0 0

√3

0 0 0

0

2 0 0

0

0√

3 0

S−| 32 32〉 =

√32( 5

2)− 32( 1

2)~ | 32 12〉 =√

3~ | 32

12〉

S−| 32 12〉 =

√32( 5

2)− 12(− 1

2)~ | 32 −12〉

= 2~ | 32− 1

2〉

S−| 32 − 12〉 =

√32( 5

2)+ 12(− 3

2)~ | 32 −32〉 =√

3~ | 32− 3

2〉

S−| 32 − 32〉 = 0




S− = ~

0

0 0 0

√3

0 0 0

0

2 0 0

0

0√

3 0

S−| 32 32〉 =

√32( 5

2)− 32( 1

2)~ | 32 12〉 =√

3~ | 32

12〉

S−| 32 12〉 =

√32( 5

2)− 12(− 1

2)~ | 32 −12〉 = 2~ | 3

2− 1

2〉

S−| 32 − 12〉 =

√32( 5

2)+ 12(− 3

2)~ | 32 −32〉 =√

3~ | 32− 3

2〉

S−| 32 − 32〉 = 0




S− = ~

0 0

0 0

√3 0

0 0

0 2

0 0

0 0

√3 0

S−| 32 32〉 =

√32( 5

2)− 32( 1

2)~ | 32 12〉 =√

3~ | 32

12〉

S−| 32 12〉 =

√32( 5

2)− 12(− 1

2)~ | 32 −12〉 = 2~ | 3

2− 1

2〉

S−| 32 − 12〉 =

√32( 5

2)+ 12(− 3

2)~ | 32 −32〉 =√

3~ | 32− 3

2〉

S−| 32 − 32〉 = 0




S− = ~

0 0

0 0

√3 0

0 0

0 2

0 0

0 0

√3 0

S−| 32 32〉 =

√32( 5

2)− 32( 1

2)~ | 32 12〉 =√

3~ | 32

12〉

S−| 32 12〉 =

√32( 5

2)− 12(− 1

2)~ | 32 −12〉 = 2~ | 3

2− 1

2〉

S−| 32 − 12〉 =

√32( 5

2)+ 12(− 3

2)~ | 32 −32〉

=√

3~ | 32− 3

2〉

S−| 32 − 32〉 = 0




S− = ~

0 0

0 0

√3 0

0 0

0 2

0 0

0 0

√3 0

S−| 32 32〉 =

√32( 5

2)− 32( 1

2)~ | 32 12〉 =√

3~ | 32

12〉

S−| 32 12〉 =

√32( 5

2)− 12(− 1

2)~ | 32 −12〉 = 2~ | 3

2− 1

2〉

S−| 32 − 12〉 =

√32( 5

2)+ 12(− 3

2)~ | 32 −32〉 =√

3~ | 32− 3

2〉

S−| 32 − 32〉 = 0




S− = ~

0 0 0

0

√3 0 0

0

0 2 0

0

0 0√

3

0

S−| 32 32〉 =

√32( 5

2)− 32( 1

2)~ | 32 12〉 =√

3~ | 32

12〉

S−| 32 12〉 =

√32( 5

2)− 12(− 1

2)~ | 32 −12〉 = 2~ | 3

2− 1

2〉

S−| 32 − 12〉 =

√32( 5

2)+ 12(− 3

2)~ | 32 −32〉 =√

3~ | 32− 3

2〉

S−| 32 − 32〉 = 0




S− = ~

0 0 0

0

√3 0 0

0

0 2 0

0

0 0√

3

0

S−| 32 32〉 =

√32( 5

2)− 32( 1

2)~ | 32 12〉 =√

3~ | 32

12〉

S−| 32 12〉 =

√32( 5

2)− 12(− 1

2)~ | 32 −12〉 = 2~ | 3

2− 1

2〉

S−| 32 − 12〉 =

√32( 5

2)+ 12(− 3

2)~ | 32 −32〉 =√

3~ | 32− 3

2〉

S−| 32 − 32〉 = 0




S− = ~

0 0 0 0√3 0 0 0

0 2 0 0

0 0√

3 0

S−| 32 32〉 =

√32( 5

2)− 32( 1

2)~ | 32 12〉 =√

3~ | 32

12〉

S−| 32 12〉 =

√32( 5

2)− 12(− 1

2)~ | 32 −12〉 = 2~ | 3

2− 1

2〉

S−| 32 − 12〉 =

√32( 5

2)+ 12(− 3

2)~ | 32 −32〉 =√

3~ | 32− 3

2〉

S−| 32 − 32〉 = 0



Sx =1

2(S+ + S−)

=~2

0√

3 0 0√3 0 2 0

0 2 0√

3

0 0√

3 0

The characteristic equation for eigenvalues is (Note that λ is really theeigenvalue divided by ~/2 for ease of notation)

0 =

∣∣∣∣∣∣∣∣− λ

√3 0 0√

3 −λ 2 0

0 2 −λ√

3

0 0√

3 −λ

∣∣∣∣∣∣∣∣ = −λ

∣∣∣∣∣∣−λ 2 0

2 −λ√

3

0√

3 −λ

∣∣∣∣∣∣−√3

∣∣∣∣∣∣√

3 2 0

0 −λ√

3

0√

3 −λ

∣∣∣∣∣∣= −λ[−λ(λ2 − 3)− 2(−2λ)]−

√3[√

3(λ2 − 3)]

= −λ[−λ3 + 7λ]− [3λ2 − 9] = λ4 − 10λ2 + 9

0 = λ4 − 10λ2 + 9 = (λ2 − 9)(λ2 − 1) = (λ+ 3)(λ− 3)(λ+ 1)(λ− 1)

the eigenvalues of Sx are : 32~, 1

2~,− 1

2~,− 3

2~



Sx =1

2(S+ + S−) =

~2

0√

3 0 0√3 0 2 0

0 2 0√

3

0 0√

3 0


0 =

∣∣∣∣∣∣∣∣− λ

√3 0 0√

3 −λ 2 0

0 2 −λ√

3

0 0√

3 −λ

∣∣∣∣∣∣∣∣ = −λ

∣∣∣∣∣∣−λ 2 0

2 −λ√

3

0√

3 −λ

∣∣∣∣∣∣−√3

∣∣∣∣∣∣√

3 2 0

0 −λ√

3

0√

3 −λ

∣∣∣∣∣∣= −λ[−λ(λ2 − 3)− 2(−2λ)]−

√3[√

3(λ2 − 3)]

= −λ[−λ3 + 7λ]− [3λ2 − 9] = λ4 − 10λ2 + 9

0 = λ4 − 10λ2 + 9 = (λ2 − 9)(λ2 − 1) = (λ+ 3)(λ− 3)(λ+ 1)(λ− 1)


2~,− 1

2~,− 3

2~



Sx =1

2(S+ + S−) =

~2

0√

3 0 0√3 0 2 0

0 2 0√

3

0 0√

3 0


0 =

∣∣∣∣∣∣∣∣− λ

√3 0 0√

3 −λ 2 0

0 2 −λ√

3

0 0√

3 −λ

∣∣∣∣∣∣∣∣ = −λ

∣∣∣∣∣∣−λ 2 0

2 −λ√

3

0√

3 −λ

∣∣∣∣∣∣−√3

∣∣∣∣∣∣√

3 2 0

0 −λ√

3

0√

3 −λ

∣∣∣∣∣∣= −λ[−λ(λ2 − 3)− 2(−2λ)]−

√3[√

3(λ2 − 3)]

= −λ[−λ3 + 7λ]− [3λ2 − 9] = λ4 − 10λ2 + 9

0 = λ4 − 10λ2 + 9 = (λ2 − 9)(λ2 − 1) = (λ+ 3)(λ− 3)(λ+ 1)(λ− 1)


2~,− 1

2~,− 3

2~



Sx =1

2(S+ + S−) =

~2

0√

3 0 0√3 0 2 0

0 2 0√

3

0 0√

3 0


0 =

∣∣∣∣∣∣∣∣− λ

√3 0 0√

3 −λ 2 0

0 2 −λ√

3

0 0√

3 −λ

∣∣∣∣∣∣∣∣

= −λ

∣∣∣∣∣∣−λ 2 0

2 −λ√

3

0√

3 −λ

∣∣∣∣∣∣−√3

∣∣∣∣∣∣√

3 2 0

0 −λ√

3

0√

3 −λ

∣∣∣∣∣∣= −λ[−λ(λ2 − 3)− 2(−2λ)]−

√3[√

3(λ2 − 3)]

= −λ[−λ3 + 7λ]− [3λ2 − 9] = λ4 − 10λ2 + 9

0 = λ4 − 10λ2 + 9 = (λ2 − 9)(λ2 − 1) = (λ+ 3)(λ− 3)(λ+ 1)(λ− 1)


2~,− 1

2~,− 3

2~



Sx =1

2(S+ + S−) =

~2

0√

3 0 0√3 0 2 0

0 2 0√

3

0 0√

3 0


0 =

∣∣∣∣∣∣∣∣− λ

√3 0 0√

3 −λ 2 0

0 2 −λ√

3

0 0√

3 −λ

∣∣∣∣∣∣∣∣ = −λ

∣∣∣∣∣∣−λ 2 0

2 −λ√

3

0√

3 −λ

∣∣∣∣∣∣

−√

3

∣∣∣∣∣∣√

3 2 0

0 −λ√

3

0√

3 −λ

∣∣∣∣∣∣= −λ[−λ(λ2 − 3)− 2(−2λ)]−

√3[√

3(λ2 − 3)]

= −λ[−λ3 + 7λ]− [3λ2 − 9] = λ4 − 10λ2 + 9

0 = λ4 − 10λ2 + 9 = (λ2 − 9)(λ2 − 1) = (λ+ 3)(λ− 3)(λ+ 1)(λ− 1)


2~,− 1

2~,− 3

2~



Sx =1

2(S+ + S−) =

~2

0√

3 0 0√3 0 2 0

0 2 0√

3

0 0√

3 0


0 =

∣∣∣∣∣∣∣∣−λ

√3 0 0√

3 −λ 2 0

0 2 −λ√

3

0 0√

3 −λ

∣∣∣∣∣∣∣∣ = −λ

∣∣∣∣∣∣−λ 2 0

2 −λ√

3

0√

3 −λ

∣∣∣∣∣∣−√3

∣∣∣∣∣∣√

3 2 0

0 −λ√

3

0√

3 −λ

∣∣∣∣∣∣

= −λ[−λ(λ2 − 3)− 2(−2λ)]−√

3[√

3(λ2 − 3)]

= −λ[−λ3 + 7λ]− [3λ2 − 9] = λ4 − 10λ2 + 9

0 = λ4 − 10λ2 + 9 = (λ2 − 9)(λ2 − 1) = (λ+ 3)(λ− 3)(λ+ 1)(λ− 1)


2~,− 1

2~,− 3

2~



Sx =1

2(S+ + S−) =

~2

0√

3 0 0√3 0 2 0

0 2 0√

3

0 0√

3 0


0 =

∣∣∣∣∣∣∣∣−λ

√3 0 0√

3 −λ 2 0

0 2 −λ√

3

0 0√

3 −λ

∣∣∣∣∣∣∣∣ = −λ

∣∣∣∣∣∣−λ 2 0

2 −λ√

3

0√

3 −λ

∣∣∣∣∣∣−√3

∣∣∣∣∣∣√

3 2 0

0 −λ√

3

0√

3 −λ

∣∣∣∣∣∣= −λ[−λ(λ2 − 3)− 2(−2λ)]−

√3[√

3(λ2 − 3)]

= −λ[−λ3 + 7λ]− [3λ2 − 9] = λ4 − 10λ2 + 9

0 = λ4 − 10λ2 + 9 = (λ2 − 9)(λ2 − 1) = (λ+ 3)(λ− 3)(λ+ 1)(λ− 1)


2~,− 1

2~,− 3

2~



Sx =1

2(S+ + S−) =

~2

0√

3 0 0√3 0 2 0

0 2 0√

3

0 0√

3 0


0 =

∣∣∣∣∣∣∣∣−λ

√3 0 0√

3 −λ 2 0

0 2 −λ√

3

0 0√

3 −λ

∣∣∣∣∣∣∣∣ = −λ

∣∣∣∣∣∣−λ 2 0

2 −λ√

3

0√

3 −λ

∣∣∣∣∣∣−√3

∣∣∣∣∣∣√

3 2 0

0 −λ√

3

0√

3 −λ

∣∣∣∣∣∣= −λ[−λ(λ2 − 3)− 2(−2λ)]−

√3[√

3(λ2 − 3)]

= −λ[−λ3 + 7λ]− [3λ2 − 9]

= λ4 − 10λ2 + 9

0 = λ4 − 10λ2 + 9 = (λ2 − 9)(λ2 − 1) = (λ+ 3)(λ− 3)(λ+ 1)(λ− 1)


2~,− 1

2~,− 3

2~



Sx =1

2(S+ + S−) =

~2

0√

3 0 0√3 0 2 0

0 2 0√

3

0 0√

3 0


0 =

∣∣∣∣∣∣∣∣−λ

√3 0 0√

3 −λ 2 0

0 2 −λ√

3

0 0√

3 −λ

∣∣∣∣∣∣∣∣ = −λ

∣∣∣∣∣∣−λ 2 0

2 −λ√

3

0√

3 −λ

∣∣∣∣∣∣−√3

∣∣∣∣∣∣√

3 2 0

0 −λ√

3

0√

3 −λ

∣∣∣∣∣∣= −λ[−λ(λ2 − 3)− 2(−2λ)]−

√3[√

3(λ2 − 3)]

= −λ[−λ3 + 7λ]− [3λ2 − 9] = λ4 − 10λ2 + 9

0 = λ4 − 10λ2 + 9 = (λ2 − 9)(λ2 − 1) = (λ+ 3)(λ− 3)(λ+ 1)(λ− 1)


2~,− 1

2~,− 3

2~



Sx =1

2(S+ + S−) =

~2

0√

3 0 0√3 0 2 0

0 2 0√

3

0 0√

3 0


0 =

∣∣∣∣∣∣∣∣−λ

√3 0 0√

3 −λ 2 0

0 2 −λ√

3

0 0√

3 −λ

∣∣∣∣∣∣∣∣ = −λ

∣∣∣∣∣∣−λ 2 0

2 −λ√

3

0√

3 −λ

∣∣∣∣∣∣−√3

∣∣∣∣∣∣√

3 2 0

0 −λ√

3

0√

3 −λ

∣∣∣∣∣∣= −λ[−λ(λ2 − 3)− 2(−2λ)]−

√3[√

3(λ2 − 3)]

= −λ[−λ3 + 7λ]− [3λ2 − 9] = λ4 − 10λ2 + 9

0 = λ4 − 10λ2 + 9

= (λ2 − 9)(λ2 − 1) = (λ+ 3)(λ− 3)(λ+ 1)(λ− 1)


2~,− 1

2~,− 3

2~



Sx =1

2(S+ + S−) =

~2

0√

3 0 0√3 0 2 0

0 2 0√

3

0 0√

3 0


0 =

∣∣∣∣∣∣∣∣−λ

√3 0 0√

3 −λ 2 0

0 2 −λ√

3

0 0√

3 −λ

∣∣∣∣∣∣∣∣ = −λ

∣∣∣∣∣∣−λ 2 0

2 −λ√

3

0√

3 −λ

∣∣∣∣∣∣−√3

∣∣∣∣∣∣√

3 2 0

0 −λ√

3

0√

3 −λ

∣∣∣∣∣∣= −λ[−λ(λ2 − 3)− 2(−2λ)]−

√3[√

3(λ2 − 3)]

= −λ[−λ3 + 7λ]− [3λ2 − 9] = λ4 − 10λ2 + 9

0 = λ4 − 10λ2 + 9 = (λ2 − 9)(λ2 − 1)

= (λ+ 3)(λ− 3)(λ+ 1)(λ− 1)


2~,− 1

2~,− 3

2~



Sx =1

2(S+ + S−) =

~2

0√

3 0 0√3 0 2 0

0 2 0√

3

0 0√

3 0


0 =

∣∣∣∣∣∣∣∣−λ

√3 0 0√

3 −λ 2 0

0 2 −λ√

3

0 0√

3 −λ

∣∣∣∣∣∣∣∣ = −λ

∣∣∣∣∣∣−λ 2 0

2 −λ√

3

0√

3 −λ

∣∣∣∣∣∣−√3

∣∣∣∣∣∣√

3 2 0

0 −λ√

3

0√

3 −λ

∣∣∣∣∣∣= −λ[−λ(λ2 − 3)− 2(−2λ)]−

√3[√

3(λ2 − 3)]

= −λ[−λ3 + 7λ]− [3λ2 − 9] = λ4 − 10λ2 + 9

0 = λ4 − 10λ2 + 9 = (λ2 − 9)(λ2 − 1) = (λ+ 3)(λ− 3)(λ+ 1)(λ− 1)


2~,− 1

2~,− 3

2~



Sx =1

2(S+ + S−) =

~2

0√

3 0 0√3 0 2 0

0 2 0√

3

0 0√

3 0


0 =

∣∣∣∣∣∣∣∣−λ

√3 0 0√

3 −λ 2 0

0 2 −λ√

3

0 0√

3 −λ

∣∣∣∣∣∣∣∣ = −λ

∣∣∣∣∣∣−λ 2 0

2 −λ√

3

0√

3 −λ

∣∣∣∣∣∣−√3

∣∣∣∣∣∣√

3 2 0

0 −λ√

3

0√

3 −λ

∣∣∣∣∣∣= −λ[−λ(λ2 − 3)− 2(−2λ)]−

√3[√

3(λ2 − 3)]

= −λ[−λ3 + 7λ]− [3λ2 − 9] = λ4 − 10λ2 + 9

0 = λ4 − 10λ2 + 9 = (λ2 − 9)(λ2 − 1) = (λ+ 3)(λ− 3)(λ+ 1)(λ− 1)


2~,− 1

2~,− 3

2~


Electron in a magnetic field

Orbital angular momentum implies a circulating charge which leads to amagnetic moment

because spin behaves in all ways like orbital angular momentum, it mustalso be associated with a magnetic dipole moment

~µ = γ~S

in the presence of a magnetic field,the dipole feels a torque

and has total energy

the Hamiltonian for a particle withspin in the presence of a magneticfield, is thus

γ is the gyromagnetic ratio

~τ = ~µ× ~B

H = −~µ · ~B

H = −γ~B · ~S





~µ = γ~S





~τ = ~µ× ~B

H = −~µ · ~B

H = −γ~B · ~S





~µ = γ~S





~τ = ~µ× ~B

H = −~µ · ~B

H = −γ~B · ~S





~µ = γ~S





~τ = ~µ× ~B

H = −~µ · ~B

H = −γ~B · ~S





~µ = γ~S





~τ = ~µ× ~B

H = −~µ · ~B

H = −γ~B · ~S





~µ = γ~S





~τ = ~µ× ~B

H = −~µ · ~B

H = −γ~B · ~S





~µ = γ~S





~τ = ~µ× ~B

H = −~µ · ~B

H = −γ~B · ~S





~µ = γ~S





~τ = ~µ× ~B

H = −~µ · ~B

H = −γ~B · ~S





~µ = γ~S





~τ = ~µ× ~B

H = −~µ · ~B

H = −γ~B · ~S





~µ = γ~S





~τ = ~µ× ~B

H = −~µ · ~B

H = −γ~B · ~S


Dipole at rest in a magnetic field

consider a particle of spin 1/2 atrest in the presence of a magneticfield in the z-direction

the Hamiltonian is

and its eigenstates are those of Szwith eigenvalues E±

the time dependent Schrodingerequation can be solved in terms ofthe stationary states

~B = B0z

H = −γB0Sz = −γB0~2

(1 00 −1

){χ+, E+ = −γB0~

2

χ−, E− = +γB0~2

i~∂χ

∂t= Hχ

χ(t) = aχ+e−E+t/~ + bχ−e

−E−t/~ =

(ae iγB0t/2

be−iγB0t/2

)




the Hamiltonian is



~B = B0z

H = −γB0Sz = −γB0~2

(1 00 −1

){χ+, E+ = −γB0~

2

χ−, E− = +γB0~2

i~∂χ

∂t= Hχ


−E−t/~ =

(ae iγB0t/2

be−iγB0t/2

)




the Hamiltonian is



~B = B0z

H = −γB0Sz = −γB0~2

(1 00 −1

){χ+, E+ = −γB0~

2

χ−, E− = +γB0~2

i~∂χ

∂t= Hχ


−E−t/~ =

(ae iγB0t/2

be−iγB0t/2

)




the Hamiltonian is



~B = B0z

H = −γB0Sz

= −γB0~2

(1 00 −1

){χ+, E+ = −γB0~

2

χ−, E− = +γB0~2

i~∂χ

∂t= Hχ


−E−t/~ =

(ae iγB0t/2

be−iγB0t/2

)




the Hamiltonian is



~B = B0z

H = −γB0Sz = −γB0~2

(1 00 −1

)

{χ+, E+ = −γB0~

2

χ−, E− = +γB0~2

i~∂χ

∂t= Hχ


−E−t/~ =

(ae iγB0t/2

be−iγB0t/2

)




the Hamiltonian is



~B = B0z

H = −γB0Sz = −γB0~2

(1 00 −1

)

{χ+, E+ = −γB0~

2

χ−, E− = +γB0~2

i~∂χ

∂t= Hχ


−E−t/~ =

(ae iγB0t/2

be−iγB0t/2

)




the Hamiltonian is



~B = B0z

H = −γB0Sz = −γB0~2

(1 00 −1

){χ+, E+ = −γB0~

2

χ−, E− = +γB0~2

i~∂χ

∂t= Hχ


−E−t/~ =

(ae iγB0t/2

be−iγB0t/2

)




the Hamiltonian is


the time dependent Schrodingerequation

can be solved in terms ofthe stationary states

~B = B0z

H = −γB0Sz = −γB0~2

(1 00 −1

){χ+, E+ = −γB0~

2

χ−, E− = +γB0~2

i~∂χ

∂t= Hχ


−E−t/~ =

(ae iγB0t/2

be−iγB0t/2

)




the Hamiltonian is


the time dependent Schrodingerequation

can be solved in terms ofthe stationary states

~B = B0z

H = −γB0Sz = −γB0~2

(1 00 −1

){χ+, E+ = −γB0~

2

χ−, E− = +γB0~2

i~∂χ

∂t= Hχ


−E−t/~ =

(ae iγB0t/2

be−iγB0t/2

)




the Hamiltonian is



~B = B0z

H = −γB0Sz = −γB0~2

(1 00 −1

){χ+, E+ = −γB0~

2

χ−, E− = +γB0~2

i~∂χ

∂t= Hχ


−E−t/~ =

(ae iγB0t/2

be−iγB0t/2

)




the Hamiltonian is



~B = B0z

H = −γB0Sz = −γB0~2

(1 00 −1

){χ+, E+ = −γB0~

2

χ−, E− = +γB0~2

i~∂χ

∂t= Hχ


−E−t/~

=

(ae iγB0t/2

be−iγB0t/2

)




the Hamiltonian is



~B = B0z

H = −γB0Sz = −γB0~2

(1 00 −1

){χ+, E+ = −γB0~

2

χ−, E− = +γB0~2

i~∂χ

∂t= Hχ


−E−t/~ =

(ae iγB0t/2

be−iγB0t/2

)


Time-dependent expectation value

The constants a and b can be de-termined from the initial conditions

giving |a|2 + |b|2 = 1 and suggest-ing the relationship

the time-dependent solution is thus

χ(0) =

(ab

)a = cos(

α

2), b = sin(

α

2)

χ(t) =

(cos(α2 )e iγB0t/2

sin(α2 )e−iγB0t/2

)the expectation value of S as a function of time can now be calculated

〈Sx〉 = χ†(t)Sxχ(t)

=(

cos(α2 )e−iγB0t/2 sin(α2 )e+iγB0t/2) ~

2

(0 11 0

)(cos(α2 )e iγB0t/2


)=

~2

(cos(α2 )e−iγB0t/2 sin(α2 )e+iγB0t/2

)(sin(α2 )e−iγB0t/2

cos(α2 )e iγB0t/2

)=

~2

cos(α/2) sin(α/2)(e−iγB0t + e iγB0t

)=

~2

sinα cos(γB0t)






χ(0) =

(ab

)

a = cos(α

2), b = sin(

α

2)

χ(t) =





=(


2

(0 11 0



)=

~2



cos(α2 )e iγB0t/2

)=

~2


)=

~2

sinα cos(γB0t)






χ(0) =

(ab

)

a = cos(α

2), b = sin(

α

2)

χ(t) =





=(


2

(0 11 0



)=

~2



cos(α2 )e iγB0t/2

)=

~2


)=

~2

sinα cos(γB0t)






χ(0) =

(ab

)a = cos(

α

2), b = sin(

α

2)

χ(t) =





=(


2

(0 11 0



)=

~2



cos(α2 )e iγB0t/2

)=

~2


)=

~2

sinα cos(γB0t)






χ(0) =

(ab

)a = cos(

α

2), b = sin(

α

2)

χ(t) =





=(


2

(0 11 0



)=

~2



cos(α2 )e iγB0t/2

)=

~2


)=

~2

sinα cos(γB0t)






χ(0) =

(ab

)a = cos(

α

2), b = sin(

α

2)

χ(t) =



)

the expectation value of S as a function of time can now be calculated


=(


2

(0 11 0



)=

~2



cos(α2 )e iγB0t/2

)=

~2


)=

~2

sinα cos(γB0t)






χ(0) =

(ab

)a = cos(

α

2), b = sin(

α

2)

χ(t) =





=(


2

(0 11 0



)=

~2



cos(α2 )e iγB0t/2

)=

~2


)=

~2

sinα cos(γB0t)






χ(0) =

(ab

)a = cos(

α

2), b = sin(

α

2)

χ(t) =





=(


2

(0 11 0



)=

~2



cos(α2 )e iγB0t/2

)=

~2


)=

~2

sinα cos(γB0t)






χ(0) =

(ab

)a = cos(

α

2), b = sin(

α

2)

χ(t) =





=(


2

(0 11 0



)

=~2



cos(α2 )e iγB0t/2

)=

~2


)=

~2

sinα cos(γB0t)






χ(0) =

(ab

)a = cos(

α

2), b = sin(

α

2)

χ(t) =





=(


2

(0 11 0



)=

~2



cos(α2 )e iγB0t/2

)

=~2


)=

~2

sinα cos(γB0t)






χ(0) =

(ab

)a = cos(

α

2), b = sin(

α

2)

χ(t) =





=(


2

(0 11 0



)=

~2



cos(α2 )e iγB0t/2

)=

~2


)

=~2

sinα cos(γB0t)






χ(0) =

(ab

)a = cos(

α

2), b = sin(

α

2)

χ(t) =





=(


2

(0 11 0



)=

~2



cos(α2 )e iγB0t/2

)=

~2


)=

~2

sinα cos(γB0t)


More expectation values

〈Sy 〉 = χ†(t)Syχ(t)

=(


2

(0 −ii 0



)=

~2


)(−i sin(α2 )e−iγB0t/2

i cos(α2 )e iγB0t/2

)=

~2

cos(α/2) sin(α/2)(−ie−iγB0t + ie iγB0t

)= −~

2sinα sin(γB0t)

〈Sz〉 = χ†(t)Szχ(t)

=(


2

(1 00 −1



)=

~2


)( cos(α2 )e iγB0t/2

− sin(α2 )e−iγB0t/2

)=

~2

(cos2(α/2)− sin2(α/2)

)=

~2

cosα




=(


2

(0 −ii 0



)

=~2




)=

~2


)= −~

2sinα sin(γB0t)


=(


2

(1 00 −1



)=

~2




)=

~2

(cos2(α/2)− sin2(α/2)

)=

~2

cosα




=(


2

(0 −ii 0



)=

~2




)

=~2


)= −~

2sinα sin(γB0t)


=(


2

(1 00 −1



)=

~2




)=

~2

(cos2(α/2)− sin2(α/2)

)=

~2

cosα




=(


2

(0 −ii 0



)=

~2




)=

~2


)

= −~2

sinα sin(γB0t)


=(


2

(1 00 −1



)=

~2




)=

~2

(cos2(α/2)− sin2(α/2)

)=

~2

cosα




=(


2

(0 −ii 0



)=

~2




)=

~2


)= −~

2sinα sin(γB0t)


=(


2

(1 00 −1



)=

~2




)=

~2

(cos2(α/2)− sin2(α/2)

)=

~2

cosα




=(


2

(0 −ii 0



)=

~2




)=

~2


)= −~

2sinα sin(γB0t)


=(


2

(1 00 −1



)=

~2




)=

~2

(cos2(α/2)− sin2(α/2)

)=

~2

cosα




=(


2

(0 −ii 0



)=

~2




)=

~2


)= −~

2sinα sin(γB0t)


=(


2

(1 00 −1



)

=~2




)=

~2

(cos2(α/2)− sin2(α/2)

)=

~2

cosα




=(


2

(0 −ii 0



)=

~2




)=

~2


)= −~

2sinα sin(γB0t)


=(


2

(1 00 −1



)=

~2




)

=~2

(cos2(α/2)− sin2(α/2)

)=

~2

cosα




=(


2

(0 −ii 0



)=

~2




)=

~2


)= −~

2sinα sin(γB0t)


=(


2

(1 00 −1



)=

~2




)=

~2

(cos2(α/2)− sin2(α/2)

)

=~2

cosα




=(


2

(0 −ii 0



)=

~2




)=

~2


)= −~

2sinα sin(γB0t)


=(


2

(1 00 −1



)=

~2




)=

~2

(cos2(α/2)− sin2(α/2)

)=

~2

cosα


Larmor precession

The expectation values for the compo-nents of S are thus

〈Sx〉 =~2

sinα cos(γB0t)

〈Sy 〉 = −~2

sinα sin(γB0t)

〈Sz〉 =~2

cosα

S precesses about the z-axis, at a fixedangle α so 〈Sz〉 is constant and the othertwo are constantly changing

the precession frequency ω = γB0 iscalled the Larmor frequency

z

y

x

S

ω

α


Larmor precession


〈Sx〉 =~2

sinα cos(γB0t)

〈Sy 〉 = −~2

sinα sin(γB0t)

〈Sz〉 =~2

cosα



z

y

x

S

ω

α


Larmor precession


〈Sx〉 =~2

sinα cos(γB0t)

〈Sy 〉 = −~2

sinα sin(γB0t)

〈Sz〉 =~2

cosα



z

y

x

S

ω

α


Larmor precession


〈Sx〉 =~2

sinα cos(γB0t)

〈Sy 〉 = −~2

sinα sin(γB0t)

〈Sz〉 =~2

cosα



z

y

x

S

ω

α


Larmor precession


〈Sx〉 =~2

sinα cos(γB0t)

〈Sy 〉 = −~2

sinα sin(γB0t)

〈Sz〉 =~2

cosα



z

y

x

S

ω

α


Larmor precession


〈Sx〉 =~2

sinα cos(γB0t)

〈Sy 〉 = −~2

sinα sin(γB0t)

〈Sz〉 =~2

cosα



z

y

x

S

ω

α


Larmor precession


〈Sx〉 =~2

sinα cos(γB0t)

〈Sy 〉 = −~2

sinα sin(γB0t)

〈Sz〉 =~2

cosα



z

y

x

S

ω

α


Total spin of the hydrogen atom

Suppose we have a hydrogen atom in its ground state (L = 0).

Theelectron, and the proton, each have spin S = 1

2. What is the total angular

momentum of the atom?

Each particle can have spin up or down, for 4 combinations.

↑↑ ↑↓ ↓↑ ↓↓

The total angular momentum is just the sum

~S ≡ ~S (1) + ~S (2)

each of the four possible states is an eigenfunction of the z-component ofthe total angular momentum

Szχ1χ2 = (S(1)z + S

(2)z )χ1χ2 = (S

(1)z χ1)χ2 + χ1(S

(2)z χ2)

= (~m1χ1)χ2 + χ1(~m2χ2) = ~(m1 + m2)χ1χ2



Suppose we have a hydrogen atom in its ground state (L = 0). Theelectron, and the proton, each have spin S = 1

2.

What is the total angularmomentum of the atom?


↑↑ ↑↓ ↓↑ ↓↓


~S ≡ ~S (1) + ~S (2)


Szχ1χ2 = (S(1)z + S

(2)z )χ1χ2 = (S

(1)z χ1)χ2 + χ1(S

(2)z χ2)

= (~m1χ1)χ2 + χ1(~m2χ2) = ~(m1 + m2)χ1χ2







↑↑ ↑↓ ↓↑ ↓↓


~S ≡ ~S (1) + ~S (2)


Szχ1χ2 = (S(1)z + S

(2)z )χ1χ2 = (S

(1)z χ1)χ2 + χ1(S

(2)z χ2)

= (~m1χ1)χ2 + χ1(~m2χ2) = ~(m1 + m2)χ1χ2







↑↑ ↑↓ ↓↑ ↓↓


~S ≡ ~S (1) + ~S (2)


Szχ1χ2 = (S(1)z + S

(2)z )χ1χ2 = (S

(1)z χ1)χ2 + χ1(S

(2)z χ2)

= (~m1χ1)χ2 + χ1(~m2χ2) = ~(m1 + m2)χ1χ2







↑↑

↑↓ ↓↑ ↓↓


~S ≡ ~S (1) + ~S (2)


Szχ1χ2 = (S(1)z + S

(2)z )χ1χ2 = (S

(1)z χ1)χ2 + χ1(S

(2)z χ2)

= (~m1χ1)χ2 + χ1(~m2χ2) = ~(m1 + m2)χ1χ2







↑↑ ↑↓

↓↑ ↓↓


~S ≡ ~S (1) + ~S (2)


Szχ1χ2 = (S(1)z + S

(2)z )χ1χ2 = (S

(1)z χ1)χ2 + χ1(S

(2)z χ2)

= (~m1χ1)χ2 + χ1(~m2χ2) = ~(m1 + m2)χ1χ2







↑↑ ↑↓ ↓↑

↓↓


~S ≡ ~S (1) + ~S (2)


Szχ1χ2 = (S(1)z + S

(2)z )χ1χ2 = (S

(1)z χ1)χ2 + χ1(S

(2)z χ2)

= (~m1χ1)χ2 + χ1(~m2χ2) = ~(m1 + m2)χ1χ2







↑↑ ↑↓ ↓↑ ↓↓


~S ≡ ~S (1) + ~S (2)


Szχ1χ2 = (S(1)z + S

(2)z )χ1χ2 = (S

(1)z χ1)χ2 + χ1(S

(2)z χ2)

= (~m1χ1)χ2 + χ1(~m2χ2) = ~(m1 + m2)χ1χ2







↑↑ ↑↓ ↓↑ ↓↓


~S ≡ ~S (1) + ~S (2)


Szχ1χ2 = (S(1)z + S

(2)z )χ1χ2 = (S

(1)z χ1)χ2 + χ1(S

(2)z χ2)

= (~m1χ1)χ2 + χ1(~m2χ2) = ~(m1 + m2)χ1χ2







↑↑ ↑↓ ↓↑ ↓↓


~S ≡ ~S (1) + ~S (2)


Szχ1χ2 = (S(1)z + S

(2)z )χ1χ2 = (S

(1)z χ1)χ2 + χ1(S

(2)z χ2)

= (~m1χ1)χ2 + χ1(~m2χ2) = ~(m1 + m2)χ1χ2







↑↑ ↑↓ ↓↑ ↓↓


~S ≡ ~S (1) + ~S (2)


Szχ1χ2 = (S(1)z + S

(2)z )χ1χ2

= (S(1)z χ1)χ2 + χ1(S

(2)z χ2)

= (~m1χ1)χ2 + χ1(~m2χ2) = ~(m1 + m2)χ1χ2







↑↑ ↑↓ ↓↑ ↓↓


~S ≡ ~S (1) + ~S (2)


Szχ1χ2 = (S(1)z + S

(2)z )χ1χ2 = (S

(1)z χ1)χ2 + χ1(S

(2)z χ2)

= (~m1χ1)χ2 + χ1(~m2χ2) = ~(m1 + m2)χ1χ2







↑↑ ↑↓ ↓↑ ↓↓


~S ≡ ~S (1) + ~S (2)


Szχ1χ2 = (S(1)z + S

(2)z )χ1χ2 = (S

(1)z χ1)χ2 + χ1(S

(2)z χ2)

= (~m1χ1)χ2 + χ1(~m2χ2)

= ~(m1 + m2)χ1χ2







↑↑ ↑↓ ↓↑ ↓↓


~S ≡ ~S (1) + ~S (2)


Szχ1χ2 = (S(1)z + S

(2)z )χ1χ2 = (S

(1)z χ1)χ2 + χ1(S

(2)z χ2)

= (~m1χ1)χ2 + χ1(~m2χ2) = ~(m1 + m2)χ1χ2


Triplet state

We can thus establish that the state designated by ↑↑ has m = 1 andconsequently must have s = 1.

Let’s check this

S2 = (~S (1) + ~S (2)) · (~S (1) + ~S (2)) = (S (1))2 + (S (2))2 + 2~S (1) · ~S (2)

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

(2)x + S

(1)y S

(2)y + S

(1)z S

(2)z )

S(1)x S

(2)x ↑↑ =

~2

(0 11 0

)(10

)~2

(0 11 0

)(10

)=

~2

4

(01

)(01

)=

~2

4↓↓

S(1)y S

(2)y ↑↑ =

~2

(0 −ii 0

)(10

)~2

(0 −ii 0

)(10

)= −~2

4

(01

)(01

)= −~2

4↓↓

S(1)z S

(2)z ↑↑ =

~2

(1 00 −1

)(10

)~2

(1 00 −1

)(10

)=

~2

4

(10

)(10

)=

~2

4↑↑

S2↑↑ = ~21

2

(3

2

)↑↑+ ~2

1

2

(3

2

)↑↑+ 2

(~2

4↓↓ − ~2

4↓↓+

~2

4↑↑)

= 2~2↑↑


Triplet state

We can thus establish that the state designated by ↑↑ has m = 1 andconsequently must have s = 1. Let’s check this

S2 = (~S (1) + ~S (2)) · (~S (1) + ~S (2)) = (S (1))2 + (S (2))2 + 2~S (1) · ~S (2)

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

(2)x + S

(1)y S

(2)y + S

(1)z S

(2)z )

S(1)x S

(2)x ↑↑ =

~2

(0 11 0

)(10

)~2

(0 11 0

)(10

)=

~2

4

(01

)(01

)=

~2

4↓↓

S(1)y S

(2)y ↑↑ =

~2

(0 −ii 0

)(10

)~2

(0 −ii 0

)(10

)= −~2

4

(01

)(01

)= −~2

4↓↓

S(1)z S

(2)z ↑↑ =

~2

(1 00 −1

)(10

)~2

(1 00 −1

)(10

)=

~2

4

(10

)(10

)=

~2

4↑↑

S2↑↑ = ~21

2

(3

2

)↑↑+ ~2

1

2

(3

2

)↑↑+ 2

(~2

4↓↓ − ~2

4↓↓+

~2

4↑↑)

= 2~2↑↑


Triplet state


S2 = (~S (1) + ~S (2)) · (~S (1) + ~S (2))

= (S (1))2 + (S (2))2 + 2~S (1) · ~S (2)

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

(2)x + S

(1)y S

(2)y + S

(1)z S

(2)z )

S(1)x S

(2)x ↑↑ =

~2

(0 11 0

)(10

)~2

(0 11 0

)(10

)=

~2

4

(01

)(01

)=

~2

4↓↓

S(1)y S

(2)y ↑↑ =

~2

(0 −ii 0

)(10

)~2

(0 −ii 0

)(10

)= −~2

4

(01

)(01

)= −~2

4↓↓

S(1)z S

(2)z ↑↑ =

~2

(1 00 −1

)(10

)~2

(1 00 −1

)(10

)=

~2

4

(10

)(10

)=

~2

4↑↑

S2↑↑ = ~21

2

(3

2

)↑↑+ ~2

1

2

(3

2

)↑↑+ 2

(~2

4↓↓ − ~2

4↓↓+

~2

4↑↑)

= 2~2↑↑


Triplet state


S2 = (~S (1) + ~S (2)) · (~S (1) + ~S (2)) = (S (1))2 + (S (2))2 + 2~S (1) · ~S (2)

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

(2)x + S

(1)y S

(2)y + S

(1)z S

(2)z )

S(1)x S

(2)x ↑↑ =

~2

(0 11 0

)(10

)~2

(0 11 0

)(10

)=

~2

4

(01

)(01

)=

~2

4↓↓

S(1)y S

(2)y ↑↑ =

~2

(0 −ii 0

)(10

)~2

(0 −ii 0

)(10

)= −~2

4

(01

)(01

)= −~2

4↓↓

S(1)z S

(2)z ↑↑ =

~2

(1 00 −1

)(10

)~2

(1 00 −1

)(10

)=

~2

4

(10

)(10

)=

~2

4↑↑

S2↑↑ = ~21

2

(3

2

)↑↑+ ~2

1

2

(3

2

)↑↑+ 2

(~2

4↓↓ − ~2

4↓↓+

~2

4↑↑)

= 2~2↑↑


Triplet state


S2 = (~S (1) + ~S (2)) · (~S (1) + ~S (2)) = (S (1))2 + (S (2))2 + 2~S (1) · ~S (2)

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

(2)x + S

(1)y S

(2)y + S

(1)z S

(2)z )

S(1)x S

(2)x ↑↑ =

~2

(0 11 0

)(10

)~2

(0 11 0

)(10

)=

~2

4

(01

)(01

)=

~2

4↓↓

S(1)y S

(2)y ↑↑ =

~2

(0 −ii 0

)(10

)~2

(0 −ii 0

)(10

)= −~2

4

(01

)(01

)= −~2

4↓↓

S(1)z S

(2)z ↑↑ =

~2

(1 00 −1

)(10

)~2

(1 00 −1

)(10

)=

~2

4

(10

)(10

)=

~2

4↑↑

S2↑↑ = ~21

2

(3

2

)↑↑+ ~2

1

2

(3

2

)↑↑+ 2

(~2

4↓↓ − ~2

4↓↓+

~2

4↑↑)

= 2~2↑↑


Triplet state


S2 = (~S (1) + ~S (2)) · (~S (1) + ~S (2)) = (S (1))2 + (S (2))2 + 2~S (1) · ~S (2)

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

(2)x + S

(1)y S

(2)y + S

(1)z S

(2)z )

S(1)x S

(2)x ↑↑

=~2

(0 11 0

)(10

)~2

(0 11 0

)(10

)=

~2

4

(01

)(01

)=

~2

4↓↓

S(1)y S

(2)y ↑↑ =

~2

(0 −ii 0

)(10

)~2

(0 −ii 0

)(10

)= −~2

4

(01

)(01

)= −~2

4↓↓

S(1)z S

(2)z ↑↑ =

~2

(1 00 −1

)(10

)~2

(1 00 −1

)(10

)=

~2

4

(10

)(10

)=

~2

4↑↑

S2↑↑ = ~21

2

(3

2

)↑↑+ ~2

1

2

(3

2

)↑↑+ 2

(~2

4↓↓ − ~2

4↓↓+

~2

4↑↑)

= 2~2↑↑


Triplet state


S2 = (~S (1) + ~S (2)) · (~S (1) + ~S (2)) = (S (1))2 + (S (2))2 + 2~S (1) · ~S (2)

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

(2)x + S

(1)y S

(2)y + S

(1)z S

(2)z )

S(1)x S

(2)x ↑↑ =

~2

(0 11 0

)(10

)~2

(0 11 0

)(10

)

=~2

4

(01

)(01

)=

~2

4↓↓

S(1)y S

(2)y ↑↑ =

~2

(0 −ii 0

)(10

)~2

(0 −ii 0

)(10

)= −~2

4

(01

)(01

)= −~2

4↓↓

S(1)z S

(2)z ↑↑ =

~2

(1 00 −1

)(10

)~2

(1 00 −1

)(10

)=

~2

4

(10

)(10

)=

~2

4↑↑

S2↑↑ = ~21

2

(3

2

)↑↑+ ~2

1

2

(3

2

)↑↑+ 2

(~2

4↓↓ − ~2

4↓↓+

~2

4↑↑)

= 2~2↑↑


Triplet state


S2 = (~S (1) + ~S (2)) · (~S (1) + ~S (2)) = (S (1))2 + (S (2))2 + 2~S (1) · ~S (2)

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

(2)x + S

(1)y S

(2)y + S

(1)z S

(2)z )

S(1)x S

(2)x ↑↑ =

~2

(0 11 0

)(10

)~2

(0 11 0

)(10

)=

~2

4

(01

)(01

)

=~2

4↓↓

S(1)y S

(2)y ↑↑ =

~2

(0 −ii 0

)(10

)~2

(0 −ii 0

)(10

)= −~2

4

(01

)(01

)= −~2

4↓↓

S(1)z S

(2)z ↑↑ =

~2

(1 00 −1

)(10

)~2

(1 00 −1

)(10

)=

~2

4

(10

)(10

)=

~2

4↑↑

S2↑↑ = ~21

2

(3

2

)↑↑+ ~2

1

2

(3

2

)↑↑+ 2

(~2

4↓↓ − ~2

4↓↓+

~2

4↑↑)

= 2~2↑↑


Triplet state


S2 = (~S (1) + ~S (2)) · (~S (1) + ~S (2)) = (S (1))2 + (S (2))2 + 2~S (1) · ~S (2)

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

(2)x + S

(1)y S

(2)y + S

(1)z S

(2)z )

S(1)x S

(2)x ↑↑ =

~2

(0 11 0

)(10

)~2

(0 11 0

)(10

)=

~2

4

(01

)(01

)=

~2

4↓↓

S(1)y S

(2)y ↑↑ =

~2

(0 −ii 0

)(10

)~2

(0 −ii 0

)(10

)= −~2

4

(01

)(01

)= −~2

4↓↓

S(1)z S

(2)z ↑↑ =

~2

(1 00 −1

)(10

)~2

(1 00 −1

)(10

)=

~2

4

(10

)(10

)=

~2

4↑↑

S2↑↑ = ~21

2

(3

2

)↑↑+ ~2

1

2

(3

2

)↑↑+ 2

(~2

4↓↓ − ~2

4↓↓+

~2

4↑↑)

= 2~2↑↑


Triplet state


S2 = (~S (1) + ~S (2)) · (~S (1) + ~S (2)) = (S (1))2 + (S (2))2 + 2~S (1) · ~S (2)

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

(2)x + S

(1)y S

(2)y + S

(1)z S

(2)z )

S(1)x S

(2)x ↑↑ =

~2

(0 11 0

)(10

)~2

(0 11 0

)(10

)=

~2

4

(01

)(01

)=

~2

4↓↓

S(1)y S

(2)y ↑↑

=~2

(0 −ii 0

)(10

)~2

(0 −ii 0

)(10

)= −~2

4

(01

)(01

)= −~2

4↓↓

S(1)z S

(2)z ↑↑ =

~2

(1 00 −1

)(10

)~2

(1 00 −1

)(10

)=

~2

4

(10

)(10

)=

~2

4↑↑

S2↑↑ = ~21

2

(3

2

)↑↑+ ~2

1

2

(3

2

)↑↑+ 2

(~2

4↓↓ − ~2

4↓↓+

~2

4↑↑)

= 2~2↑↑


Triplet state


S2 = (~S (1) + ~S (2)) · (~S (1) + ~S (2)) = (S (1))2 + (S (2))2 + 2~S (1) · ~S (2)

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

(2)x + S

(1)y S

(2)y + S

(1)z S

(2)z )

S(1)x S

(2)x ↑↑ =

~2

(0 11 0

)(10

)~2

(0 11 0

)(10

)=

~2

4

(01

)(01

)=

~2

4↓↓

S(1)y S

(2)y ↑↑ =

~2

(0 −ii 0

)(10

)~2

(0 −ii 0

)(10

)

= −~2

4

(01

)(01

)= −~2

4↓↓

S(1)z S

(2)z ↑↑ =

~2

(1 00 −1

)(10

)~2

(1 00 −1

)(10

)=

~2

4

(10

)(10

)=

~2

4↑↑

S2↑↑ = ~21

2

(3

2

)↑↑+ ~2

1

2

(3

2

)↑↑+ 2

(~2

4↓↓ − ~2

4↓↓+

~2

4↑↑)

= 2~2↑↑


Triplet state


S2 = (~S (1) + ~S (2)) · (~S (1) + ~S (2)) = (S (1))2 + (S (2))2 + 2~S (1) · ~S (2)

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

(2)x + S

(1)y S

(2)y + S

(1)z S

(2)z )

S(1)x S

(2)x ↑↑ =

~2

(0 11 0

)(10

)~2

(0 11 0

)(10

)=

~2

4

(01

)(01

)=

~2

4↓↓

S(1)y S

(2)y ↑↑ =

~2

(0 −ii 0

)(10

)~2

(0 −ii 0

)(10

)= −~2

4

(01

)(01

)

= −~2

4↓↓

S(1)z S

(2)z ↑↑ =

~2

(1 00 −1

)(10

)~2

(1 00 −1

)(10

)=

~2

4

(10

)(10

)=

~2

4↑↑

S2↑↑ = ~21

2

(3

2

)↑↑+ ~2

1

2

(3

2

)↑↑+ 2

(~2

4↓↓ − ~2

4↓↓+

~2

4↑↑)

= 2~2↑↑


Triplet state


S2 = (~S (1) + ~S (2)) · (~S (1) + ~S (2)) = (S (1))2 + (S (2))2 + 2~S (1) · ~S (2)

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

(2)x + S

(1)y S

(2)y + S

(1)z S

(2)z )

S(1)x S

(2)x ↑↑ =

~2

(0 11 0

)(10

)~2

(0 11 0

)(10

)=

~2

4

(01

)(01

)=

~2

4↓↓

S(1)y S

(2)y ↑↑ =

~2

(0 −ii 0

)(10

)~2

(0 −ii 0

)(10

)= −~2

4

(01

)(01

)= −~2

4↓↓

S(1)z S

(2)z ↑↑ =

~2

(1 00 −1

)(10

)~2

(1 00 −1

)(10

)=

~2

4

(10

)(10

)=

~2

4↑↑

S2↑↑ = ~21

2

(3

2

)↑↑+ ~2

1

2

(3

2

)↑↑+ 2

(~2

4↓↓ − ~2

4↓↓+

~2

4↑↑)

= 2~2↑↑


Triplet state


S2 = (~S (1) + ~S (2)) · (~S (1) + ~S (2)) = (S (1))2 + (S (2))2 + 2~S (1) · ~S (2)

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

(2)x + S

(1)y S

(2)y + S

(1)z S

(2)z )

S(1)x S

(2)x ↑↑ =

~2

(0 11 0

)(10

)~2

(0 11 0

)(10

)=

~2

4

(01

)(01

)=

~2

4↓↓

S(1)y S

(2)y ↑↑ =

~2

(0 −ii 0

)(10

)~2

(0 −ii 0

)(10

)= −~2

4

(01

)(01

)= −~2

4↓↓

S(1)z S

(2)z ↑↑

=~2

(1 00 −1

)(10

)~2

(1 00 −1

)(10

)=

~2

4

(10

)(10

)=

~2

4↑↑

S2↑↑ = ~21

2

(3

2

)↑↑+ ~2

1

2

(3

2

)↑↑+ 2

(~2

4↓↓ − ~2

4↓↓+

~2

4↑↑)

= 2~2↑↑


Triplet state


S2 = (~S (1) + ~S (2)) · (~S (1) + ~S (2)) = (S (1))2 + (S (2))2 + 2~S (1) · ~S (2)

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

(2)x + S

(1)y S

(2)y + S

(1)z S

(2)z )

S(1)x S

(2)x ↑↑ =

~2

(0 11 0

)(10

)~2

(0 11 0

)(10

)=

~2

4

(01

)(01

)=

~2

4↓↓

S(1)y S

(2)y ↑↑ =

~2

(0 −ii 0

)(10

)~2

(0 −ii 0

)(10

)= −~2

4

(01

)(01

)= −~2

4↓↓

S(1)z S

(2)z ↑↑ =

~2

(1 00 −1

)(10

)~2

(1 00 −1

)(10

)

=~2

4

(10

)(10

)=

~2

4↑↑

S2↑↑ = ~21

2

(3

2

)↑↑+ ~2

1

2

(3

2

)↑↑+ 2

(~2

4↓↓ − ~2

4↓↓+

~2

4↑↑)

= 2~2↑↑


Triplet state


S2 = (~S (1) + ~S (2)) · (~S (1) + ~S (2)) = (S (1))2 + (S (2))2 + 2~S (1) · ~S (2)

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

(2)x + S

(1)y S

(2)y + S

(1)z S

(2)z )

S(1)x S

(2)x ↑↑ =

~2

(0 11 0

)(10

)~2

(0 11 0

)(10

)=

~2

4

(01

)(01

)=

~2

4↓↓

S(1)y S

(2)y ↑↑ =

~2

(0 −ii 0

)(10

)~2

(0 −ii 0

)(10

)= −~2

4

(01

)(01

)= −~2

4↓↓

S(1)z S

(2)z ↑↑ =

~2

(1 00 −1

)(10

)~2

(1 00 −1

)(10

)=

~2

4

(10

)(10

)

=~2

4↑↑

S2↑↑ = ~21

2

(3

2

)↑↑+ ~2

1

2

(3

2

)↑↑+ 2

(~2

4↓↓ − ~2

4↓↓+

~2

4↑↑)

= 2~2↑↑


Triplet state


S2 = (~S (1) + ~S (2)) · (~S (1) + ~S (2)) = (S (1))2 + (S (2))2 + 2~S (1) · ~S (2)

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

(2)x + S

(1)y S

(2)y + S

(1)z S

(2)z )

S(1)x S

(2)x ↑↑ =

~2

(0 11 0

)(10

)~2

(0 11 0

)(10

)=

~2

4

(01

)(01

)=

~2

4↓↓

S(1)y S

(2)y ↑↑ =

~2

(0 −ii 0

)(10

)~2

(0 −ii 0

)(10

)= −~2

4

(01

)(01

)= −~2

4↓↓

S(1)z S

(2)z ↑↑ =

~2

(1 00 −1

)(10

)~2

(1 00 −1

)(10

)=

~2

4

(10

)(10

)=

~2

4↑↑

S2↑↑ = ~21

2

(3

2

)↑↑+ ~2

1

2

(3

2

)↑↑+ 2

(~2

4↓↓ − ~2

4↓↓+

~2

4↑↑)

= 2~2↑↑


Triplet state


S2 = (~S (1) + ~S (2)) · (~S (1) + ~S (2)) = (S (1))2 + (S (2))2 + 2~S (1) · ~S (2)

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

(2)x + S

(1)y S

(2)y + S

(1)z S

(2)z )

S(1)x S

(2)x ↑↑ =

~2

(0 11 0

)(10

)~2

(0 11 0

)(10

)=

~2

4

(01

)(01

)=

~2

4↓↓

S(1)y S

(2)y ↑↑ =

~2

(0 −ii 0

)(10

)~2

(0 −ii 0

)(10

)= −~2

4

(01

)(01

)= −~2

4↓↓

S(1)z S

(2)z ↑↑ =

~2

(1 00 −1

)(10

)~2

(1 00 −1

)(10

)=

~2

4

(10

)(10

)=

~2

4↑↑

S2↑↑ =

~21

2

(3

2

)↑↑+ ~2

1

2

(3

2

)↑↑+ 2

(~2

4↓↓ − ~2

4↓↓+

~2

4↑↑)

= 2~2↑↑


Triplet state


S2 = (~S (1) + ~S (2)) · (~S (1) + ~S (2)) = (S (1))2 + (S (2))2 + 2~S (1) · ~S (2)

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

(2)x + S

(1)y S

(2)y + S

(1)z S

(2)z )

S(1)x S

(2)x ↑↑ =

~2

(0 11 0

)(10

)~2

(0 11 0

)(10

)=

~2

4

(01

)(01

)=

~2

4↓↓

S(1)y S

(2)y ↑↑ =

~2

(0 −ii 0

)(10

)~2

(0 −ii 0

)(10

)= −~2

4

(01

)(01

)= −~2

4↓↓

S(1)z S

(2)z ↑↑ =

~2

(1 00 −1

)(10

)~2

(1 00 −1

)(10

)=

~2

4

(10

)(10

)=

~2

4↑↑

S2↑↑ = ~21

2

(3

2

)↑↑

+ ~21

2

(3

2

)↑↑+ 2

(~2

4↓↓ − ~2

4↓↓+

~2

4↑↑)

= 2~2↑↑


Triplet state


S2 = (~S (1) + ~S (2)) · (~S (1) + ~S (2)) = (S (1))2 + (S (2))2 + 2~S (1) · ~S (2)

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

(2)x + S

(1)y S

(2)y + S

(1)z S

(2)z )

S(1)x S

(2)x ↑↑ =

~2

(0 11 0

)(10

)~2

(0 11 0

)(10

)=

~2

4

(01

)(01

)=

~2

4↓↓

S(1)y S

(2)y ↑↑ =

~2

(0 −ii 0

)(10

)~2

(0 −ii 0

)(10

)= −~2

4

(01

)(01

)= −~2

4↓↓

S(1)z S

(2)z ↑↑ =

~2

(1 00 −1

)(10

)~2

(1 00 −1

)(10

)=

~2

4

(10

)(10

)=

~2

4↑↑

S2↑↑ = ~21

2

(3

2

)↑↑+ ~2

1

2

(3

2

)↑↑

+ 2

(~2

4↓↓ − ~2

4↓↓+

~2

4↑↑)

= 2~2↑↑


Triplet state


S2 = (~S (1) + ~S (2)) · (~S (1) + ~S (2)) = (S (1))2 + (S (2))2 + 2~S (1) · ~S (2)

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

(2)x + S

(1)y S

(2)y + S

(1)z S

(2)z )

S(1)x S

(2)x ↑↑ =

~2

(0 11 0

)(10

)~2

(0 11 0

)(10

)=

~2

4

(01

)(01

)=

~2

4↓↓

S(1)y S

(2)y ↑↑ =

~2

(0 −ii 0

)(10

)~2

(0 −ii 0

)(10

)= −~2

4

(01

)(01

)= −~2

4↓↓

S(1)z S

(2)z ↑↑ =

~2

(1 00 −1

)(10

)~2

(1 00 −1

)(10

)=

~2

4

(10

)(10

)=

~2

4↑↑

S2↑↑ = ~21

2

(3

2

)↑↑+ ~2

1

2

(3

2

)↑↑+ 2

(~2

4↓↓ − ~2

4↓↓+

~2

4↑↑)

= 2~2↑↑


Triplet state


S2 = (~S (1) + ~S (2)) · (~S (1) + ~S (2)) = (S (1))2 + (S (2))2 + 2~S (1) · ~S (2)

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

(2)x + S

(1)y S

(2)y + S

(1)z S

(2)z )

S(1)x S

(2)x ↑↑ =

~2

(0 11 0

)(10

)~2

(0 11 0

)(10

)=

~2

4

(01

)(01

)=

~2

4↓↓

S(1)y S

(2)y ↑↑ =

~2

(0 −ii 0

)(10

)~2

(0 −ii 0

)(10

)= −~2

4

(01

)(01

)= −~2

4↓↓

S(1)z S

(2)z ↑↑ =

~2

(1 00 −1

)(10

)~2

(1 00 −1

)(10

)=

~2

4

(10

)(10

)=

~2

4↑↑

S2↑↑ = ~21

2

(3

2

)↑↑+ ~2

1

2

(3

2

)↑↑+ 2

(~2

4↓↓ − ~2

4↓↓+

~2

4↑↑)

= 2~2↑↑


Triplet state (cont.)

Use the lowering operator to generate the state with total s = 1 and m = 0

S−↑↑ = (S(1)− + S

(2)− )↑↑ = (S

(1)− ↑)↑+ ↑(S (2)

− ↑)

= ~

√1

2

(3

2

)− 1

2

(−1

2

)(↓↑+ ↑↓) = ~(↓↑+ ↑↓) =

√2~

1√2

(↓↑+ ↑↓)

Apply the lowering operator again to get s = 1, m = −1

S−1√2

(↓↑+ ↑↓) = (S(1)− + S

(2)− )

1√2

(↓↑+ ↑↓)

=1√2

[S(1)− (↓↑+ ↑↓) + S

(2)− (↓↑+ ↑↓)

]=

1√2

[~↓↓+ ~↓↓]

=√

2~↓↓

The triplet state is thus

|1 1〉 = ↑↑, |1 0〉 =1√2

(↓↑+ ↑↓), |1−1〉 = ↓↓




S−↑↑ = (S(1)− + S

(2)− )↑↑

= (S(1)− ↑)↑+ ↑(S (2)

− ↑)

= ~

√1

2

(3

2

)− 1

2

(−1

2

)(↓↑+ ↑↓) = ~(↓↑+ ↑↓) =

√2~

1√2

(↓↑+ ↑↓)


S−1√2

(↓↑+ ↑↓) = (S(1)− + S

(2)− )

1√2

(↓↑+ ↑↓)

=1√2

[S(1)− (↓↑+ ↑↓) + S

(2)− (↓↑+ ↑↓)

]=

1√2

[~↓↓+ ~↓↓]

=√

2~↓↓


|1 1〉 = ↑↑, |1 0〉 =1√2

(↓↑+ ↑↓), |1−1〉 = ↓↓




S−↑↑ = (S(1)− + S

(2)− )↑↑ = (S

(1)− ↑)↑+ ↑(S (2)

− ↑)

= ~

√1

2

(3

2

)− 1

2

(−1

2

)(↓↑+ ↑↓) = ~(↓↑+ ↑↓) =

√2~

1√2

(↓↑+ ↑↓)


S−1√2

(↓↑+ ↑↓) = (S(1)− + S

(2)− )

1√2

(↓↑+ ↑↓)

=1√2

[S(1)− (↓↑+ ↑↓) + S

(2)− (↓↑+ ↑↓)

]=

1√2

[~↓↓+ ~↓↓]

=√

2~↓↓


|1 1〉 = ↑↑, |1 0〉 =1√2

(↓↑+ ↑↓), |1−1〉 = ↓↓




S−↑↑ = (S(1)− + S

(2)− )↑↑ = (S

(1)− ↑)↑+ ↑(S (2)

− ↑)

= ~

√1

2

(3

2

)− 1

2

(−1

2

)(↓↑+ ↑↓)

= ~(↓↑+ ↑↓) =√

2~1√2

(↓↑+ ↑↓)


S−1√2

(↓↑+ ↑↓) = (S(1)− + S

(2)− )

1√2

(↓↑+ ↑↓)

=1√2

[S(1)− (↓↑+ ↑↓) + S

(2)− (↓↑+ ↑↓)

]=

1√2

[~↓↓+ ~↓↓]

=√

2~↓↓


|1 1〉 = ↑↑, |1 0〉 =1√2

(↓↑+ ↑↓), |1−1〉 = ↓↓




S−↑↑ = (S(1)− + S

(2)− )↑↑ = (S

(1)− ↑)↑+ ↑(S (2)

− ↑)

= ~

√1

2

(3

2

)− 1

2

(−1

2

)(↓↑+ ↑↓) = ~(↓↑+ ↑↓)

=√

2~1√2

(↓↑+ ↑↓)


S−1√2

(↓↑+ ↑↓) = (S(1)− + S

(2)− )

1√2

(↓↑+ ↑↓)

=1√2

[S(1)− (↓↑+ ↑↓) + S

(2)− (↓↑+ ↑↓)

]=

1√2

[~↓↓+ ~↓↓]

=√

2~↓↓


|1 1〉 = ↑↑, |1 0〉 =1√2

(↓↑+ ↑↓), |1−1〉 = ↓↓




S−↑↑ = (S(1)− + S

(2)− )↑↑ = (S

(1)− ↑)↑+ ↑(S (2)

− ↑)

= ~

√1

2

(3

2

)− 1

2

(−1

2

)(↓↑+ ↑↓) = ~(↓↑+ ↑↓) =

√2~

1√2

(↓↑+ ↑↓)


S−1√2

(↓↑+ ↑↓) = (S(1)− + S

(2)− )

1√2

(↓↑+ ↑↓)

=1√2

[S(1)− (↓↑+ ↑↓) + S

(2)− (↓↑+ ↑↓)

]=

1√2

[~↓↓+ ~↓↓]

=√

2~↓↓


|1 1〉 = ↑↑, |1 0〉 =1√2

(↓↑+ ↑↓), |1−1〉 = ↓↓




S−↑↑ = (S(1)− + S

(2)− )↑↑ = (S

(1)− ↑)↑+ ↑(S (2)

− ↑)

= ~

√1

2

(3

2

)− 1

2

(−1

2

)(↓↑+ ↑↓) = ~(↓↑+ ↑↓) =

√2~

1√2

(↓↑+ ↑↓)


S−1√2

(↓↑+ ↑↓) = (S(1)− + S

(2)− )

1√2

(↓↑+ ↑↓)

=1√2

[S(1)− (↓↑+ ↑↓) + S

(2)− (↓↑+ ↑↓)

]=

1√2

[~↓↓+ ~↓↓]

=√

2~↓↓


|1 1〉 = ↑↑, |1 0〉 =1√2

(↓↑+ ↑↓), |1−1〉 = ↓↓




S−↑↑ = (S(1)− + S

(2)− )↑↑ = (S

(1)− ↑)↑+ ↑(S (2)

− ↑)

= ~

√1

2

(3

2

)− 1

2

(−1

2

)(↓↑+ ↑↓) = ~(↓↑+ ↑↓) =

√2~

1√2

(↓↑+ ↑↓)


S−1√2

(↓↑+ ↑↓) = (S(1)− + S

(2)− )

1√2

(↓↑+ ↑↓)

=1√2

[S(1)− (↓↑+ ↑↓) + S

(2)− (↓↑+ ↑↓)

]=

1√2

[~↓↓+ ~↓↓]

=√

2~↓↓


|1 1〉 = ↑↑, |1 0〉 =1√2

(↓↑+ ↑↓), |1−1〉 = ↓↓




S−↑↑ = (S(1)− + S

(2)− )↑↑ = (S

(1)− ↑)↑+ ↑(S (2)

− ↑)

= ~

√1

2

(3

2

)− 1

2

(−1

2

)(↓↑+ ↑↓) = ~(↓↑+ ↑↓) =

√2~

1√2

(↓↑+ ↑↓)


S−1√2

(↓↑+ ↑↓) = (S(1)− + S

(2)− )

1√2

(↓↑+ ↑↓)

=1√2

[S(1)− (↓↑+ ↑↓) + S

(2)− (↓↑+ ↑↓)

]

=1√2

[~↓↓+ ~↓↓]

=√

2~↓↓


|1 1〉 = ↑↑, |1 0〉 =1√2

(↓↑+ ↑↓), |1−1〉 = ↓↓




S−↑↑ = (S(1)− + S

(2)− )↑↑ = (S

(1)− ↑)↑+ ↑(S (2)

− ↑)

= ~

√1

2

(3

2

)− 1

2

(−1

2

)(↓↑+ ↑↓) = ~(↓↑+ ↑↓) =

√2~

1√2

(↓↑+ ↑↓)


S−1√2

(↓↑+ ↑↓) = (S(1)− + S

(2)− )

1√2

(↓↑+ ↑↓)

=1√2

[S(1)− (↓↑+ ↑↓) + S

(2)− (↓↑+ ↑↓)

]=

1√2

[~↓↓+ ~↓↓]

=√

2~↓↓


|1 1〉 = ↑↑, |1 0〉 =1√2

(↓↑+ ↑↓), |1−1〉 = ↓↓




S−↑↑ = (S(1)− + S

(2)− )↑↑ = (S

(1)− ↑)↑+ ↑(S (2)

− ↑)

= ~

√1

2

(3

2

)− 1

2

(−1

2

)(↓↑+ ↑↓) = ~(↓↑+ ↑↓) =

√2~

1√2

(↓↑+ ↑↓)


S−1√2

(↓↑+ ↑↓) = (S(1)− + S

(2)− )

1√2

(↓↑+ ↑↓)

=1√2

[S(1)− (↓↑+ ↑↓) + S

(2)− (↓↑+ ↑↓)

]=

1√2

[~↓↓+ ~↓↓]

=√

2~↓↓


|1 1〉 = ↑↑, |1 0〉 =1√2

(↓↑+ ↑↓), |1−1〉 = ↓↓




S−↑↑ = (S(1)− + S

(2)− )↑↑ = (S

(1)− ↑)↑+ ↑(S (2)

− ↑)

= ~

√1

2

(3

2

)− 1

2

(−1

2

)(↓↑+ ↑↓) = ~(↓↑+ ↑↓) =

√2~

1√2

(↓↑+ ↑↓)


S−1√2

(↓↑+ ↑↓) = (S(1)− + S

(2)− )

1√2

(↓↑+ ↑↓)

=1√2

[S(1)− (↓↑+ ↑↓) + S

(2)− (↓↑+ ↑↓)

]=

1√2

[~↓↓+ ~↓↓]

=√

2~↓↓


|1 1〉 = ↑↑, |1 0〉 =1√2

(↓↑+ ↑↓), |1−1〉 = ↓↓




S−↑↑ = (S(1)− + S

(2)− )↑↑ = (S

(1)− ↑)↑+ ↑(S (2)

− ↑)

= ~

√1

2

(3

2

)− 1

2

(−1

2

)(↓↑+ ↑↓) = ~(↓↑+ ↑↓) =

√2~

1√2

(↓↑+ ↑↓)


S−1√2

(↓↑+ ↑↓) = (S(1)− + S

(2)− )

1√2

(↓↑+ ↑↓)

=1√2

[S(1)− (↓↑+ ↑↓) + S

(2)− (↓↑+ ↑↓)

]=

1√2

[~↓↓+ ~↓↓]

=√

2~↓↓


|1 1〉 = ↑↑, |1 0〉 =1√2

(↓↑+ ↑↓), |1−1〉 = ↓↓


Documents

Today’s Outline - October 30, 2013 - IITphys.iit.edu/~segre/phys405/13F/lecture_18.pdf · 2013. 10. 31. · C. Segre (IIT) PHYS 405 - Fall 2013 October 30, 2013 1 / 17. Today’s