Today’s Outline - October 17, 2017csrri.iit.edu/~segre/phys405/17F/lecture_16.pdf · Today’s...

Today’s Outline - October 17, 2017

• Angular momentum operator review

• Spin

• Pauli matrices

• Problems from Chapter 4

Reading Assignment: Chapter 4.4

Homework Assignment #07:Chapter 4: 1,2,4,5,7,8due Tuesday, October 24, 2017

Homework Assignment #08:Chapter 4: 10,13,14,15,16,38due Tuesday, October 31, 2017

C. Segre (IIT) PHYS 405 - Fall 2017 October 17, 2017 1 / 21

Angular momentum operators

We know the properties of the angular momentum eigenfunctions but westill need to determine their functional form

~L =~i

[φ∂

∂θ− θ 1

sin θ

∂

∂φ

]Lx =

~i

(− sinφ

∂

∂θ− cot θ cosφ

∂

∂φ

)Ly =

~i

(+ cosφ

∂

∂θ− cot θ sinφ

∂

∂φ

)Lz =

~i

∂

∂φ

L± = ±~e±iφ[∂

∂θ± i cot θ

∂

∂φ

]use the result of problem 4.21 to solve this




~L =~i

[φ∂

∂θ− θ 1

sin θ

∂

∂φ

]

Lx =~i

(− sinφ

∂


∂

∂φ

)Ly =

~i

(+ cosφ

∂


∂

∂φ

)Lz =

~i

∂

∂φ

L± = ±~e±iφ[∂

∂θ± i cot θ

∂

∂φ





~L =~i

[φ∂

∂θ− θ 1

sin θ

∂

∂φ

]Lx =

~i

(− sinφ

∂


∂

∂φ

)

Ly =~i

(+ cosφ

∂


∂

∂φ

)Lz =

~i

∂

∂φ

L± = ±~e±iφ[∂

∂θ± i cot θ

∂

∂φ





~L =~i

[φ∂

∂θ− θ 1

sin θ

∂

∂φ

]Lx =

~i

(− sinφ

∂


∂

∂φ

)Ly =

~i

(+ cosφ

∂


∂

∂φ

)

Lz =~i

∂

∂φ

L± = ±~e±iφ[∂

∂θ± i cot θ

∂

∂φ





~L =~i

[φ∂

∂θ− θ 1

sin θ

∂

∂φ

]Lx =

~i

(− sinφ

∂


∂

∂φ

)Ly =

~i

(+ cosφ

∂


∂

∂φ

)Lz =

~i

∂

∂φ

L± = ±~e±iφ[∂

∂θ± i cot θ

∂

∂φ





~L =~i

[φ∂

∂θ− θ 1

sin θ

∂

∂φ

]Lx =

~i

(− sinφ

∂


∂

∂φ

)Ly =

~i

(+ cosφ

∂


∂

∂φ

)Lz =

~i

∂

∂φ

L± = ±~e±iφ[∂

∂θ± i cot θ

∂

∂φ

]

use the result of problem 4.21 to solve this




~L =~i

[φ∂

∂θ− θ 1

sin θ

∂

∂φ

]Lx =

~i

(− sinφ

∂


∂

∂φ

)Ly =

~i

(+ cosφ

∂


∂

∂φ

)Lz =

~i

∂

∂φ

L± = ±~e±iφ[∂

∂θ± i cot θ

∂

∂φ



Problem 4.21

(a) Derive

L+L− = −~2(∂2

∂θ2+ cot θ

∂

∂θ+ cot2 θ

∂2

∂φ2+ i

∂

∂φ

)

(b) Using the result of part (a) derive

L2 = −~2[

1

sin θ

∂

∂θ

(sin θ

∂

∂θ

)+

1

sin2 θ

∂2

∂φ2

]


Problem 4.21(a) – solution

L+L− =− ~2e iφ(∂

∂θ+ i cot θ

∂

∂φ

)[e−iφ

(∂

∂θ− i cot θ

∂

∂φ

)]

=− ~2[

∂2

∂θ2+ i csc2θ

∂

∂φ− i cot θ

∂

∂θ

∂

∂φ

+ cot θ

(∂

∂θ− i cot θ

∂

∂φ

)+ i cot θ

∂

∂φ

∂

∂θ+ cot2θ

∂2

∂φ2

]=− ~2

[

∂2

∂θ2+ i(csc2θ − cot2θ)

∂

∂φ+ cot θ

∂

∂θ+ cot2θ

∂2

∂φ2

]=− ~2

(∂2

∂θ2+ i

∂

∂φ+ cot θ

∂

∂θ+ cot2θ

∂2

∂φ2

)Now it is possible to construct the spherical coordinates form of L2



L+L− =− ~2e iφ(∂

∂θ+ i cot θ

∂

∂φ

)[e−iφ

(∂

∂θ− i cot θ

∂

∂φ

)]=− ~2

[

∂2

∂θ2+ i csc2θ

∂

∂φ− i cot θ

∂

∂θ

∂

∂φ

+ cot θ

(∂

∂θ− i cot θ

∂

∂φ

)+ i cot θ

∂

∂φ

∂

∂θ+ cot2θ

∂2

∂φ2

]

=− ~2[

∂2


∂

∂φ+ cot θ

∂

∂θ+ cot2θ

∂2

∂φ2

]=− ~2

(∂2

∂θ2+ i

∂

∂φ+ cot θ

∂

∂θ+ cot2θ

∂2

∂φ2




L+L− =− ~2e iφ(∂

∂θ+ i cot θ

∂

∂φ

)[e−iφ

(∂

∂θ− i cot θ

∂

∂φ

)]=− ~2

[∂2

∂θ2

+ i csc2θ∂

∂φ− i cot θ

∂

∂θ

∂

∂φ

+ cot θ

(∂

∂θ− i cot θ

∂

∂φ

)+ i cot θ

∂

∂φ

∂

∂θ+ cot2θ

∂2

∂φ2

]

=− ~2[

∂2


∂

∂φ+ cot θ

∂

∂θ+ cot2θ

∂2

∂φ2

]=− ~2

(∂2

∂θ2+ i

∂

∂φ+ cot θ

∂

∂θ+ cot2θ

∂2

∂φ2




L+L− =− ~2e iφ(∂

∂θ+ i cot θ

∂

∂φ

)[e−iφ

(∂

∂θ− i cot θ

∂

∂φ

)]=− ~2

[∂2

∂θ2+ i csc2θ

∂

∂φ

− i cot θ∂

∂θ

∂

∂φ

+ cot θ

(∂

∂θ− i cot θ

∂

∂φ

)+ i cot θ

∂

∂φ

∂

∂θ+ cot2θ

∂2

∂φ2

]

=− ~2[

∂2


∂

∂φ+ cot θ

∂

∂θ+ cot2θ

∂2

∂φ2

]=− ~2

(∂2

∂θ2+ i

∂

∂φ+ cot θ

∂

∂θ+ cot2θ

∂2

∂φ2




L+L− =− ~2e iφ(∂

∂θ+ i cot θ

∂

∂φ

)[e−iφ

(∂

∂θ− i cot θ

∂

∂φ

)]=− ~2

[∂2

∂θ2+ i csc2θ

∂

∂φ− i cot θ

∂

∂θ

∂

∂φ

+ cot θ

(∂

∂θ− i cot θ

∂

∂φ

)+ i cot θ

∂

∂φ

∂

∂θ+ cot2θ

∂2

∂φ2

]

=− ~2[

∂2


∂

∂φ+ cot θ

∂

∂θ+ cot2θ

∂2

∂φ2

]=− ~2

(∂2

∂θ2+ i

∂

∂φ+ cot θ

∂

∂θ+ cot2θ

∂2

∂φ2




L+L− =− ~2e iφ(∂

∂θ+ i cot θ

∂

∂φ

)[e−iφ

(∂

∂θ− i cot θ

∂

∂φ

)]=− ~2

[∂2

∂θ2+ i csc2θ

∂

∂φ− i cot θ

∂

∂θ

∂

∂φ

+ cot θ

(∂

∂θ− i cot θ

∂

∂φ

)

+ i cot θ∂

∂φ

∂

∂θ+ cot2θ

∂2

∂φ2

]

=− ~2[

∂2


∂

∂φ+ cot θ

∂

∂θ+ cot2θ

∂2

∂φ2

]=− ~2

(∂2

∂θ2+ i

∂

∂φ+ cot θ

∂

∂θ+ cot2θ

∂2

∂φ2




L+L− =− ~2e iφ(∂

∂θ+ i cot θ

∂

∂φ

)[e−iφ

(∂

∂θ− i cot θ

∂

∂φ

)]=− ~2

[∂2

∂θ2+ i csc2θ

∂

∂φ− i cot θ

∂

∂θ

∂

∂φ

+ cot θ

(∂

∂θ− i cot θ

∂

∂φ

)+ i cot θ

∂

∂φ

∂

∂θ

+ cot2θ∂2

∂φ2

]

=− ~2[

∂2


∂

∂φ+ cot θ

∂

∂θ+ cot2θ

∂2

∂φ2

]=− ~2

(∂2

∂θ2+ i

∂

∂φ+ cot θ

∂

∂θ+ cot2θ

∂2

∂φ2




L+L− =− ~2e iφ(∂

∂θ+ i cot θ

∂

∂φ

)[e−iφ

(∂

∂θ− i cot θ

∂

∂φ

)]=− ~2

[∂2

∂θ2+ i csc2θ

∂

∂φ− i cot θ

∂

∂θ

∂

∂φ

+ cot θ

(∂

∂θ− i cot θ

∂

∂φ

)+ i cot θ

∂

∂φ

∂

∂θ+ cot2θ

∂2

∂φ2

]

=− ~2[

∂2


∂

∂φ+ cot θ

∂

∂θ+ cot2θ

∂2

∂φ2

]=− ~2

(∂2

∂θ2+ i

∂

∂φ+ cot θ

∂

∂θ+ cot2θ

∂2

∂φ2




L+L− =− ~2e iφ(∂

∂θ+ i cot θ

∂

∂φ

)[e−iφ

(∂

∂θ− i cot θ

∂

∂φ

)]=− ~2

[∂2

∂θ2+ i csc2θ

∂

∂φ− i cot θ

∂

∂θ

∂

∂φ

+ cot θ

(∂

∂θ− i cot θ

∂

∂φ

)+ i cot θ

∂

∂φ

∂

∂θ+ cot2θ

∂2

∂φ2

]=− ~2

[

∂2


∂

∂φ+ cot θ

∂

∂θ+ cot2θ

∂2

∂φ2

]

=− ~2(∂2

∂θ2+ i

∂

∂φ+ cot θ

∂

∂θ+ cot2θ

∂2

∂φ2




L+L− =− ~2e iφ(∂

∂θ+ i cot θ

∂

∂φ

)[e−iφ

(∂

∂θ− i cot θ

∂

∂φ

)]=− ~2

[∂2

∂θ2+ i csc2θ

∂

∂φ−��

��i cot θ

∂

∂θ

∂

∂φ

+ cot θ

(∂

∂θ− i cot θ

∂

∂φ

)+��

��

i cot θ∂

∂φ

∂

∂θ+ cot2θ

∂2

∂φ2

]

=− ~2[

∂2


∂

∂φ+ cot θ

∂

∂θ+ cot2θ

∂2

∂φ2

]

=− ~2(∂2

∂θ2+ i

∂

∂φ+ cot θ

∂

∂θ+ cot2θ

∂2

∂φ2




L+L− =− ~2e iφ(∂

∂θ+ i cot θ

∂

∂φ

)[e−iφ

(∂

∂θ− i cot θ

∂

∂φ

)]=− ~2

[∂2

∂θ2+ i csc2θ

∂

∂φ−��

��i cot θ

∂

∂θ

∂

∂φ

+ cot θ

(∂

∂θ− i cot θ

∂

∂φ

)+��

��

i cot θ∂

∂φ

∂

∂θ+ cot2θ

∂2

∂φ2

]

=− ~2[∂2

∂θ2

+ i(csc2θ − cot2θ)∂

∂φ+ cot θ

∂

∂θ+ cot2θ

∂2

∂φ2

]

=− ~2(∂2

∂θ2+ i

∂

∂φ+ cot θ

∂

∂θ+ cot2θ

∂2

∂φ2




L+L− =− ~2e iφ(∂

∂θ+ i cot θ

∂

∂φ

)[e−iφ

(∂

∂θ− i cot θ

∂

∂φ

)]=− ~2

[∂2

∂θ2+ i csc2θ

∂

∂φ−��

��i cot θ

∂

∂θ

∂

∂φ

+ cot θ

(∂

∂θ− i cot θ

∂

∂φ

)+��

��

i cot θ∂

∂φ

∂

∂θ+ cot2θ

∂2

∂φ2

]

=− ~2[∂2


∂

∂φ

+ cot θ∂

∂θ+ cot2θ

∂2

∂φ2

]

=− ~2(∂2

∂θ2+ i

∂

∂φ+ cot θ

∂

∂θ+ cot2θ

∂2

∂φ2




L+L− =− ~2e iφ(∂

∂θ+ i cot θ

∂

∂φ

)[e−iφ

(∂

∂θ− i cot θ

∂

∂φ

)]=− ~2

[∂2

∂θ2+ i csc2θ

∂

∂φ−��

��i cot θ

∂

∂θ

∂

∂φ

+ cot θ

(∂

∂θ− i cot θ

∂

∂φ

)+��

��

i cot θ∂

∂φ

∂

∂θ+ cot2θ

∂2

∂φ2

]

=− ~2[∂2


∂

∂φ+ cot θ

∂

∂θ

+ cot2θ∂2

∂φ2

]

=− ~2(∂2

∂θ2+ i

∂

∂φ+ cot θ

∂

∂θ+ cot2θ

∂2

∂φ2




L+L− =− ~2e iφ(∂

∂θ+ i cot θ

∂

∂φ

)[e−iφ

(∂

∂θ− i cot θ

∂

∂φ

)]=− ~2

[∂2

∂θ2+ i csc2θ

∂

∂φ−��

��i cot θ

∂

∂θ

∂

∂φ

+ cot θ

(∂

∂θ− i cot θ

∂

∂φ

)+��

��

i cot θ∂

∂φ

∂

∂θ+ cot2θ

∂2

∂φ2

]

=− ~2[∂2


∂

∂φ+ cot θ

∂

∂θ+ cot2θ

∂2

∂φ2

]

=− ~2(∂2

∂θ2+ i

∂

∂φ+ cot θ

∂

∂θ+ cot2θ

∂2

∂φ2




L+L− =− ~2e iφ(∂

∂θ+ i cot θ

∂

∂φ

)[e−iφ

(∂

∂θ− i cot θ

∂

∂φ

)]=− ~2

[∂2

∂θ2+ i csc2θ

∂

∂φ−��

��i cot θ

∂

∂θ

∂

∂φ

+ cot θ

(∂

∂θ− i cot θ

∂

∂φ

)+��

��

i cot θ∂

∂φ

∂

∂θ+ cot2θ

∂2

∂φ2

]

=− ~2[∂2


∂

∂φ+ cot θ

∂

∂θ+ cot2θ

∂2

∂φ2

]=− ~2

(∂2

∂θ2+ i

∂

∂φ+ cot θ

∂

∂θ+ cot2θ

∂2

∂φ2

)

Now it is possible to construct the spherical coordinates form of L2



L+L− =− ~2e iφ(∂

∂θ+ i cot θ

∂

∂φ

)[e−iφ

(∂

∂θ− i cot θ

∂

∂φ

)]=− ~2

[∂2

∂θ2+ i csc2θ

∂

∂φ−��

��i cot θ

∂

∂θ

∂

∂φ

+ cot θ

(∂

∂θ− i cot θ

∂

∂φ

)+��

��

i cot θ∂

∂φ

∂

∂θ+ cot2θ

∂2

∂φ2

]

=− ~2[∂2


∂

∂φ+ cot θ

∂

∂θ+ cot2θ

∂2

∂φ2

]=− ~2

(∂2

∂θ2+ i

∂

∂φ+ cot θ

∂

∂θ+ cot2θ

∂2

∂φ2



Problem 4.21(b) – solution

From eq 4.112

L2 = L+L− + L2z − ~Lz

= − ~2(∂2

∂θ2+ i

∂

∂φ+ cot θ

∂

∂θ+ cot2θ

∂2

∂φ2

)

− ~2∂2

∂φ2− ~

(~i

)∂

∂φ

= −~2(∂2

∂θ2+ cot θ

∂

∂θ+ (cot2θ + 1)

∂2

∂φ2

)= −~2

(∂2

∂θ2+ cot θ

∂

∂θ+

1

sin2θ

∂2

∂φ2

)= −~2

[1

sin θ

∂

∂θ

(sin θ

∂

∂θ

)+

1

sin2θ

∂2

∂φ2

]but this is just the separated angular equation for the hydrogen atom andthe solutions are the spherical harmonics

L2Yml = ~2l(l + 1)Ym

l

LzYml = ~mYm

l



From eq 4.112

L2 = L+L− + L2z − ~Lz

= − ~2(∂2

∂θ2+ i

∂

∂φ+ cot θ

∂

∂θ+ cot2θ

∂2

∂φ2

)− ~2

∂2

∂φ2

− ~(~i

)∂

∂φ

= −~2(∂2

∂θ2+ cot θ

∂

∂θ+ (cot2θ + 1)

∂2

∂φ2

)= −~2

(∂2

∂θ2+ cot θ

∂

∂θ+

1

sin2θ

∂2

∂φ2

)= −~2

[1

sin θ

∂

∂θ

(sin θ

∂

∂θ

)+

1

sin2θ

∂2

∂φ2



l

LzYml = ~mYm

l



From eq 4.112

L2 = L+L− + L2z − ~Lz

= − ~2(∂2

∂θ2+ i

∂

∂φ+ cot θ

∂

∂θ+ cot2θ

∂2

∂φ2

)− ~2

∂2

∂φ2− ~

(~i

)∂

∂φ

= −~2(∂2

∂θ2+ cot θ

∂

∂θ+ (cot2θ + 1)

∂2

∂φ2

)= −~2

(∂2

∂θ2+ cot θ

∂

∂θ+

1

sin2θ

∂2

∂φ2

)= −~2

[1

sin θ

∂

∂θ

(sin θ

∂

∂θ

)+

1

sin2θ

∂2

∂φ2



l

LzYml = ~mYm

l



From eq 4.112

L2 = L+L− + L2z − ~Lz

= − ~2(∂2

∂θ2+

��

i∂

∂φ+ cot θ

∂

∂θ+ cot2θ

∂2

∂φ2

)− ~2

∂2

∂φ2−��

��~(~i

)∂

∂φ

= −~2(∂2

∂θ2+ cot θ

∂

∂θ+ (cot2θ + 1)

∂2

∂φ2

)= −~2

(∂2

∂θ2+ cot θ

∂

∂θ+

1

sin2θ

∂2

∂φ2

)= −~2

[1

sin θ

∂

∂θ

(sin θ

∂

∂θ

)+

1

sin2θ

∂2

∂φ2



l

LzYml = ~mYm

l



From eq 4.112

L2 = L+L− + L2z − ~Lz

= − ~2(∂2

∂θ2+

��

i∂

∂φ+ cot θ

∂

∂θ+ cot2θ

∂2

∂φ2

)− ~2

∂2

∂φ2−��

��~(~i

)∂

∂φ

= −~2(∂2

∂θ2+ cot θ

∂

∂θ+ (cot2θ + 1)

∂2

∂φ2

)

= −~2(∂2

∂θ2+ cot θ

∂

∂θ+

1

sin2θ

∂2

∂φ2

)= −~2

[1

sin θ

∂

∂θ

(sin θ

∂

∂θ

)+

1

sin2θ

∂2

∂φ2



l

LzYml = ~mYm

l



From eq 4.112

L2 = L+L− + L2z − ~Lz

= − ~2(∂2

∂θ2+

��

i∂

∂φ+ cot θ

∂

∂θ+ cot2θ

∂2

∂φ2

)− ~2

∂2

∂φ2−��

��~(~i

)∂

∂φ

= −~2(∂2

∂θ2+ cot θ

∂

∂θ+ (cot2θ + 1)

∂2

∂φ2

)= −~2

(∂2

∂θ2+ cot θ

∂

∂θ+

1

sin2θ

∂2

∂φ2

)

= −~2[

1

sin θ

∂

∂θ

(sin θ

∂

∂θ

)+

1

sin2θ

∂2

∂φ2



l

LzYml = ~mYm

l



From eq 4.112

L2 = L+L− + L2z − ~Lz

= − ~2(∂2

∂θ2+

��

i∂

∂φ+ cot θ

∂

∂θ+ cot2θ

∂2

∂φ2

)− ~2

∂2

∂φ2−��

��~(~i

)∂

∂φ

= −~2(∂2

∂θ2+ cot θ

∂

∂θ+ (cot2θ + 1)

∂2

∂φ2

)= −~2

(∂2

∂θ2+ cot θ

∂

∂θ+

1

sin2θ

∂2

∂φ2

)= −~2

[1

sin θ

∂

∂θ

(sin θ

∂

∂θ

)+

1

sin2θ

∂2

∂φ2

]

but this is just the separated angular equation for the hydrogen atom andthe solutions are the spherical harmonics


l

LzYml = ~mYm

l



From eq 4.112

L2 = L+L− + L2z − ~Lz

= − ~2(∂2

∂θ2+

��

i∂

∂φ+ cot θ

∂

∂θ+ cot2θ

∂2

∂φ2

)− ~2

∂2

∂φ2−��

��~(~i

)∂

∂φ

= −~2(∂2

∂θ2+ cot θ

∂

∂θ+ (cot2θ + 1)

∂2

∂φ2

)= −~2

(∂2

∂θ2+ cot θ

∂

∂θ+

1

sin2θ

∂2

∂φ2

)= −~2

[1

sin θ

∂

∂θ

(sin θ

∂

∂θ

)+

1

sin2θ

∂2

∂φ2



l

LzYml = ~mYm

l



From eq 4.112

L2 = L+L− + L2z − ~Lz

= − ~2(∂2

∂θ2+

��

i∂

∂φ+ cot θ

∂

∂θ+ cot2θ

∂2

∂φ2

)− ~2

∂2

∂φ2−��

��~(~i

)∂

∂φ

= −~2(∂2

∂θ2+ cot θ

∂

∂θ+ (cot2θ + 1)

∂2

∂φ2

)= −~2

(∂2

∂θ2+ cot θ

∂

∂θ+

1

sin2θ

∂2

∂φ2

)= −~2

[1

sin θ

∂

∂θ

(sin θ

∂

∂θ

)+

1

sin2θ

∂2

∂φ2



l LzYml = ~mYm

l


Intrinsic angular momentum

In quantum mechanics, we deal with small particles, which can, like anelectron be described as “point” particles. We have seen the orbitalangular momentum, characterized by l and m. These particles also carryanother intrinsic form of angular momentum which we call “spin” angularmomentum even though it is not associated with any physical variables ofrotation

Spin behaves in all ways as orbital angular momentum

the spin operator obeyscommutation relations

there exist eigenvectors ofthe spin operator

and there are ladder oper-ators

there are no restrictionsforcing integer spin

[Sx ,Sy ] = i~SzS2|s,m〉 = ~2s(s + 1)|s,m〉Sz |s,m〉 = ~m|s,m〉

S±|s,m〉 = ~√

s(s + 1)−m(m ± 1)|s,m ± 1〉s = 0, 12 , 1,

32 , . . .

m = −s,−s + 1, . . . , s − 1, s










S±|s,m〉 = ~√

s(s + 1)−m(m ± 1)|s,m ± 1〉s = 0, 12 , 1,

32 , . . .

m = −s,−s + 1, . . . , s − 1, s










S±|s,m〉 = ~√

s(s + 1)−m(m ± 1)|s,m ± 1〉

s = 0, 12 , 1,32 , . . .

m = −s,−s + 1, . . . , s − 1, s










S±|s,m〉 = ~√

s(s + 1)−m(m ± 1)|s,m ± 1〉s = 0, 12 , 1,

32 , . . .

m = −s,−s + 1, . . . , s − 1, s


Spin 12

Simplest system to work out, only2 states

a better representation is that ofa two element vector called a“spinor”

with a general state represented by

operators are 2× 2 matrices

consider the S2 operator

S2 =

(c de f

)

∣∣12 + 1

2

⟩, spin up∣∣1

2 −12

⟩, spin down

χ+ =

(10

), χ− =

(01

)

χ =

(ab

)= aχ+ + bχ−

S2χ± = s(s + 1)~2χ± = 34~

2χ±


Spin 12






S2 =

(c de f

)

∣∣12 + 1

2

⟩, spin up

∣∣12 −

12

⟩, spin down

χ+ =

(10

), χ− =

(01

)

χ =

(ab

)= aχ+ + bχ−

S2χ± = s(s + 1)~2χ± = 34~

2χ±


Spin 12






S2 =

(c de f

)

∣∣12 + 1

2

⟩, spin up∣∣1

2 −12

⟩, spin down

χ+ =

(10

), χ− =

(01

)

χ =

(ab

)= aχ+ + bχ−

S2χ± = s(s + 1)~2χ± = 34~

2χ±


Spin 12






S2 =

(c de f

)

∣∣12 + 1

2

⟩, spin up∣∣1

2 −12

⟩, spin down

χ+ =

(10

)

, χ− =

(01

)

χ =

(ab

)= aχ+ + bχ−

S2χ± = s(s + 1)~2χ± = 34~

2χ±


Spin 12






S2 =

(c de f

)

∣∣12 + 1

2

⟩, spin up∣∣1

2 −12

⟩, spin down

χ+ =

(10

), χ− =

(01

)

χ =

(ab

)= aχ+ + bχ−

S2χ± = s(s + 1)~2χ± = 34~

2χ±


Spin 12






S2 =

(c de f

)

∣∣12 + 1

2

⟩, spin up∣∣1

2 −12

⟩, spin down

χ+ =

(10

), χ− =

(01

)

χ =

(ab

)= aχ+ + bχ−

S2χ± = s(s + 1)~2χ±

= 34~

2χ±


Spin 12






S2 =

(c de f

)

∣∣12 + 1

2

⟩, spin up∣∣1

2 −12

⟩, spin down

χ+ =

(10

), χ− =

(01

)

χ =

(ab

)= aχ+ + bχ−

S2χ± = s(s + 1)~2χ± = 34~

2χ±


S2 matrix

S2χ+ =3

4~2χ+

(c de f

)(10

)=

3

4~2(

10

)(

ce

)=

(34~

2

0

)c =

3

4~2, e = 0

S2χ− =3

4~2χ−

(c de f

)(01

)=

3

4~2(

01

)(

df

)=

(0

34~

2

)d = 0, f =

3

4~2

S2 =3

4~2(

1 00 1

)


S2 matrix

S2χ+ =3

4~2χ+

(c de f

)(10

)=

3

4~2(

10

)

(ce

)=

(34~

2

0

)c =

3

4~2, e = 0

S2χ− =3

4~2χ−

(c de f

)(01

)=

3

4~2(

01

)(

df

)=

(0

34~

2

)d = 0, f =

3

4~2

S2 =3

4~2(

1 00 1

)


S2 matrix

S2χ+ =3

4~2χ+

(c de f

)(10

)=

3

4~2(

10

)(

ce

)=

(34~

2

0

)

c =3

4~2, e = 0

S2χ− =3

4~2χ−

(c de f

)(01

)=

3

4~2(

01

)(

df

)=

(0

34~

2

)d = 0, f =

3

4~2

S2 =3

4~2(

1 00 1

)


S2 matrix

S2χ+ =3

4~2χ+

(c de f

)(10

)=

3

4~2(

10

)(

ce

)=

(34~

2

0

)c =

3

4~2, e = 0

S2χ− =3

4~2χ−

(c de f

)(01

)=

3

4~2(

01

)(

df

)=

(0

34~

2

)d = 0, f =

3

4~2

S2 =3

4~2(

1 00 1

)


S2 matrix

S2χ+ =3

4~2χ+

(c de f

)(10

)=

3

4~2(

10

)(

ce

)=

(34~

2

0

)c =

3

4~2, e = 0

S2χ− =3

4~2χ−

(c de f

)(01

)=

3

4~2(

01

)

(df

)=

(0

34~

2

)d = 0, f =

3

4~2

S2 =3

4~2(

1 00 1

)


S2 matrix

S2χ+ =3

4~2χ+

(c de f

)(10

)=

3

4~2(

10

)(

ce

)=

(34~

2

0

)c =

3

4~2, e = 0

S2χ− =3

4~2χ−

(c de f

)(01

)=

3

4~2(

01

)(

df

)=

(0

34~

2

)

d = 0, f =3

4~2

S2 =3

4~2(

1 00 1

)


S2 matrix

S2χ+ =3

4~2χ+

(c de f

)(10

)=

3

4~2(

10

)(

ce

)=

(34~

2

0

)c =

3

4~2, e = 0

S2χ− =3

4~2χ−

(c de f

)(01

)=

3

4~2(

01

)(

df

)=

(0

34~

2

)d = 0, f =

3

4~2

S2 =3

4~2(

1 00 1

)


Sz matrix

Szχ+ =~2χ+

(c de f

)(10

)=

~2

(10

)(

ce

)=

( ~20

)c =

~2, e = 0

Szχ− = −~2χ−

(c de f

)(01

)= −~

2

(01

)(

df

)=

(0

−~2

)d = 0, f = −~

2

Sz =~2

(1 00 −1

)


Sz matrix

Szχ+ =~2χ+

(c de f

)(10

)=

~2

(10

)

(ce

)=

( ~20

)c =

~2, e = 0

Szχ− = −~2χ−

(c de f

)(01

)= −~

2

(01

)(

df

)=

(0

−~2

)d = 0, f = −~

2

Sz =~2

(1 00 −1

)


Sz matrix

Szχ+ =~2χ+

(c de f

)(10

)=

~2

(10

)(

ce

)=

( ~20

)

c =~2, e = 0

Szχ− = −~2χ−

(c de f

)(01

)= −~

2

(01

)(

df

)=

(0

−~2

)d = 0, f = −~

2

Sz =~2

(1 00 −1

)


Sz matrix

Szχ+ =~2χ+

(c de f

)(10

)=

~2

(10

)(

ce

)=

( ~20

)c =

~2, e = 0

Szχ− = −~2χ−

(c de f

)(01

)= −~

2

(01

)(

df

)=

(0

−~2

)d = 0, f = −~

2

Sz =~2

(1 00 −1

)


Sz matrix

Szχ+ =~2χ+

(c de f

)(10

)=

~2

(10

)(

ce

)=

( ~20

)c =

~2, e = 0

Szχ− = −~2χ−

(c de f

)(01

)= −~

2

(01

)

(df

)=

(0

−~2

)d = 0, f = −~

2

Sz =~2

(1 00 −1

)


Sz matrix

Szχ+ =~2χ+

(c de f

)(10

)=

~2

(10

)(

ce

)=

( ~20

)c =

~2, e = 0

Szχ− = −~2χ−

(c de f

)(01

)= −~

2

(01

)(

df

)=

(0

−~2

)

d = 0, f = −~2

Sz =~2

(1 00 −1

)


Sz matrix

Szχ+ =~2χ+

(c de f

)(10

)=

~2

(10

)(

ce

)=

( ~20

)c =

~2, e = 0

Szχ− = −~2χ−

(c de f

)(01

)= −~

2

(01

)(

df

)=

(0

−~2

)d = 0, f = −~

2

Sz =~2

(1 00 −1

)


Spin ladder matrices

We can now proceed to find the matrix representation of the spin raisingand lowering operators. We take the general definition and apply it to thecase where S = 1

2

S±|s,m〉 = ~√

s(s + 1)−m(m ± 1)|s,m ± 1〉

S+χ− = ~√

12

(12 + 1

)−(−1

2

) (−1

2 + 1)χ+

= ~√

34 + 1

4χ+ = ~χ+

S−χ+ = ~√

12

(12 + 1

)−(12

) (12 − 1

)χ−

= ~√

34 + 1

4χ− = ~χ−

(c de f

)(01

)= ~

(10

)(

c de f

)(10

)= ~

(00

)d = ~, c , e, f = 0

thus the two operators become

S+ = ~(

0 10 0

), S− = ~

(0 01 0

)




2

S±|s,m〉 = ~√

s(s + 1)−m(m ± 1)|s,m ± 1〉

S+χ− = ~√

12

(12 + 1

)−(−1

2

) (−1

2 + 1)χ+ = ~

√34 + 1

4χ+

= ~χ+

S−χ+ = ~√

12

(12 + 1

)−(12

) (12 − 1

)χ−

= ~√

34 + 1

4χ− = ~χ−

(c de f

)(01

)= ~

(10

)(

c de f

)(10

)= ~

(00

)d = ~, c , e, f = 0


S+ = ~(

0 10 0

), S− = ~

(0 01 0

)




2

S±|s,m〉 = ~√

s(s + 1)−m(m ± 1)|s,m ± 1〉

S+χ− = ~√

12

(12 + 1

)−(−1

2

) (−1

2 + 1)χ+ = ~

√34 + 1

4χ+ = ~χ+

S−χ+ = ~√

12

(12 + 1

)−(12

) (12 − 1

)χ−

= ~√

34 + 1

4χ− = ~χ−

(c de f

)(01

)= ~

(10

)(

c de f

)(10

)= ~

(00

)d = ~, c , e, f = 0


S+ = ~(

0 10 0

), S− = ~

(0 01 0

)




2

S±|s,m〉 = ~√

s(s + 1)−m(m ± 1)|s,m ± 1〉

S+χ− = ~√

12

(12 + 1

)−(−1

2

) (−1

2 + 1)χ+ = ~

√34 + 1

4χ+ = ~χ+

S−χ+ = ~√

12

(12 + 1

)−(12

) (12 − 1

)χ− = ~

√34 + 1

4χ−

= ~χ−

(c de f

)(01

)= ~

(10

)(

c de f

)(10

)= ~

(00

)d = ~, c , e, f = 0


S+ = ~(

0 10 0

), S− = ~

(0 01 0

)




2

S±|s,m〉 = ~√

s(s + 1)−m(m ± 1)|s,m ± 1〉

S+χ− = ~√

12

(12 + 1

)−(−1

2

) (−1

2 + 1)χ+ = ~

√34 + 1

4χ+ = ~χ+

S−χ+ = ~√

12

(12 + 1

)−(12

) (12 − 1

)χ− = ~

√34 + 1

4χ− = ~χ−

(c de f

)(01

)= ~

(10

)(

c de f

)(10

)= ~

(00

)d = ~, c , e, f = 0


S+ = ~(

0 10 0

), S− = ~

(0 01 0

)




2

S±|s,m〉 = ~√

s(s + 1)−m(m ± 1)|s,m ± 1〉

S+χ− = ~√

12

(12 + 1

)−(−1

2

) (−1

2 + 1)χ+ = ~

√34 + 1

4χ+ = ~χ+

S−χ+ = ~√

12

(12 + 1

)−(12

) (12 − 1

)χ− = ~

√34 + 1

4χ− = ~χ−

(c de f

)(01

)= ~

(10

)

(c de f

)(10

)= ~

(00

)d = ~, c , e, f = 0


S+ = ~(

0 10 0

), S− = ~

(0 01 0

)




2

S±|s,m〉 = ~√

s(s + 1)−m(m ± 1)|s,m ± 1〉

S+χ− = ~√

12

(12 + 1

)−(−1

2

) (−1

2 + 1)χ+ = ~

√34 + 1

4χ+ = ~χ+

S−χ+ = ~√

12

(12 + 1

)−(12

) (12 − 1

)χ− = ~

√34 + 1

4χ− = ~χ−

(c de f

)(01

)= ~

(10

)(

c de f

)(10

)= ~

(00

)

d = ~, c , e, f = 0


S+ = ~(

0 10 0

), S− = ~

(0 01 0

)




2

S±|s,m〉 = ~√

s(s + 1)−m(m ± 1)|s,m ± 1〉

S+χ− = ~√

12

(12 + 1

)−(−1

2

) (−1

2 + 1)χ+ = ~

√34 + 1

4χ+ = ~χ+

S−χ+ = ~√

12

(12 + 1

)−(12

) (12 − 1

)χ− = ~

√34 + 1

4χ− = ~χ−

(c de f

)(01

)= ~

(10

)(

c de f

)(10

)= ~

(00

)d = ~, c , e, f = 0


S+ = ~(

0 10 0

), S− = ~

(0 01 0

)




2

S±|s,m〉 = ~√

s(s + 1)−m(m ± 1)|s,m ± 1〉

S+χ− = ~√

12

(12 + 1

)−(−1

2

) (−1

2 + 1)χ+ = ~

√34 + 1

4χ+ = ~χ+

S−χ+ = ~√

12

(12 + 1

)−(12

) (12 − 1

)χ− = ~

√34 + 1

4χ− = ~χ−

(c de f

)(01

)= ~

(10

)(

c de f

)(10

)= ~

(00

)d = ~, c , e, f = 0


S+ = ~(

0 10 0

)

, S− = ~(

0 01 0

)




2

S±|s,m〉 = ~√

s(s + 1)−m(m ± 1)|s,m ± 1〉

S+χ− = ~√

12

(12 + 1

)−(−1

2

) (−1

2 + 1)χ+ = ~

√34 + 1

4χ+ = ~χ+

S−χ+ = ~√

12

(12 + 1

)−(12

) (12 − 1

)χ− = ~

√34 + 1

4χ− = ~χ−

(c de f

)(01

)= ~

(10

)(

c de f

)(10

)= ~

(00

)d = ~, c , e, f = 0


S+ = ~(

0 10 0

), S− = ~

(0 01 0

)


Pauli matrices

Recall that the raising and lower-ing operators were related to the x-and y-components of angular mo-mentum. This is the same for spin.

the matrix representations are thus

Since each component of ~S has aprefactor of ~/2, it is more commonto write the components of the spinusing the Pauli spin matrices

S± = Sx ± iSy

Sx =1

2(S+ + S−)

=~2

(0 11 0

)

Sy =1

2i(S+ − S−)

=~2

(0 −ii 0

)

~S =~2~σ

σx ≡(

0 11 0

)σy ≡

(0 −ii 0

)σz ≡

(1 00 −1

)


Pauli matrices




S± = Sx ± iSy

Sx =1

2(S+ + S−)

=~2

(0 11 0

)Sy =

1

2i(S+ − S−)

=~2

(0 −ii 0

)

~S =~2~σ

σx ≡(

0 11 0

)σy ≡

(0 −ii 0

)σz ≡

(1 00 −1

)


Pauli matrices




S± = Sx ± iSy

Sx =1

2(S+ + S−)

=~2

(0 11 0

)

Sy =1

2i(S+ − S−)

=~2

(0 −ii 0

)~S =

~2~σ

σx ≡(

0 11 0

)σy ≡

(0 −ii 0

)σz ≡

(1 00 −1

)


Pauli matrices




S± = Sx ± iSy

Sx =1

2(S+ + S−) =

~2

(0 11 0

)Sy =

1

2i(S+ − S−)

=~2

(0 −ii 0

)~S =

~2~σ

σx ≡(

0 11 0

)σy ≡

(0 −ii 0

)σz ≡

(1 00 −1

)


Pauli matrices




S± = Sx ± iSy

Sx =1

2(S+ + S−) =

~2

(0 11 0

)Sy =

1

2i(S+ − S−) =

~2

(0 −ii 0

)

~S =~2~σ

σx ≡(

0 11 0

)σy ≡

(0 −ii 0

)σz ≡

(1 00 −1

)


Pauli matrices




S± = Sx ± iSy

Sx =1

2(S+ + S−) =

~2

(0 11 0

)Sy =

1

2i(S+ − S−) =

~2

(0 −ii 0

)~S =

~2~σ

σx ≡(

0 11 0

)σy ≡

(0 −ii 0

)σz ≡

(1 00 −1

)


Pauli matrices




S± = Sx ± iSy

Sx =1

2(S+ + S−) =

~2

(0 11 0

)Sy =

1

2i(S+ − S−) =

~2

(0 −ii 0

)~S =

~2~σ

σx ≡(

0 11 0

)

σy ≡(

0 −ii 0

)σz ≡

(1 00 −1

)


Pauli matrices




S± = Sx ± iSy

Sx =1

2(S+ + S−) =

~2

(0 11 0

)Sy =

1

2i(S+ − S−) =

~2

(0 −ii 0

)~S =

~2~σ

σx ≡(

0 11 0

)σy ≡

(0 −ii 0

)

σz ≡(

1 00 −1

)


Pauli matrices




S± = Sx ± iSy

Sx =1

2(S+ + S−) =

~2

(0 11 0

)Sy =

1

2i(S+ − S−) =

~2

(0 −ii 0

)~S =

~2~σ

σx ≡(

0 11 0

)σy ≡

(0 −ii 0

)σz ≡

(1 00 −1

)


Sx eigenvalues and eigenvectors

What will be observed when measuring Sx? By a symmetry argument, onemight expect to get the same values as for Sz . Let’s see...

starting with the eigenvalue equation

Sxχ = λχ

the eigenvectors are obtained by as-suming

χ =

(αβ

)0 = det

∣∣∣∣ −λ ~/2~/2 −λ

∣∣∣∣0 = λ2 −

(~2

)2

λ = ±~2

~2

(0 11 0

)(αβ

)= ±~

2

(αβ

)→

(βα

)= ±

(αβ

)

β = ±α → χ(x)+ =

1√2

(11

), χ

(x)− =

1√2

(1−1

)





Sxχ = λχ


χ =

(αβ

)

0 = det

∣∣∣∣ −λ ~/2~/2 −λ

∣∣∣∣

0 = λ2 −(~2

)2

λ = ±~2

~2

(0 11 0

)(αβ

)= ±~

2

(αβ

)→

(βα

)= ±

(αβ

)

β = ±α → χ(x)+ =

1√2

(11

), χ

(x)− =

1√2

(1−1

)





Sxχ = λχ


χ =

(αβ

)

0 = det

∣∣∣∣ −λ ~/2~/2 −λ

∣∣∣∣0 = λ2 −

(~2

)2

λ = ±~2

~2

(0 11 0

)(αβ

)= ±~

2

(αβ

)→

(βα

)= ±

(αβ

)

β = ±α → χ(x)+ =

1√2

(11

), χ

(x)− =

1√2

(1−1

)





Sxχ = λχ


χ =

(αβ

)0 = det

∣∣∣∣ −λ ~/2~/2 −λ

∣∣∣∣0 = λ2 −

(~2

)2

λ = ±~2

~2

(0 11 0

)(αβ

)= ±~

2

(αβ

)→

(βα

)= ±

(αβ

)

β = ±α → χ(x)+ =

1√2

(11

), χ

(x)− =

1√2

(1−1

)





Sxχ = λχ


χ =

(αβ

)0 = det

∣∣∣∣ −λ ~/2~/2 −λ

∣∣∣∣0 = λ2 −

(~2

)2

λ = ±~2

~2

(0 11 0

)(αβ

)= ±~

2

(αβ

)

→(βα

)= ±

(αβ

)

β = ±α → χ(x)+ =

1√2

(11

), χ

(x)− =

1√2

(1−1

)





Sxχ = λχ


χ =

(αβ

)0 = det

∣∣∣∣ −λ ~/2~/2 −λ

∣∣∣∣0 = λ2 −

(~2

)2

λ = ±~2

~2

(0 11 0

)(αβ

)= ±~

2

(αβ

)→

(βα

)= ±

(αβ

)

β = ±α → χ(x)+ =

1√2

(11

), χ

(x)− =

1√2

(1−1

)





Sxχ = λχ


χ =

(αβ

)0 = det

∣∣∣∣ −λ ~/2~/2 −λ

∣∣∣∣0 = λ2 −

(~2

)2

λ = ±~2

~2

(0 11 0

)(αβ

)= ±~

2

(αβ

)→

(βα

)= ±

(αβ

)

β = ±α

→ χ(x)+ =

1√2

(11

), χ

(x)− =

1√2

(1−1

)





Sxχ = λχ


χ =

(αβ

)0 = det

∣∣∣∣ −λ ~/2~/2 −λ

∣∣∣∣0 = λ2 −

(~2

)2

λ = ±~2

~2

(0 11 0

)(αβ

)= ±~

2

(αβ

)→

(βα

)= ±

(αβ

)

β = ±α → χ(x)+ =

1√2

(11

)

, χ(x)− =

1√2

(1−1

)





Sxχ = λχ


χ =

(αβ

)0 = det

∣∣∣∣ −λ ~/2~/2 −λ

∣∣∣∣0 = λ2 −

(~2

)2

λ = ±~2

~2

(0 11 0

)(αβ

)= ±~

2

(αβ

)→

(βα

)= ±

(αβ

)

β = ±α → χ(x)+ =

1√2

(11

), χ

(x)− =

1√2

(1−1

)C. Segre (IIT) PHYS 405 - Fall 2017 October 17, 2017 12 / 21

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−iE+t/~ + bχ−e

−iE−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


−iE−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


−iE−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


−iE−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


−iE−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


−iE−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


−iE−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


−iE−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

(α2

), b = sin

(α2

)

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

(α2

)

χ(t) =





=(


2

(0 11 0



)=

~2



cos(α2 )e iγB0t/2

)=

~2


)=

~2

sinα cos(γB0t)






χ(0) =

(ab

)a = cos

(α2

), b = sin

(α2

)

χ(t) =





=(


2

(0 11 0



)=

~2



cos(α2 )e iγB0t/2

)=

~2


)=

~2

sinα cos(γB0t)






χ(0) =

(ab

)a = cos

(α2

), b = sin

(α2

)

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

(α2

), b = sin

(α2

)

χ(t) =





=(


2

(0 11 0



)=

~2



cos(α2 )e iγB0t/2

)=

~2


)=

~2

sinα cos(γB0t)






χ(0) =

(ab

)a = cos

(α2

), b = sin

(α2

)

χ(t) =





=(


2

(0 11 0



)

=~2



cos(α2 )e iγB0t/2

)=

~2


)=

~2

sinα cos(γB0t)






χ(0) =

(ab

)a = cos

(α2

), b = sin

(α2

)

χ(t) =





=(


2

(0 11 0



)=

~2



cos(α2 )e iγB0t/2

)

=~2


)=

~2

sinα cos(γB0t)






χ(0) =

(ab

)a = cos

(α2

), b = sin

(α2

)

χ(t) =





=(


2

(0 11 0



)=

~2



cos(α2 )e iγB0t/2

)=

~2


)

=~2

sinα cos(γB0t)






χ(0) =

(ab

)a = cos

(α2

), b = sin

(α2

)

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


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?



Starting with the eigenvalue equation

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

)




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

)




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

)




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

)




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

)




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

)




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

)




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

)




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

)



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 probabilities as follows.

Theprobabilities sum to 1 if χ is normalized.

+~2, P+ =

1

2|a− ib|2 − ~

2, P− =

1

2|a + ib|2




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




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




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




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




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




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




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)

Thus we will measure the eigenvalues with probabilities as follows. Theprobabilities sum to 1 if χ is normalized.

+~2, P+ =

1

2|a− ib|2 − ~

2, P− =

1

2|a + ib|2


Problem 4.29(c) – solution

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






(+~2c+χ

(y)+ −

~2c−χ

(y)−

)= χ†

(~2

4c+χ

(y)+ +

~2

4c−χ

(y)−

)

= χ†~2

4χ =

~2

4







(+~2c+χ

(y)+ −

~2c−χ

(y)−

)= χ†

(~2

4c+χ

(y)+ +

~2

4c−χ

(y)−

)= χ†

~2

4χ

=~2

4







(+~2c+χ

(y)+ −

~2c−χ

(y)−

)= χ†

(~2

4c+χ

(y)+ +

~2

4c−χ

(y)−

)= χ†

~2

4χ =

~2

4



Today’s Outline - October 17, 2017csrri.iit.edu/~segre/phys405/17F/lecture_16.pdf · Today’s...

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