today’s outline - november 05, 2012 - iitphys.iit.edu/~segre/phys405/12f/lecture_20.pdf · 2012....

Today’s Outline - November 05, 2012

• Problem 4.52

• Electrons in a magnetic field

• Adding angular momentum

C. Segre (IIT) PHYS 405 - Fall 2012 November 05, 2012 1 / 10

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

−12〉 =

0010

| 32

−32〉 =

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

−12〉




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

−12〉




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

−12〉




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

−12〉




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

−12〉




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

−12〉




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

−12〉




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

−12〉




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

−12〉




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

−12〉




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

−12〉




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

−12〉




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

−12〉



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

2−12〉

S−| 32 −12〉 =

√32( 5

2)+ 12(− 3

2)~ | 32 −32〉 =√

3~ | 32

−32〉

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

2−12〉

S−| 32 −12〉 =

√32( 5

2)+ 12(− 3

2)~ | 32 −32〉 =√

3~ | 32

−32〉

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

2−12〉

S−| 32 −12〉 =

√32( 5

2)+ 12(− 3

2)~ | 32 −32〉 =√

3~ | 32

−32〉

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

2−12〉

S−| 32 −12〉 =

√32( 5

2)+ 12(− 3

2)~ | 32 −32〉 =√

3~ | 32

−32〉

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

−12〉

S−| 32 −12〉 =

√32( 5

2)+ 12(− 3

2)~ | 32 −32〉 =√

3~ | 32

−32〉

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

2−12〉

S−| 32 −12〉 =

√32( 5

2)+ 12(− 3

2)~ | 32 −32〉 =√

3~ | 32

−32〉

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

2−12〉

S−| 32 −12〉 =

√32( 5

2)+ 12(− 3

2)~ | 32 −32〉 =√

3~ | 32

−32〉

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

2−12〉

S−| 32 −12〉 =

√32( 5

2)+ 12(− 3

2)~ | 32 −32〉

=√

3~ | 32

−32〉

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

2−12〉

S−| 32 −12〉 =

√32( 5

2)+ 12(− 3

2)~ | 32 −32〉 =√

3~ | 32

−32〉

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

2−12〉

S−| 32 −12〉 =

√32( 5

2)+ 12(− 3

2)~ | 32 −32〉 =√

3~ | 32

−32〉

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

2−12〉

S−| 32 −12〉 =

√32( 5

2)+ 12(− 3

2)~ | 32 −32〉 =√

3~ | 32

−32〉

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

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~


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


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


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) =



)

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)


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


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


today’s outline - november 05, 2012 - iitphys.iit.edu/~segre/phys405/12f/lecture_20.pdf · 2012....

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