Post on 06-May-2018
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
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
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
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
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
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
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
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
C. Segre (IIT) PHYS 405 - Fall 2017 October 17, 2017 2 / 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
C. Segre (IIT) PHYS 405 - Fall 2017 October 17, 2017 2 / 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
C. Segre (IIT) PHYS 405 - Fall 2017 October 17, 2017 2 / 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
C. Segre (IIT) PHYS 405 - Fall 2017 October 17, 2017 2 / 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
C. Segre (IIT) PHYS 405 - Fall 2017 October 17, 2017 2 / 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
C. Segre (IIT) PHYS 405 - Fall 2017 October 17, 2017 2 / 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
C. Segre (IIT) PHYS 405 - Fall 2017 October 17, 2017 2 / 21
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
]
C. Segre (IIT) PHYS 405 - Fall 2017 October 17, 2017 3 / 21
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
C. Segre (IIT) PHYS 405 - Fall 2017 October 17, 2017 4 / 21
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
C. Segre (IIT) PHYS 405 - Fall 2017 October 17, 2017 4 / 21
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
C. Segre (IIT) PHYS 405 - Fall 2017 October 17, 2017 4 / 21
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
C. Segre (IIT) PHYS 405 - Fall 2017 October 17, 2017 4 / 21
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
C. Segre (IIT) PHYS 405 - Fall 2017 October 17, 2017 4 / 21
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
C. Segre (IIT) PHYS 405 - Fall 2017 October 17, 2017 4 / 21
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
C. Segre (IIT) PHYS 405 - Fall 2017 October 17, 2017 4 / 21
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
C. Segre (IIT) PHYS 405 - Fall 2017 October 17, 2017 4 / 21
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
C. Segre (IIT) PHYS 405 - Fall 2017 October 17, 2017 4 / 21
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
C. Segre (IIT) PHYS 405 - Fall 2017 October 17, 2017 4 / 21
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
C. Segre (IIT) PHYS 405 - Fall 2017 October 17, 2017 4 / 21
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
C. Segre (IIT) PHYS 405 - Fall 2017 October 17, 2017 4 / 21
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
C. Segre (IIT) PHYS 405 - Fall 2017 October 17, 2017 4 / 21
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
C. Segre (IIT) PHYS 405 - Fall 2017 October 17, 2017 4 / 21
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
C. Segre (IIT) PHYS 405 - Fall 2017 October 17, 2017 4 / 21
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
C. Segre (IIT) PHYS 405 - Fall 2017 October 17, 2017 4 / 21
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
C. Segre (IIT) PHYS 405 - Fall 2017 October 17, 2017 5 / 21
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
C. Segre (IIT) PHYS 405 - Fall 2017 October 17, 2017 5 / 21
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
C. Segre (IIT) PHYS 405 - Fall 2017 October 17, 2017 5 / 21
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
C. Segre (IIT) PHYS 405 - Fall 2017 October 17, 2017 5 / 21
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
C. Segre (IIT) PHYS 405 - Fall 2017 October 17, 2017 5 / 21
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
C. Segre (IIT) PHYS 405 - Fall 2017 October 17, 2017 5 / 21
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
C. Segre (IIT) PHYS 405 - Fall 2017 October 17, 2017 5 / 21
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
C. Segre (IIT) PHYS 405 - Fall 2017 October 17, 2017 5 / 21
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
C. Segre (IIT) PHYS 405 - Fall 2017 October 17, 2017 5 / 21
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
C. Segre (IIT) PHYS 405 - Fall 2017 October 17, 2017 5 / 21
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
C. Segre (IIT) PHYS 405 - Fall 2017 October 17, 2017 5 / 21
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
C. Segre (IIT) PHYS 405 - Fall 2017 October 17, 2017 6 / 21
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
C. Segre (IIT) PHYS 405 - Fall 2017 October 17, 2017 6 / 21
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
C. Segre (IIT) PHYS 405 - Fall 2017 October 17, 2017 6 / 21
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~Sz
S2|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
C. Segre (IIT) PHYS 405 - Fall 2017 October 17, 2017 6 / 21
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~Sz
S2|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
C. Segre (IIT) PHYS 405 - Fall 2017 October 17, 2017 6 / 21
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
C. Segre (IIT) PHYS 405 - Fall 2017 October 17, 2017 6 / 21
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
C. Segre (IIT) PHYS 405 - Fall 2017 October 17, 2017 6 / 21
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
C. Segre (IIT) PHYS 405 - Fall 2017 October 17, 2017 6 / 21
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
C. Segre (IIT) PHYS 405 - Fall 2017 October 17, 2017 6 / 21
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
C. Segre (IIT) PHYS 405 - Fall 2017 October 17, 2017 6 / 21
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
C. Segre (IIT) PHYS 405 - Fall 2017 October 17, 2017 6 / 21
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±
C. Segre (IIT) PHYS 405 - Fall 2017 October 17, 2017 7 / 21
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
∣∣12 −
12
⟩, spin down
χ+ =
(10
), χ− =
(01
)
χ =
(ab
)= aχ+ + bχ−
S2χ± = s(s + 1)~2χ± = 34~
2χ±
C. Segre (IIT) PHYS 405 - Fall 2017 October 17, 2017 7 / 21
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±
C. Segre (IIT) PHYS 405 - Fall 2017 October 17, 2017 7 / 21
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±
C. Segre (IIT) PHYS 405 - Fall 2017 October 17, 2017 7 / 21
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±
C. Segre (IIT) PHYS 405 - Fall 2017 October 17, 2017 7 / 21
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±
C. Segre (IIT) PHYS 405 - Fall 2017 October 17, 2017 7 / 21
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±
C. Segre (IIT) PHYS 405 - Fall 2017 October 17, 2017 7 / 21
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±
C. Segre (IIT) PHYS 405 - Fall 2017 October 17, 2017 7 / 21
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±
C. Segre (IIT) PHYS 405 - Fall 2017 October 17, 2017 7 / 21
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±
C. Segre (IIT) PHYS 405 - Fall 2017 October 17, 2017 7 / 21
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±
C. Segre (IIT) PHYS 405 - Fall 2017 October 17, 2017 7 / 21
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±
C. Segre (IIT) PHYS 405 - Fall 2017 October 17, 2017 7 / 21
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
)
C. Segre (IIT) PHYS 405 - Fall 2017 October 17, 2017 8 / 21
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
)
C. Segre (IIT) PHYS 405 - Fall 2017 October 17, 2017 8 / 21
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
)
C. Segre (IIT) PHYS 405 - Fall 2017 October 17, 2017 8 / 21
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
)
C. Segre (IIT) PHYS 405 - Fall 2017 October 17, 2017 8 / 21
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
)
C. Segre (IIT) PHYS 405 - Fall 2017 October 17, 2017 8 / 21
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
)
C. Segre (IIT) PHYS 405 - Fall 2017 October 17, 2017 8 / 21
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
)
C. Segre (IIT) PHYS 405 - Fall 2017 October 17, 2017 8 / 21
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
)
C. Segre (IIT) PHYS 405 - Fall 2017 October 17, 2017 8 / 21
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
)
C. Segre (IIT) PHYS 405 - Fall 2017 October 17, 2017 9 / 21
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
)
C. Segre (IIT) PHYS 405 - Fall 2017 October 17, 2017 9 / 21
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
)
C. Segre (IIT) PHYS 405 - Fall 2017 October 17, 2017 9 / 21
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
)
C. Segre (IIT) PHYS 405 - Fall 2017 October 17, 2017 9 / 21
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
)
C. Segre (IIT) PHYS 405 - Fall 2017 October 17, 2017 9 / 21
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
)
C. Segre (IIT) PHYS 405 - Fall 2017 October 17, 2017 9 / 21
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
)
C. Segre (IIT) PHYS 405 - Fall 2017 October 17, 2017 9 / 21
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
)
C. Segre (IIT) PHYS 405 - Fall 2017 October 17, 2017 9 / 21
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
)
C. Segre (IIT) PHYS 405 - Fall 2017 October 17, 2017 10 / 21
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
)
C. Segre (IIT) PHYS 405 - Fall 2017 October 17, 2017 10 / 21
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
)
C. Segre (IIT) PHYS 405 - Fall 2017 October 17, 2017 10 / 21
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
)
C. Segre (IIT) PHYS 405 - Fall 2017 October 17, 2017 10 / 21
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
)
C. Segre (IIT) PHYS 405 - Fall 2017 October 17, 2017 10 / 21
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
)
C. Segre (IIT) PHYS 405 - Fall 2017 October 17, 2017 10 / 21
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
)
C. Segre (IIT) PHYS 405 - Fall 2017 October 17, 2017 10 / 21
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
)
C. Segre (IIT) PHYS 405 - Fall 2017 October 17, 2017 10 / 21
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
)
C. Segre (IIT) PHYS 405 - Fall 2017 October 17, 2017 10 / 21
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
)
C. Segre (IIT) PHYS 405 - Fall 2017 October 17, 2017 10 / 21
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
)
C. Segre (IIT) PHYS 405 - Fall 2017 October 17, 2017 10 / 21
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
)
C. Segre (IIT) PHYS 405 - Fall 2017 October 17, 2017 10 / 21
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
)
C. Segre (IIT) PHYS 405 - Fall 2017 October 17, 2017 10 / 21
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
)
C. Segre (IIT) PHYS 405 - Fall 2017 October 17, 2017 10 / 21
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
)
C. Segre (IIT) PHYS 405 - Fall 2017 October 17, 2017 11 / 21
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
)
C. Segre (IIT) PHYS 405 - Fall 2017 October 17, 2017 11 / 21
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
)
C. Segre (IIT) PHYS 405 - Fall 2017 October 17, 2017 11 / 21
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
)
C. Segre (IIT) PHYS 405 - Fall 2017 October 17, 2017 11 / 21
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
)
C. Segre (IIT) PHYS 405 - Fall 2017 October 17, 2017 11 / 21
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
)
C. Segre (IIT) PHYS 405 - Fall 2017 October 17, 2017 11 / 21
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
)
C. Segre (IIT) PHYS 405 - Fall 2017 October 17, 2017 11 / 21
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
)
C. Segre (IIT) PHYS 405 - Fall 2017 October 17, 2017 11 / 21
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
)
C. Segre (IIT) PHYS 405 - Fall 2017 October 17, 2017 11 / 21
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
)
C. Segre (IIT) PHYS 405 - Fall 2017 October 17, 2017 11 / 21
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
)
C. Segre (IIT) PHYS 405 - Fall 2017 October 17, 2017 11 / 21
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
)
C. Segre (IIT) PHYS 405 - Fall 2017 October 17, 2017 11 / 21
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
)
C. Segre (IIT) PHYS 405 - Fall 2017 October 17, 2017 12 / 21
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
)
C. Segre (IIT) PHYS 405 - Fall 2017 October 17, 2017 12 / 21
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
)
C. Segre (IIT) PHYS 405 - Fall 2017 October 17, 2017 12 / 21
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
)
C. Segre (IIT) PHYS 405 - Fall 2017 October 17, 2017 12 / 21
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
)
C. Segre (IIT) PHYS 405 - Fall 2017 October 17, 2017 12 / 21
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
)
C. Segre (IIT) PHYS 405 - Fall 2017 October 17, 2017 12 / 21
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
)
C. Segre (IIT) PHYS 405 - Fall 2017 October 17, 2017 12 / 21
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
)
C. Segre (IIT) PHYS 405 - Fall 2017 October 17, 2017 12 / 21
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
)
C. Segre (IIT) PHYS 405 - Fall 2017 October 17, 2017 12 / 21
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
)
C. Segre (IIT) PHYS 405 - Fall 2017 October 17, 2017 12 / 21
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
)
C. Segre (IIT) PHYS 405 - Fall 2017 October 17, 2017 12 / 21
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
)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
C. Segre (IIT) PHYS 405 - Fall 2017 October 17, 2017 13 / 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
C. Segre (IIT) PHYS 405 - Fall 2017 October 17, 2017 13 / 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
C. Segre (IIT) PHYS 405 - Fall 2017 October 17, 2017 13 / 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
C. Segre (IIT) PHYS 405 - Fall 2017 October 17, 2017 13 / 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
C. Segre (IIT) PHYS 405 - Fall 2017 October 17, 2017 13 / 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
C. Segre (IIT) PHYS 405 - Fall 2017 October 17, 2017 13 / 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
C. Segre (IIT) PHYS 405 - Fall 2017 October 17, 2017 13 / 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
C. Segre (IIT) PHYS 405 - Fall 2017 October 17, 2017 13 / 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
C. Segre (IIT) PHYS 405 - Fall 2017 October 17, 2017 13 / 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
C. Segre (IIT) PHYS 405 - Fall 2017 October 17, 2017 13 / 21
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
)
C. Segre (IIT) PHYS 405 - Fall 2017 October 17, 2017 14 / 21
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
)
C. Segre (IIT) PHYS 405 - Fall 2017 October 17, 2017 14 / 21
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
)
C. Segre (IIT) PHYS 405 - Fall 2017 October 17, 2017 14 / 21
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
)
C. Segre (IIT) PHYS 405 - Fall 2017 October 17, 2017 14 / 21
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
)
C. Segre (IIT) PHYS 405 - Fall 2017 October 17, 2017 14 / 21
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
)
C. Segre (IIT) PHYS 405 - Fall 2017 October 17, 2017 14 / 21
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
)
C. Segre (IIT) PHYS 405 - Fall 2017 October 17, 2017 14 / 21
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
)
C. Segre (IIT) PHYS 405 - Fall 2017 October 17, 2017 14 / 21
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
)
C. Segre (IIT) PHYS 405 - Fall 2017 October 17, 2017 14 / 21
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
)
C. Segre (IIT) PHYS 405 - Fall 2017 October 17, 2017 14 / 21
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
)
C. Segre (IIT) PHYS 405 - Fall 2017 October 17, 2017 14 / 21
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
)
C. Segre (IIT) PHYS 405 - Fall 2017 October 17, 2017 14 / 21
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
sin(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)
C. Segre (IIT) PHYS 405 - Fall 2017 October 17, 2017 15 / 21
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
(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
sin(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)
C. Segre (IIT) PHYS 405 - Fall 2017 October 17, 2017 15 / 21
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
(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
sin(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)
C. Segre (IIT) PHYS 405 - Fall 2017 October 17, 2017 15 / 21
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
sin(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)
C. Segre (IIT) PHYS 405 - Fall 2017 October 17, 2017 15 / 21
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
sin(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)
C. Segre (IIT) PHYS 405 - Fall 2017 October 17, 2017 15 / 21
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
sin(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)
C. Segre (IIT) PHYS 405 - Fall 2017 October 17, 2017 15 / 21
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
sin(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)
C. Segre (IIT) PHYS 405 - Fall 2017 October 17, 2017 15 / 21
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
sin(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)
C. Segre (IIT) PHYS 405 - Fall 2017 October 17, 2017 15 / 21
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
sin(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)
C. Segre (IIT) PHYS 405 - Fall 2017 October 17, 2017 15 / 21
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
sin(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)
C. Segre (IIT) PHYS 405 - Fall 2017 October 17, 2017 15 / 21
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
sin(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)
C. Segre (IIT) PHYS 405 - Fall 2017 October 17, 2017 15 / 21
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
sin(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)
C. Segre (IIT) PHYS 405 - Fall 2017 October 17, 2017 15 / 21
More expectation values
〈Sy 〉 = χ†(t)Syχ(t)
=(
cos(α2 )e−iγB0t/2 sin(α2 )e+iγB0t/2) ~
2
(0 −ii 0
)(cos(α2 )e iγB0t/2
sin(α2 )e−iγB0t/2
)=
~2
(cos(α2 )e−iγB0t/2 sin(α2 )e+iγB0t/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)
=(
cos(α2 )e−iγB0t/2 sin(α2 )e+iγB0t/2) ~
2
(1 00 −1
)(cos(α2 )e iγB0t/2
sin(α2 )e−iγB0t/2
)=
~2
(cos(α2 )e−iγB0t/2 sin(α2 )e+iγB0t/2
)( cos(α2 )e iγB0t/2
− sin(α2 )e−iγB0t/2
)=
~2
(cos2(α/2)− sin2(α/2)
)=
~2
cosα
C. Segre (IIT) PHYS 405 - Fall 2017 October 17, 2017 16 / 21
More expectation values
〈Sy 〉 = χ†(t)Syχ(t)
=(
cos(α2 )e−iγB0t/2 sin(α2 )e+iγB0t/2) ~
2
(0 −ii 0
)(cos(α2 )e iγB0t/2
sin(α2 )e−iγB0t/2
)
=~2
(cos(α2 )e−iγB0t/2 sin(α2 )e+iγB0t/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)
=(
cos(α2 )e−iγB0t/2 sin(α2 )e+iγB0t/2) ~
2
(1 00 −1
)(cos(α2 )e iγB0t/2
sin(α2 )e−iγB0t/2
)=
~2
(cos(α2 )e−iγB0t/2 sin(α2 )e+iγB0t/2
)( cos(α2 )e iγB0t/2
− sin(α2 )e−iγB0t/2
)=
~2
(cos2(α/2)− sin2(α/2)
)=
~2
cosα
C. Segre (IIT) PHYS 405 - Fall 2017 October 17, 2017 16 / 21
More expectation values
〈Sy 〉 = χ†(t)Syχ(t)
=(
cos(α2 )e−iγB0t/2 sin(α2 )e+iγB0t/2) ~
2
(0 −ii 0
)(cos(α2 )e iγB0t/2
sin(α2 )e−iγB0t/2
)=
~2
(cos(α2 )e−iγB0t/2 sin(α2 )e+iγB0t/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)
=(
cos(α2 )e−iγB0t/2 sin(α2 )e+iγB0t/2) ~
2
(1 00 −1
)(cos(α2 )e iγB0t/2
sin(α2 )e−iγB0t/2
)=
~2
(cos(α2 )e−iγB0t/2 sin(α2 )e+iγB0t/2
)( cos(α2 )e iγB0t/2
− sin(α2 )e−iγB0t/2
)=
~2
(cos2(α/2)− sin2(α/2)
)=
~2
cosα
C. Segre (IIT) PHYS 405 - Fall 2017 October 17, 2017 16 / 21
More expectation values
〈Sy 〉 = χ†(t)Syχ(t)
=(
cos(α2 )e−iγB0t/2 sin(α2 )e+iγB0t/2) ~
2
(0 −ii 0
)(cos(α2 )e iγB0t/2
sin(α2 )e−iγB0t/2
)=
~2
(cos(α2 )e−iγB0t/2 sin(α2 )e+iγB0t/2
)(−i sin(α2 )e−iγB0t/2
i cos(α2 )e iγB0t/2
)=
~2
cos(α/2) sin(α/2)(−ie−iγB0t + ie iγB0t
)
= −~2
sinα sin(γB0t)
〈Sz〉 = χ†(t)Szχ(t)
=(
cos(α2 )e−iγB0t/2 sin(α2 )e+iγB0t/2) ~
2
(1 00 −1
)(cos(α2 )e iγB0t/2
sin(α2 )e−iγB0t/2
)=
~2
(cos(α2 )e−iγB0t/2 sin(α2 )e+iγB0t/2
)( cos(α2 )e iγB0t/2
− sin(α2 )e−iγB0t/2
)=
~2
(cos2(α/2)− sin2(α/2)
)=
~2
cosα
C. Segre (IIT) PHYS 405 - Fall 2017 October 17, 2017 16 / 21
More expectation values
〈Sy 〉 = χ†(t)Syχ(t)
=(
cos(α2 )e−iγB0t/2 sin(α2 )e+iγB0t/2) ~
2
(0 −ii 0
)(cos(α2 )e iγB0t/2
sin(α2 )e−iγB0t/2
)=
~2
(cos(α2 )e−iγB0t/2 sin(α2 )e+iγB0t/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)
=(
cos(α2 )e−iγB0t/2 sin(α2 )e+iγB0t/2) ~
2
(1 00 −1
)(cos(α2 )e iγB0t/2
sin(α2 )e−iγB0t/2
)=
~2
(cos(α2 )e−iγB0t/2 sin(α2 )e+iγB0t/2
)( cos(α2 )e iγB0t/2
− sin(α2 )e−iγB0t/2
)=
~2
(cos2(α/2)− sin2(α/2)
)=
~2
cosα
C. Segre (IIT) PHYS 405 - Fall 2017 October 17, 2017 16 / 21
More expectation values
〈Sy 〉 = χ†(t)Syχ(t)
=(
cos(α2 )e−iγB0t/2 sin(α2 )e+iγB0t/2) ~
2
(0 −ii 0
)(cos(α2 )e iγB0t/2
sin(α2 )e−iγB0t/2
)=
~2
(cos(α2 )e−iγB0t/2 sin(α2 )e+iγB0t/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)
=(
cos(α2 )e−iγB0t/2 sin(α2 )e+iγB0t/2) ~
2
(1 00 −1
)(cos(α2 )e iγB0t/2
sin(α2 )e−iγB0t/2
)=
~2
(cos(α2 )e−iγB0t/2 sin(α2 )e+iγB0t/2
)( cos(α2 )e iγB0t/2
− sin(α2 )e−iγB0t/2
)=
~2
(cos2(α/2)− sin2(α/2)
)=
~2
cosα
C. Segre (IIT) PHYS 405 - Fall 2017 October 17, 2017 16 / 21
More expectation values
〈Sy 〉 = χ†(t)Syχ(t)
=(
cos(α2 )e−iγB0t/2 sin(α2 )e+iγB0t/2) ~
2
(0 −ii 0
)(cos(α2 )e iγB0t/2
sin(α2 )e−iγB0t/2
)=
~2
(cos(α2 )e−iγB0t/2 sin(α2 )e+iγB0t/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)
=(
cos(α2 )e−iγB0t/2 sin(α2 )e+iγB0t/2) ~
2
(1 00 −1
)(cos(α2 )e iγB0t/2
sin(α2 )e−iγB0t/2
)
=~2
(cos(α2 )e−iγB0t/2 sin(α2 )e+iγB0t/2
)( cos(α2 )e iγB0t/2
− sin(α2 )e−iγB0t/2
)=
~2
(cos2(α/2)− sin2(α/2)
)=
~2
cosα
C. Segre (IIT) PHYS 405 - Fall 2017 October 17, 2017 16 / 21
More expectation values
〈Sy 〉 = χ†(t)Syχ(t)
=(
cos(α2 )e−iγB0t/2 sin(α2 )e+iγB0t/2) ~
2
(0 −ii 0
)(cos(α2 )e iγB0t/2
sin(α2 )e−iγB0t/2
)=
~2
(cos(α2 )e−iγB0t/2 sin(α2 )e+iγB0t/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)
=(
cos(α2 )e−iγB0t/2 sin(α2 )e+iγB0t/2) ~
2
(1 00 −1
)(cos(α2 )e iγB0t/2
sin(α2 )e−iγB0t/2
)=
~2
(cos(α2 )e−iγB0t/2 sin(α2 )e+iγB0t/2
)( cos(α2 )e iγB0t/2
− sin(α2 )e−iγB0t/2
)
=~2
(cos2(α/2)− sin2(α/2)
)=
~2
cosα
C. Segre (IIT) PHYS 405 - Fall 2017 October 17, 2017 16 / 21
More expectation values
〈Sy 〉 = χ†(t)Syχ(t)
=(
cos(α2 )e−iγB0t/2 sin(α2 )e+iγB0t/2) ~
2
(0 −ii 0
)(cos(α2 )e iγB0t/2
sin(α2 )e−iγB0t/2
)=
~2
(cos(α2 )e−iγB0t/2 sin(α2 )e+iγB0t/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)
=(
cos(α2 )e−iγB0t/2 sin(α2 )e+iγB0t/2) ~
2
(1 00 −1
)(cos(α2 )e iγB0t/2
sin(α2 )e−iγB0t/2
)=
~2
(cos(α2 )e−iγB0t/2 sin(α2 )e+iγB0t/2
)( cos(α2 )e iγB0t/2
− sin(α2 )e−iγB0t/2
)=
~2
(cos2(α/2)− sin2(α/2)
)
=~2
cosα
C. Segre (IIT) PHYS 405 - Fall 2017 October 17, 2017 16 / 21
More expectation values
〈Sy 〉 = χ†(t)Syχ(t)
=(
cos(α2 )e−iγB0t/2 sin(α2 )e+iγB0t/2) ~
2
(0 −ii 0
)(cos(α2 )e iγB0t/2
sin(α2 )e−iγB0t/2
)=
~2
(cos(α2 )e−iγB0t/2 sin(α2 )e+iγB0t/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)
=(
cos(α2 )e−iγB0t/2 sin(α2 )e+iγB0t/2) ~
2
(1 00 −1
)(cos(α2 )e iγB0t/2
sin(α2 )e−iγB0t/2
)=
~2
(cos(α2 )e−iγB0t/2 sin(α2 )e+iγB0t/2
)( cos(α2 )e iγB0t/2
− sin(α2 )e−iγB0t/2
)=
~2
(cos2(α/2)− sin2(α/2)
)=
~2
cosα
C. Segre (IIT) PHYS 405 - Fall 2017 October 17, 2017 16 / 21
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
C. Segre (IIT) PHYS 405 - Fall 2017 October 17, 2017 17 / 21
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
C. Segre (IIT) PHYS 405 - Fall 2017 October 17, 2017 17 / 21
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
C. Segre (IIT) PHYS 405 - Fall 2017 October 17, 2017 17 / 21
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
C. Segre (IIT) PHYS 405 - Fall 2017 October 17, 2017 17 / 21
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
C. Segre (IIT) PHYS 405 - Fall 2017 October 17, 2017 17 / 21
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
C. Segre (IIT) PHYS 405 - Fall 2017 October 17, 2017 17 / 21
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
C. Segre (IIT) PHYS 405 - Fall 2017 October 17, 2017 17 / 21
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?
C. Segre (IIT) PHYS 405 - Fall 2017 October 17, 2017 18 / 21
Problem 4.29(a) – solution
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
)
C. Segre (IIT) PHYS 405 - Fall 2017 October 17, 2017 19 / 21
Problem 4.29(a) – solution
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
)
C. Segre (IIT) PHYS 405 - Fall 2017 October 17, 2017 19 / 21
Problem 4.29(a) – solution
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
)
C. Segre (IIT) PHYS 405 - Fall 2017 October 17, 2017 19 / 21
Problem 4.29(a) – solution
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
)
C. Segre (IIT) PHYS 405 - Fall 2017 October 17, 2017 19 / 21
Problem 4.29(a) – solution
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
)
C. Segre (IIT) PHYS 405 - Fall 2017 October 17, 2017 19 / 21
Problem 4.29(a) – solution
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
)
C. Segre (IIT) PHYS 405 - Fall 2017 October 17, 2017 19 / 21
Problem 4.29(a) – solution
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
)
C. Segre (IIT) PHYS 405 - Fall 2017 October 17, 2017 19 / 21
Problem 4.29(a) – solution
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
)
C. Segre (IIT) PHYS 405 - Fall 2017 October 17, 2017 19 / 21
Problem 4.29(a) – solution
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
)
C. Segre (IIT) PHYS 405 - Fall 2017 October 17, 2017 19 / 21
Problem 4.29(a) – solution
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
)
C. Segre (IIT) PHYS 405 - Fall 2017 October 17, 2017 19 / 21
Problem 4.29(a) – solution
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
)
C. Segre (IIT) PHYS 405 - Fall 2017 October 17, 2017 19 / 21
Problem 4.29(a) – solution
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
)
C. Segre (IIT) PHYS 405 - Fall 2017 October 17, 2017 19 / 21
Problem 4.29(b) – solution
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
C. Segre (IIT) PHYS 405 - Fall 2017 October 17, 2017 20 / 21
Problem 4.29(b) – solution
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
C. Segre (IIT) PHYS 405 - Fall 2017 October 17, 2017 20 / 21
Problem 4.29(b) – solution
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
C. Segre (IIT) PHYS 405 - Fall 2017 October 17, 2017 20 / 21
Problem 4.29(b) – solution
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
C. Segre (IIT) PHYS 405 - Fall 2017 October 17, 2017 20 / 21
Problem 4.29(b) – solution
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
C. Segre (IIT) PHYS 405 - Fall 2017 October 17, 2017 20 / 21
Problem 4.29(b) – solution
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
C. Segre (IIT) PHYS 405 - Fall 2017 October 17, 2017 20 / 21
Problem 4.29(b) – solution
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
C. Segre (IIT) PHYS 405 - Fall 2017 October 17, 2017 20 / 21
Problem 4.29(b) – solution
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
C. Segre (IIT) PHYS 405 - Fall 2017 October 17, 2017 20 / 21
Problem 4.29(b) – solution
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
C. Segre (IIT) PHYS 405 - Fall 2017 October 17, 2017 20 / 21
Problem 4.29(b) – solution
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
C. Segre (IIT) PHYS 405 - Fall 2017 October 17, 2017 20 / 21
Problem 4.29(b) – solution
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
C. Segre (IIT) PHYS 405 - Fall 2017 October 17, 2017 20 / 21
Problem 4.29(b) – solution
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
C. Segre (IIT) PHYS 405 - Fall 2017 October 17, 2017 20 / 21
Problem 4.29(b) – solution
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
C. Segre (IIT) PHYS 405 - Fall 2017 October 17, 2017 20 / 21
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
C. Segre (IIT) PHYS 405 - Fall 2017 October 17, 2017 21 / 21
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
C. Segre (IIT) PHYS 405 - Fall 2017 October 17, 2017 21 / 21
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
C. Segre (IIT) PHYS 405 - Fall 2017 October 17, 2017 21 / 21
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
C. Segre (IIT) PHYS 405 - Fall 2017 October 17, 2017 21 / 21
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
C. Segre (IIT) PHYS 405 - Fall 2017 October 17, 2017 21 / 21
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
C. Segre (IIT) PHYS 405 - Fall 2017 October 17, 2017 21 / 21