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Today’s Outline - November 07, 2012
• Total spin of hydrogen
• Adding angular momentum
• Problem 4.55
C. Segre (IIT) PHYS 405 - Fall 2012 November 07, 2012 1 / 12
Today’s Outline - November 07, 2012
• Total spin of hydrogen
• Adding angular momentum
• Problem 4.55
C. Segre (IIT) PHYS 405 - Fall 2012 November 07, 2012 1 / 12
Today’s Outline - November 07, 2012
• Total spin of hydrogen
• Adding angular momentum
• Problem 4.55
C. Segre (IIT) PHYS 405 - Fall 2012 November 07, 2012 1 / 12
Today’s Outline - November 07, 2012
• Total spin of hydrogen
• Adding angular momentum
• Problem 4.55
C. Segre (IIT) PHYS 405 - Fall 2012 November 07, 2012 1 / 12
Total spin of the hydrogen atom
Suppose we have a hydrogen atom in its ground state (L = 0).
Theelectron, and the proton, each have spin S = 1
2. What is the total angular
momentum of the atom?
Each particle can have spin up or down, for 4 combinations.
↑↑ ↑↓ ↓↑ ↓↓
The total angular momentum is just the sum
~S ≡ ~S (1) + ~S (2)
each of the four possible states is an eigenfunction of the z-component ofthe total angular momentum
Szχ1χ2 = (S(1)z + S
(2)z )χ1χ2 = (S
(1)z χ1)χ2 + χ1(S
(2)z χ2)
= (~m1χ1)χ2 + χ1(~m2χ2) = ~(m1 + m2)χ1χ2
C. Segre (IIT) PHYS 405 - Fall 2012 November 07, 2012 2 / 12
Total spin of the hydrogen atom
Suppose we have a hydrogen atom in its ground state (L = 0). Theelectron, and the proton, each have spin S = 1
2.
What is the total angularmomentum of the atom?
Each particle can have spin up or down, for 4 combinations.
↑↑ ↑↓ ↓↑ ↓↓
The total angular momentum is just the sum
~S ≡ ~S (1) + ~S (2)
each of the four possible states is an eigenfunction of the z-component ofthe total angular momentum
Szχ1χ2 = (S(1)z + S
(2)z )χ1χ2 = (S
(1)z χ1)χ2 + χ1(S
(2)z χ2)
= (~m1χ1)χ2 + χ1(~m2χ2) = ~(m1 + m2)χ1χ2
C. Segre (IIT) PHYS 405 - Fall 2012 November 07, 2012 2 / 12
Total spin of the hydrogen atom
Suppose we have a hydrogen atom in its ground state (L = 0). Theelectron, and the proton, each have spin S = 1
2. What is the total angular
momentum of the atom?
Each particle can have spin up or down, for 4 combinations.
↑↑ ↑↓ ↓↑ ↓↓
The total angular momentum is just the sum
~S ≡ ~S (1) + ~S (2)
each of the four possible states is an eigenfunction of the z-component ofthe total angular momentum
Szχ1χ2 = (S(1)z + S
(2)z )χ1χ2 = (S
(1)z χ1)χ2 + χ1(S
(2)z χ2)
= (~m1χ1)χ2 + χ1(~m2χ2) = ~(m1 + m2)χ1χ2
C. Segre (IIT) PHYS 405 - Fall 2012 November 07, 2012 2 / 12
Total spin of the hydrogen atom
Suppose we have a hydrogen atom in its ground state (L = 0). Theelectron, and the proton, each have spin S = 1
2. What is the total angular
momentum of the atom?
Each particle can have spin up or down, for 4 combinations.
↑↑ ↑↓ ↓↑ ↓↓
The total angular momentum is just the sum
~S ≡ ~S (1) + ~S (2)
each of the four possible states is an eigenfunction of the z-component ofthe total angular momentum
Szχ1χ2 = (S(1)z + S
(2)z )χ1χ2 = (S
(1)z χ1)χ2 + χ1(S
(2)z χ2)
= (~m1χ1)χ2 + χ1(~m2χ2) = ~(m1 + m2)χ1χ2
C. Segre (IIT) PHYS 405 - Fall 2012 November 07, 2012 2 / 12
Total spin of the hydrogen atom
Suppose we have a hydrogen atom in its ground state (L = 0). Theelectron, and the proton, each have spin S = 1
2. What is the total angular
momentum of the atom?
Each particle can have spin up or down, for 4 combinations.
↑↑
↑↓ ↓↑ ↓↓
The total angular momentum is just the sum
~S ≡ ~S (1) + ~S (2)
each of the four possible states is an eigenfunction of the z-component ofthe total angular momentum
Szχ1χ2 = (S(1)z + S
(2)z )χ1χ2 = (S
(1)z χ1)χ2 + χ1(S
(2)z χ2)
= (~m1χ1)χ2 + χ1(~m2χ2) = ~(m1 + m2)χ1χ2
C. Segre (IIT) PHYS 405 - Fall 2012 November 07, 2012 2 / 12
Total spin of the hydrogen atom
Suppose we have a hydrogen atom in its ground state (L = 0). Theelectron, and the proton, each have spin S = 1
2. What is the total angular
momentum of the atom?
Each particle can have spin up or down, for 4 combinations.
↑↑ ↑↓
↓↑ ↓↓
The total angular momentum is just the sum
~S ≡ ~S (1) + ~S (2)
each of the four possible states is an eigenfunction of the z-component ofthe total angular momentum
Szχ1χ2 = (S(1)z + S
(2)z )χ1χ2 = (S
(1)z χ1)χ2 + χ1(S
(2)z χ2)
= (~m1χ1)χ2 + χ1(~m2χ2) = ~(m1 + m2)χ1χ2
C. Segre (IIT) PHYS 405 - Fall 2012 November 07, 2012 2 / 12
Total spin of the hydrogen atom
Suppose we have a hydrogen atom in its ground state (L = 0). Theelectron, and the proton, each have spin S = 1
2. What is the total angular
momentum of the atom?
Each particle can have spin up or down, for 4 combinations.
↑↑ ↑↓ ↓↑
↓↓
The total angular momentum is just the sum
~S ≡ ~S (1) + ~S (2)
each of the four possible states is an eigenfunction of the z-component ofthe total angular momentum
Szχ1χ2 = (S(1)z + S
(2)z )χ1χ2 = (S
(1)z χ1)χ2 + χ1(S
(2)z χ2)
= (~m1χ1)χ2 + χ1(~m2χ2) = ~(m1 + m2)χ1χ2
C. Segre (IIT) PHYS 405 - Fall 2012 November 07, 2012 2 / 12
Total spin of the hydrogen atom
Suppose we have a hydrogen atom in its ground state (L = 0). Theelectron, and the proton, each have spin S = 1
2. What is the total angular
momentum of the atom?
Each particle can have spin up or down, for 4 combinations.
↑↑ ↑↓ ↓↑ ↓↓
The total angular momentum is just the sum
~S ≡ ~S (1) + ~S (2)
each of the four possible states is an eigenfunction of the z-component ofthe total angular momentum
Szχ1χ2 = (S(1)z + S
(2)z )χ1χ2 = (S
(1)z χ1)χ2 + χ1(S
(2)z χ2)
= (~m1χ1)χ2 + χ1(~m2χ2) = ~(m1 + m2)χ1χ2
C. Segre (IIT) PHYS 405 - Fall 2012 November 07, 2012 2 / 12
Total spin of the hydrogen atom
Suppose we have a hydrogen atom in its ground state (L = 0). Theelectron, and the proton, each have spin S = 1
2. What is the total angular
momentum of the atom?
Each particle can have spin up or down, for 4 combinations.
↑↑ ↑↓ ↓↑ ↓↓
The total angular momentum is just the sum
~S ≡ ~S (1) + ~S (2)
each of the four possible states is an eigenfunction of the z-component ofthe total angular momentum
Szχ1χ2 = (S(1)z + S
(2)z )χ1χ2 = (S
(1)z χ1)χ2 + χ1(S
(2)z χ2)
= (~m1χ1)χ2 + χ1(~m2χ2) = ~(m1 + m2)χ1χ2
C. Segre (IIT) PHYS 405 - Fall 2012 November 07, 2012 2 / 12
Total spin of the hydrogen atom
Suppose we have a hydrogen atom in its ground state (L = 0). Theelectron, and the proton, each have spin S = 1
2. What is the total angular
momentum of the atom?
Each particle can have spin up or down, for 4 combinations.
↑↑ ↑↓ ↓↑ ↓↓
The total angular momentum is just the sum
~S ≡ ~S (1) + ~S (2)
each of the four possible states is an eigenfunction of the z-component ofthe total angular momentum
Szχ1χ2 = (S(1)z + S
(2)z )χ1χ2 = (S
(1)z χ1)χ2 + χ1(S
(2)z χ2)
= (~m1χ1)χ2 + χ1(~m2χ2) = ~(m1 + m2)χ1χ2
C. Segre (IIT) PHYS 405 - Fall 2012 November 07, 2012 2 / 12
Total spin of the hydrogen atom
Suppose we have a hydrogen atom in its ground state (L = 0). Theelectron, and the proton, each have spin S = 1
2. What is the total angular
momentum of the atom?
Each particle can have spin up or down, for 4 combinations.
↑↑ ↑↓ ↓↑ ↓↓
The total angular momentum is just the sum
~S ≡ ~S (1) + ~S (2)
each of the four possible states is an eigenfunction of the z-component ofthe total angular momentum
Szχ1χ2 = (S(1)z + S
(2)z )χ1χ2
= (S(1)z χ1)χ2 + χ1(S
(2)z χ2)
= (~m1χ1)χ2 + χ1(~m2χ2) = ~(m1 + m2)χ1χ2
C. Segre (IIT) PHYS 405 - Fall 2012 November 07, 2012 2 / 12
Total spin of the hydrogen atom
Suppose we have a hydrogen atom in its ground state (L = 0). Theelectron, and the proton, each have spin S = 1
2. What is the total angular
momentum of the atom?
Each particle can have spin up or down, for 4 combinations.
↑↑ ↑↓ ↓↑ ↓↓
The total angular momentum is just the sum
~S ≡ ~S (1) + ~S (2)
each of the four possible states is an eigenfunction of the z-component ofthe total angular momentum
Szχ1χ2 = (S(1)z + S
(2)z )χ1χ2 = (S
(1)z χ1)χ2 + χ1(S
(2)z χ2)
= (~m1χ1)χ2 + χ1(~m2χ2) = ~(m1 + m2)χ1χ2
C. Segre (IIT) PHYS 405 - Fall 2012 November 07, 2012 2 / 12
Total spin of the hydrogen atom
Suppose we have a hydrogen atom in its ground state (L = 0). Theelectron, and the proton, each have spin S = 1
2. What is the total angular
momentum of the atom?
Each particle can have spin up or down, for 4 combinations.
↑↑ ↑↓ ↓↑ ↓↓
The total angular momentum is just the sum
~S ≡ ~S (1) + ~S (2)
each of the four possible states is an eigenfunction of the z-component ofthe total angular momentum
Szχ1χ2 = (S(1)z + S
(2)z )χ1χ2 = (S
(1)z χ1)χ2 + χ1(S
(2)z χ2)
= (~m1χ1)χ2 + χ1(~m2χ2)
= ~(m1 + m2)χ1χ2
C. Segre (IIT) PHYS 405 - Fall 2012 November 07, 2012 2 / 12
Total spin of the hydrogen atom
Suppose we have a hydrogen atom in its ground state (L = 0). Theelectron, and the proton, each have spin S = 1
2. What is the total angular
momentum of the atom?
Each particle can have spin up or down, for 4 combinations.
↑↑ ↑↓ ↓↑ ↓↓
The total angular momentum is just the sum
~S ≡ ~S (1) + ~S (2)
each of the four possible states is an eigenfunction of the z-component ofthe total angular momentum
Szχ1χ2 = (S(1)z + S
(2)z )χ1χ2 = (S
(1)z χ1)χ2 + χ1(S
(2)z χ2)
= (~m1χ1)χ2 + χ1(~m2χ2) = ~(m1 + m2)χ1χ2
C. Segre (IIT) PHYS 405 - Fall 2012 November 07, 2012 2 / 12
Triplet state
We can thus establish that the state designated by ↑↑ has m = 1 andconsequently must have s = 1.
Let’s check this
S2 = (~S (1) + ~S (2)) · (~S (1) + ~S (2)) = (S (1))2 + (S (2))2 + 2~S (1) · ~S (2)
= (S (1))2 + (S (2))2 + 2(S(1)x S
(2)x + S
(1)y S
(2)y + S
(1)z S
(2)z )
S(1)x S
(2)x ↑↑ =
~2
(0 11 0
)(10
)~2
(0 11 0
)(10
)=
~2
4
(01
)(01
)=
~2
4↓↓
S(1)y S
(2)y ↑↑ =
~2
(0 −ii 0
)(10
)~2
(0 −ii 0
)(10
)= −~2
4
(01
)(01
)= −~2
4↓↓
S(1)z S
(2)z ↑↑ =
~2
(1 00 −1
)(10
)~2
(1 00 −1
)(10
)=
~2
4
(10
)(10
)=
~2
4↑↑
S2↑↑ = ~21
2
(3
2
)↑↑+ ~2
1
2
(3
2
)↑↑+ 2
(~2
4↓↓ − ~2
4↓↓+
~2
4↑↑)
= 2~2↑↑
C. Segre (IIT) PHYS 405 - Fall 2012 November 07, 2012 3 / 12
Triplet state
We can thus establish that the state designated by ↑↑ has m = 1 andconsequently must have s = 1. Let’s check this
S2 = (~S (1) + ~S (2)) · (~S (1) + ~S (2)) = (S (1))2 + (S (2))2 + 2~S (1) · ~S (2)
= (S (1))2 + (S (2))2 + 2(S(1)x S
(2)x + S
(1)y S
(2)y + S
(1)z S
(2)z )
S(1)x S
(2)x ↑↑ =
~2
(0 11 0
)(10
)~2
(0 11 0
)(10
)=
~2
4
(01
)(01
)=
~2
4↓↓
S(1)y S
(2)y ↑↑ =
~2
(0 −ii 0
)(10
)~2
(0 −ii 0
)(10
)= −~2
4
(01
)(01
)= −~2
4↓↓
S(1)z S
(2)z ↑↑ =
~2
(1 00 −1
)(10
)~2
(1 00 −1
)(10
)=
~2
4
(10
)(10
)=
~2
4↑↑
S2↑↑ = ~21
2
(3
2
)↑↑+ ~2
1
2
(3
2
)↑↑+ 2
(~2
4↓↓ − ~2
4↓↓+
~2
4↑↑)
= 2~2↑↑
C. Segre (IIT) PHYS 405 - Fall 2012 November 07, 2012 3 / 12
Triplet state
We can thus establish that the state designated by ↑↑ has m = 1 andconsequently must have s = 1. Let’s check this
S2 = (~S (1) + ~S (2)) · (~S (1) + ~S (2))
= (S (1))2 + (S (2))2 + 2~S (1) · ~S (2)
= (S (1))2 + (S (2))2 + 2(S(1)x S
(2)x + S
(1)y S
(2)y + S
(1)z S
(2)z )
S(1)x S
(2)x ↑↑ =
~2
(0 11 0
)(10
)~2
(0 11 0
)(10
)=
~2
4
(01
)(01
)=
~2
4↓↓
S(1)y S
(2)y ↑↑ =
~2
(0 −ii 0
)(10
)~2
(0 −ii 0
)(10
)= −~2
4
(01
)(01
)= −~2
4↓↓
S(1)z S
(2)z ↑↑ =
~2
(1 00 −1
)(10
)~2
(1 00 −1
)(10
)=
~2
4
(10
)(10
)=
~2
4↑↑
S2↑↑ = ~21
2
(3
2
)↑↑+ ~2
1
2
(3
2
)↑↑+ 2
(~2
4↓↓ − ~2
4↓↓+
~2
4↑↑)
= 2~2↑↑
C. Segre (IIT) PHYS 405 - Fall 2012 November 07, 2012 3 / 12
Triplet state
We can thus establish that the state designated by ↑↑ has m = 1 andconsequently must have s = 1. Let’s check this
S2 = (~S (1) + ~S (2)) · (~S (1) + ~S (2)) = (S (1))2 + (S (2))2 + 2~S (1) · ~S (2)
= (S (1))2 + (S (2))2 + 2(S(1)x S
(2)x + S
(1)y S
(2)y + S
(1)z S
(2)z )
S(1)x S
(2)x ↑↑ =
~2
(0 11 0
)(10
)~2
(0 11 0
)(10
)=
~2
4
(01
)(01
)=
~2
4↓↓
S(1)y S
(2)y ↑↑ =
~2
(0 −ii 0
)(10
)~2
(0 −ii 0
)(10
)= −~2
4
(01
)(01
)= −~2
4↓↓
S(1)z S
(2)z ↑↑ =
~2
(1 00 −1
)(10
)~2
(1 00 −1
)(10
)=
~2
4
(10
)(10
)=
~2
4↑↑
S2↑↑ = ~21
2
(3
2
)↑↑+ ~2
1
2
(3
2
)↑↑+ 2
(~2
4↓↓ − ~2
4↓↓+
~2
4↑↑)
= 2~2↑↑
C. Segre (IIT) PHYS 405 - Fall 2012 November 07, 2012 3 / 12
Triplet state
We can thus establish that the state designated by ↑↑ has m = 1 andconsequently must have s = 1. Let’s check this
S2 = (~S (1) + ~S (2)) · (~S (1) + ~S (2)) = (S (1))2 + (S (2))2 + 2~S (1) · ~S (2)
= (S (1))2 + (S (2))2 + 2(S(1)x S
(2)x + S
(1)y S
(2)y + S
(1)z S
(2)z )
S(1)x S
(2)x ↑↑ =
~2
(0 11 0
)(10
)~2
(0 11 0
)(10
)=
~2
4
(01
)(01
)=
~2
4↓↓
S(1)y S
(2)y ↑↑ =
~2
(0 −ii 0
)(10
)~2
(0 −ii 0
)(10
)= −~2
4
(01
)(01
)= −~2
4↓↓
S(1)z S
(2)z ↑↑ =
~2
(1 00 −1
)(10
)~2
(1 00 −1
)(10
)=
~2
4
(10
)(10
)=
~2
4↑↑
S2↑↑ = ~21
2
(3
2
)↑↑+ ~2
1
2
(3
2
)↑↑+ 2
(~2
4↓↓ − ~2
4↓↓+
~2
4↑↑)
= 2~2↑↑
C. Segre (IIT) PHYS 405 - Fall 2012 November 07, 2012 3 / 12
Triplet state
We can thus establish that the state designated by ↑↑ has m = 1 andconsequently must have s = 1. Let’s check this
S2 = (~S (1) + ~S (2)) · (~S (1) + ~S (2)) = (S (1))2 + (S (2))2 + 2~S (1) · ~S (2)
= (S (1))2 + (S (2))2 + 2(S(1)x S
(2)x + S
(1)y S
(2)y + S
(1)z S
(2)z )
S(1)x S
(2)x ↑↑
=~2
(0 11 0
)(10
)~2
(0 11 0
)(10
)=
~2
4
(01
)(01
)=
~2
4↓↓
S(1)y S
(2)y ↑↑ =
~2
(0 −ii 0
)(10
)~2
(0 −ii 0
)(10
)= −~2
4
(01
)(01
)= −~2
4↓↓
S(1)z S
(2)z ↑↑ =
~2
(1 00 −1
)(10
)~2
(1 00 −1
)(10
)=
~2
4
(10
)(10
)=
~2
4↑↑
S2↑↑ = ~21
2
(3
2
)↑↑+ ~2
1
2
(3
2
)↑↑+ 2
(~2
4↓↓ − ~2
4↓↓+
~2
4↑↑)
= 2~2↑↑
C. Segre (IIT) PHYS 405 - Fall 2012 November 07, 2012 3 / 12
Triplet state
We can thus establish that the state designated by ↑↑ has m = 1 andconsequently must have s = 1. Let’s check this
S2 = (~S (1) + ~S (2)) · (~S (1) + ~S (2)) = (S (1))2 + (S (2))2 + 2~S (1) · ~S (2)
= (S (1))2 + (S (2))2 + 2(S(1)x S
(2)x + S
(1)y S
(2)y + S
(1)z S
(2)z )
S(1)x S
(2)x ↑↑ =
~2
(0 11 0
)(10
)~2
(0 11 0
)(10
)
=~2
4
(01
)(01
)=
~2
4↓↓
S(1)y S
(2)y ↑↑ =
~2
(0 −ii 0
)(10
)~2
(0 −ii 0
)(10
)= −~2
4
(01
)(01
)= −~2
4↓↓
S(1)z S
(2)z ↑↑ =
~2
(1 00 −1
)(10
)~2
(1 00 −1
)(10
)=
~2
4
(10
)(10
)=
~2
4↑↑
S2↑↑ = ~21
2
(3
2
)↑↑+ ~2
1
2
(3
2
)↑↑+ 2
(~2
4↓↓ − ~2
4↓↓+
~2
4↑↑)
= 2~2↑↑
C. Segre (IIT) PHYS 405 - Fall 2012 November 07, 2012 3 / 12
Triplet state
We can thus establish that the state designated by ↑↑ has m = 1 andconsequently must have s = 1. Let’s check this
S2 = (~S (1) + ~S (2)) · (~S (1) + ~S (2)) = (S (1))2 + (S (2))2 + 2~S (1) · ~S (2)
= (S (1))2 + (S (2))2 + 2(S(1)x S
(2)x + S
(1)y S
(2)y + S
(1)z S
(2)z )
S(1)x S
(2)x ↑↑ =
~2
(0 11 0
)(10
)~2
(0 11 0
)(10
)=
~2
4
(01
)(01
)
=~2
4↓↓
S(1)y S
(2)y ↑↑ =
~2
(0 −ii 0
)(10
)~2
(0 −ii 0
)(10
)= −~2
4
(01
)(01
)= −~2
4↓↓
S(1)z S
(2)z ↑↑ =
~2
(1 00 −1
)(10
)~2
(1 00 −1
)(10
)=
~2
4
(10
)(10
)=
~2
4↑↑
S2↑↑ = ~21
2
(3
2
)↑↑+ ~2
1
2
(3
2
)↑↑+ 2
(~2
4↓↓ − ~2
4↓↓+
~2
4↑↑)
= 2~2↑↑
C. Segre (IIT) PHYS 405 - Fall 2012 November 07, 2012 3 / 12
Triplet state
We can thus establish that the state designated by ↑↑ has m = 1 andconsequently must have s = 1. Let’s check this
S2 = (~S (1) + ~S (2)) · (~S (1) + ~S (2)) = (S (1))2 + (S (2))2 + 2~S (1) · ~S (2)
= (S (1))2 + (S (2))2 + 2(S(1)x S
(2)x + S
(1)y S
(2)y + S
(1)z S
(2)z )
S(1)x S
(2)x ↑↑ =
~2
(0 11 0
)(10
)~2
(0 11 0
)(10
)=
~2
4
(01
)(01
)=
~2
4↓↓
S(1)y S
(2)y ↑↑ =
~2
(0 −ii 0
)(10
)~2
(0 −ii 0
)(10
)= −~2
4
(01
)(01
)= −~2
4↓↓
S(1)z S
(2)z ↑↑ =
~2
(1 00 −1
)(10
)~2
(1 00 −1
)(10
)=
~2
4
(10
)(10
)=
~2
4↑↑
S2↑↑ = ~21
2
(3
2
)↑↑+ ~2
1
2
(3
2
)↑↑+ 2
(~2
4↓↓ − ~2
4↓↓+
~2
4↑↑)
= 2~2↑↑
C. Segre (IIT) PHYS 405 - Fall 2012 November 07, 2012 3 / 12
Triplet state
We can thus establish that the state designated by ↑↑ has m = 1 andconsequently must have s = 1. Let’s check this
S2 = (~S (1) + ~S (2)) · (~S (1) + ~S (2)) = (S (1))2 + (S (2))2 + 2~S (1) · ~S (2)
= (S (1))2 + (S (2))2 + 2(S(1)x S
(2)x + S
(1)y S
(2)y + S
(1)z S
(2)z )
S(1)x S
(2)x ↑↑ =
~2
(0 11 0
)(10
)~2
(0 11 0
)(10
)=
~2
4
(01
)(01
)=
~2
4↓↓
S(1)y S
(2)y ↑↑
=~2
(0 −ii 0
)(10
)~2
(0 −ii 0
)(10
)= −~2
4
(01
)(01
)= −~2
4↓↓
S(1)z S
(2)z ↑↑ =
~2
(1 00 −1
)(10
)~2
(1 00 −1
)(10
)=
~2
4
(10
)(10
)=
~2
4↑↑
S2↑↑ = ~21
2
(3
2
)↑↑+ ~2
1
2
(3
2
)↑↑+ 2
(~2
4↓↓ − ~2
4↓↓+
~2
4↑↑)
= 2~2↑↑
C. Segre (IIT) PHYS 405 - Fall 2012 November 07, 2012 3 / 12
Triplet state
We can thus establish that the state designated by ↑↑ has m = 1 andconsequently must have s = 1. Let’s check this
S2 = (~S (1) + ~S (2)) · (~S (1) + ~S (2)) = (S (1))2 + (S (2))2 + 2~S (1) · ~S (2)
= (S (1))2 + (S (2))2 + 2(S(1)x S
(2)x + S
(1)y S
(2)y + S
(1)z S
(2)z )
S(1)x S
(2)x ↑↑ =
~2
(0 11 0
)(10
)~2
(0 11 0
)(10
)=
~2
4
(01
)(01
)=
~2
4↓↓
S(1)y S
(2)y ↑↑ =
~2
(0 −ii 0
)(10
)~2
(0 −ii 0
)(10
)
= −~2
4
(01
)(01
)= −~2
4↓↓
S(1)z S
(2)z ↑↑ =
~2
(1 00 −1
)(10
)~2
(1 00 −1
)(10
)=
~2
4
(10
)(10
)=
~2
4↑↑
S2↑↑ = ~21
2
(3
2
)↑↑+ ~2
1
2
(3
2
)↑↑+ 2
(~2
4↓↓ − ~2
4↓↓+
~2
4↑↑)
= 2~2↑↑
C. Segre (IIT) PHYS 405 - Fall 2012 November 07, 2012 3 / 12
Triplet state
We can thus establish that the state designated by ↑↑ has m = 1 andconsequently must have s = 1. Let’s check this
S2 = (~S (1) + ~S (2)) · (~S (1) + ~S (2)) = (S (1))2 + (S (2))2 + 2~S (1) · ~S (2)
= (S (1))2 + (S (2))2 + 2(S(1)x S
(2)x + S
(1)y S
(2)y + S
(1)z S
(2)z )
S(1)x S
(2)x ↑↑ =
~2
(0 11 0
)(10
)~2
(0 11 0
)(10
)=
~2
4
(01
)(01
)=
~2
4↓↓
S(1)y S
(2)y ↑↑ =
~2
(0 −ii 0
)(10
)~2
(0 −ii 0
)(10
)= −~2
4
(01
)(01
)
= −~2
4↓↓
S(1)z S
(2)z ↑↑ =
~2
(1 00 −1
)(10
)~2
(1 00 −1
)(10
)=
~2
4
(10
)(10
)=
~2
4↑↑
S2↑↑ = ~21
2
(3
2
)↑↑+ ~2
1
2
(3
2
)↑↑+ 2
(~2
4↓↓ − ~2
4↓↓+
~2
4↑↑)
= 2~2↑↑
C. Segre (IIT) PHYS 405 - Fall 2012 November 07, 2012 3 / 12
Triplet state
We can thus establish that the state designated by ↑↑ has m = 1 andconsequently must have s = 1. Let’s check this
S2 = (~S (1) + ~S (2)) · (~S (1) + ~S (2)) = (S (1))2 + (S (2))2 + 2~S (1) · ~S (2)
= (S (1))2 + (S (2))2 + 2(S(1)x S
(2)x + S
(1)y S
(2)y + S
(1)z S
(2)z )
S(1)x S
(2)x ↑↑ =
~2
(0 11 0
)(10
)~2
(0 11 0
)(10
)=
~2
4
(01
)(01
)=
~2
4↓↓
S(1)y S
(2)y ↑↑ =
~2
(0 −ii 0
)(10
)~2
(0 −ii 0
)(10
)= −~2
4
(01
)(01
)= −~2
4↓↓
S(1)z S
(2)z ↑↑ =
~2
(1 00 −1
)(10
)~2
(1 00 −1
)(10
)=
~2
4
(10
)(10
)=
~2
4↑↑
S2↑↑ = ~21
2
(3
2
)↑↑+ ~2
1
2
(3
2
)↑↑+ 2
(~2
4↓↓ − ~2
4↓↓+
~2
4↑↑)
= 2~2↑↑
C. Segre (IIT) PHYS 405 - Fall 2012 November 07, 2012 3 / 12
Triplet state
We can thus establish that the state designated by ↑↑ has m = 1 andconsequently must have s = 1. Let’s check this
S2 = (~S (1) + ~S (2)) · (~S (1) + ~S (2)) = (S (1))2 + (S (2))2 + 2~S (1) · ~S (2)
= (S (1))2 + (S (2))2 + 2(S(1)x S
(2)x + S
(1)y S
(2)y + S
(1)z S
(2)z )
S(1)x S
(2)x ↑↑ =
~2
(0 11 0
)(10
)~2
(0 11 0
)(10
)=
~2
4
(01
)(01
)=
~2
4↓↓
S(1)y S
(2)y ↑↑ =
~2
(0 −ii 0
)(10
)~2
(0 −ii 0
)(10
)= −~2
4
(01
)(01
)= −~2
4↓↓
S(1)z S
(2)z ↑↑
=~2
(1 00 −1
)(10
)~2
(1 00 −1
)(10
)=
~2
4
(10
)(10
)=
~2
4↑↑
S2↑↑ = ~21
2
(3
2
)↑↑+ ~2
1
2
(3
2
)↑↑+ 2
(~2
4↓↓ − ~2
4↓↓+
~2
4↑↑)
= 2~2↑↑
C. Segre (IIT) PHYS 405 - Fall 2012 November 07, 2012 3 / 12
Triplet state
We can thus establish that the state designated by ↑↑ has m = 1 andconsequently must have s = 1. Let’s check this
S2 = (~S (1) + ~S (2)) · (~S (1) + ~S (2)) = (S (1))2 + (S (2))2 + 2~S (1) · ~S (2)
= (S (1))2 + (S (2))2 + 2(S(1)x S
(2)x + S
(1)y S
(2)y + S
(1)z S
(2)z )
S(1)x S
(2)x ↑↑ =
~2
(0 11 0
)(10
)~2
(0 11 0
)(10
)=
~2
4
(01
)(01
)=
~2
4↓↓
S(1)y S
(2)y ↑↑ =
~2
(0 −ii 0
)(10
)~2
(0 −ii 0
)(10
)= −~2
4
(01
)(01
)= −~2
4↓↓
S(1)z S
(2)z ↑↑ =
~2
(1 00 −1
)(10
)~2
(1 00 −1
)(10
)
=~2
4
(10
)(10
)=
~2
4↑↑
S2↑↑ = ~21
2
(3
2
)↑↑+ ~2
1
2
(3
2
)↑↑+ 2
(~2
4↓↓ − ~2
4↓↓+
~2
4↑↑)
= 2~2↑↑
C. Segre (IIT) PHYS 405 - Fall 2012 November 07, 2012 3 / 12
Triplet state
We can thus establish that the state designated by ↑↑ has m = 1 andconsequently must have s = 1. Let’s check this
S2 = (~S (1) + ~S (2)) · (~S (1) + ~S (2)) = (S (1))2 + (S (2))2 + 2~S (1) · ~S (2)
= (S (1))2 + (S (2))2 + 2(S(1)x S
(2)x + S
(1)y S
(2)y + S
(1)z S
(2)z )
S(1)x S
(2)x ↑↑ =
~2
(0 11 0
)(10
)~2
(0 11 0
)(10
)=
~2
4
(01
)(01
)=
~2
4↓↓
S(1)y S
(2)y ↑↑ =
~2
(0 −ii 0
)(10
)~2
(0 −ii 0
)(10
)= −~2
4
(01
)(01
)= −~2
4↓↓
S(1)z S
(2)z ↑↑ =
~2
(1 00 −1
)(10
)~2
(1 00 −1
)(10
)=
~2
4
(10
)(10
)
=~2
4↑↑
S2↑↑ = ~21
2
(3
2
)↑↑+ ~2
1
2
(3
2
)↑↑+ 2
(~2
4↓↓ − ~2
4↓↓+
~2
4↑↑)
= 2~2↑↑
C. Segre (IIT) PHYS 405 - Fall 2012 November 07, 2012 3 / 12
Triplet state
We can thus establish that the state designated by ↑↑ has m = 1 andconsequently must have s = 1. Let’s check this
S2 = (~S (1) + ~S (2)) · (~S (1) + ~S (2)) = (S (1))2 + (S (2))2 + 2~S (1) · ~S (2)
= (S (1))2 + (S (2))2 + 2(S(1)x S
(2)x + S
(1)y S
(2)y + S
(1)z S
(2)z )
S(1)x S
(2)x ↑↑ =
~2
(0 11 0
)(10
)~2
(0 11 0
)(10
)=
~2
4
(01
)(01
)=
~2
4↓↓
S(1)y S
(2)y ↑↑ =
~2
(0 −ii 0
)(10
)~2
(0 −ii 0
)(10
)= −~2
4
(01
)(01
)= −~2
4↓↓
S(1)z S
(2)z ↑↑ =
~2
(1 00 −1
)(10
)~2
(1 00 −1
)(10
)=
~2
4
(10
)(10
)=
~2
4↑↑
S2↑↑ = ~21
2
(3
2
)↑↑+ ~2
1
2
(3
2
)↑↑+ 2
(~2
4↓↓ − ~2
4↓↓+
~2
4↑↑)
= 2~2↑↑
C. Segre (IIT) PHYS 405 - Fall 2012 November 07, 2012 3 / 12
Triplet state
We can thus establish that the state designated by ↑↑ has m = 1 andconsequently must have s = 1. Let’s check this
S2 = (~S (1) + ~S (2)) · (~S (1) + ~S (2)) = (S (1))2 + (S (2))2 + 2~S (1) · ~S (2)
= (S (1))2 + (S (2))2 + 2(S(1)x S
(2)x + S
(1)y S
(2)y + S
(1)z S
(2)z )
S(1)x S
(2)x ↑↑ =
~2
(0 11 0
)(10
)~2
(0 11 0
)(10
)=
~2
4
(01
)(01
)=
~2
4↓↓
S(1)y S
(2)y ↑↑ =
~2
(0 −ii 0
)(10
)~2
(0 −ii 0
)(10
)= −~2
4
(01
)(01
)= −~2
4↓↓
S(1)z S
(2)z ↑↑ =
~2
(1 00 −1
)(10
)~2
(1 00 −1
)(10
)=
~2
4
(10
)(10
)=
~2
4↑↑
S2↑↑ =
~21
2
(3
2
)↑↑+ ~2
1
2
(3
2
)↑↑+ 2
(~2
4↓↓ − ~2
4↓↓+
~2
4↑↑)
= 2~2↑↑
C. Segre (IIT) PHYS 405 - Fall 2012 November 07, 2012 3 / 12
Triplet state
We can thus establish that the state designated by ↑↑ has m = 1 andconsequently must have s = 1. Let’s check this
S2 = (~S (1) + ~S (2)) · (~S (1) + ~S (2)) = (S (1))2 + (S (2))2 + 2~S (1) · ~S (2)
= (S (1))2 + (S (2))2 + 2(S(1)x S
(2)x + S
(1)y S
(2)y + S
(1)z S
(2)z )
S(1)x S
(2)x ↑↑ =
~2
(0 11 0
)(10
)~2
(0 11 0
)(10
)=
~2
4
(01
)(01
)=
~2
4↓↓
S(1)y S
(2)y ↑↑ =
~2
(0 −ii 0
)(10
)~2
(0 −ii 0
)(10
)= −~2
4
(01
)(01
)= −~2
4↓↓
S(1)z S
(2)z ↑↑ =
~2
(1 00 −1
)(10
)~2
(1 00 −1
)(10
)=
~2
4
(10
)(10
)=
~2
4↑↑
S2↑↑ = ~21
2
(3
2
)↑↑
+ ~21
2
(3
2
)↑↑+ 2
(~2
4↓↓ − ~2
4↓↓+
~2
4↑↑)
= 2~2↑↑
C. Segre (IIT) PHYS 405 - Fall 2012 November 07, 2012 3 / 12
Triplet state
We can thus establish that the state designated by ↑↑ has m = 1 andconsequently must have s = 1. Let’s check this
S2 = (~S (1) + ~S (2)) · (~S (1) + ~S (2)) = (S (1))2 + (S (2))2 + 2~S (1) · ~S (2)
= (S (1))2 + (S (2))2 + 2(S(1)x S
(2)x + S
(1)y S
(2)y + S
(1)z S
(2)z )
S(1)x S
(2)x ↑↑ =
~2
(0 11 0
)(10
)~2
(0 11 0
)(10
)=
~2
4
(01
)(01
)=
~2
4↓↓
S(1)y S
(2)y ↑↑ =
~2
(0 −ii 0
)(10
)~2
(0 −ii 0
)(10
)= −~2
4
(01
)(01
)= −~2
4↓↓
S(1)z S
(2)z ↑↑ =
~2
(1 00 −1
)(10
)~2
(1 00 −1
)(10
)=
~2
4
(10
)(10
)=
~2
4↑↑
S2↑↑ = ~21
2
(3
2
)↑↑+ ~2
1
2
(3
2
)↑↑
+ 2
(~2
4↓↓ − ~2
4↓↓+
~2
4↑↑)
= 2~2↑↑
C. Segre (IIT) PHYS 405 - Fall 2012 November 07, 2012 3 / 12
Triplet state
We can thus establish that the state designated by ↑↑ has m = 1 andconsequently must have s = 1. Let’s check this
S2 = (~S (1) + ~S (2)) · (~S (1) + ~S (2)) = (S (1))2 + (S (2))2 + 2~S (1) · ~S (2)
= (S (1))2 + (S (2))2 + 2(S(1)x S
(2)x + S
(1)y S
(2)y + S
(1)z S
(2)z )
S(1)x S
(2)x ↑↑ =
~2
(0 11 0
)(10
)~2
(0 11 0
)(10
)=
~2
4
(01
)(01
)=
~2
4↓↓
S(1)y S
(2)y ↑↑ =
~2
(0 −ii 0
)(10
)~2
(0 −ii 0
)(10
)= −~2
4
(01
)(01
)= −~2
4↓↓
S(1)z S
(2)z ↑↑ =
~2
(1 00 −1
)(10
)~2
(1 00 −1
)(10
)=
~2
4
(10
)(10
)=
~2
4↑↑
S2↑↑ = ~21
2
(3
2
)↑↑+ ~2
1
2
(3
2
)↑↑+ 2
(~2
4↓↓ − ~2
4↓↓+
~2
4↑↑)
= 2~2↑↑
C. Segre (IIT) PHYS 405 - Fall 2012 November 07, 2012 3 / 12
Triplet state
We can thus establish that the state designated by ↑↑ has m = 1 andconsequently must have s = 1. Let’s check this
S2 = (~S (1) + ~S (2)) · (~S (1) + ~S (2)) = (S (1))2 + (S (2))2 + 2~S (1) · ~S (2)
= (S (1))2 + (S (2))2 + 2(S(1)x S
(2)x + S
(1)y S
(2)y + S
(1)z S
(2)z )
S(1)x S
(2)x ↑↑ =
~2
(0 11 0
)(10
)~2
(0 11 0
)(10
)=
~2
4
(01
)(01
)=
~2
4↓↓
S(1)y S
(2)y ↑↑ =
~2
(0 −ii 0
)(10
)~2
(0 −ii 0
)(10
)= −~2
4
(01
)(01
)= −~2
4↓↓
S(1)z S
(2)z ↑↑ =
~2
(1 00 −1
)(10
)~2
(1 00 −1
)(10
)=
~2
4
(10
)(10
)=
~2
4↑↑
S2↑↑ = ~21
2
(3
2
)↑↑+ ~2
1
2
(3
2
)↑↑+ 2
(~2
4↓↓ − ~2
4↓↓+
~2
4↑↑)
= 2~2↑↑
C. Segre (IIT) PHYS 405 - Fall 2012 November 07, 2012 3 / 12
Triplet state (cont.)
Use the lowering operator to generate the state with total s = 1 and m = 0
S−↑↑ = (S(1)− + S
(2)− )↑↑ = (S
(1)− ↑)↑+ ↑(S (2)
− ↑)
= ~
√1
2
(3
2
)− 1
2
(−1
2
)(↓↑+ ↑↓) = ~(↓↑+ ↑↓) =
√2~
1√2
(↓↑+ ↑↓)
Apply the lowering operator again to get s = 1, m = −1
S−1√2
(↓↑+ ↑↓) = (S(1)− + S
(2)− )
1√2
(↓↑+ ↑↓)
=1√2
[S(1)− (↓↑+ ↑↓) + S
(2)− (↓↑+ ↑↓)
]=
1√2
[~↓↓+ ~↓↓]
=√
2~↓↓
The triplet state is thus
|1 1〉 = ↑↑, |1 0〉 =1√2
(↓↑+ ↑↓), |1−1〉 = ↓↓
C. Segre (IIT) PHYS 405 - Fall 2012 November 07, 2012 4 / 12
Triplet state (cont.)
Use the lowering operator to generate the state with total s = 1 and m = 0
S−↑↑ = (S(1)− + S
(2)− )↑↑
= (S(1)− ↑)↑+ ↑(S (2)
− ↑)
= ~
√1
2
(3
2
)− 1
2
(−1
2
)(↓↑+ ↑↓) = ~(↓↑+ ↑↓) =
√2~
1√2
(↓↑+ ↑↓)
Apply the lowering operator again to get s = 1, m = −1
S−1√2
(↓↑+ ↑↓) = (S(1)− + S
(2)− )
1√2
(↓↑+ ↑↓)
=1√2
[S(1)− (↓↑+ ↑↓) + S
(2)− (↓↑+ ↑↓)
]=
1√2
[~↓↓+ ~↓↓]
=√
2~↓↓
The triplet state is thus
|1 1〉 = ↑↑, |1 0〉 =1√2
(↓↑+ ↑↓), |1−1〉 = ↓↓
C. Segre (IIT) PHYS 405 - Fall 2012 November 07, 2012 4 / 12
Triplet state (cont.)
Use the lowering operator to generate the state with total s = 1 and m = 0
S−↑↑ = (S(1)− + S
(2)− )↑↑ = (S
(1)− ↑)↑+ ↑(S (2)
− ↑)
= ~
√1
2
(3
2
)− 1
2
(−1
2
)(↓↑+ ↑↓) = ~(↓↑+ ↑↓) =
√2~
1√2
(↓↑+ ↑↓)
Apply the lowering operator again to get s = 1, m = −1
S−1√2
(↓↑+ ↑↓) = (S(1)− + S
(2)− )
1√2
(↓↑+ ↑↓)
=1√2
[S(1)− (↓↑+ ↑↓) + S
(2)− (↓↑+ ↑↓)
]=
1√2
[~↓↓+ ~↓↓]
=√
2~↓↓
The triplet state is thus
|1 1〉 = ↑↑, |1 0〉 =1√2
(↓↑+ ↑↓), |1−1〉 = ↓↓
C. Segre (IIT) PHYS 405 - Fall 2012 November 07, 2012 4 / 12
Triplet state (cont.)
Use the lowering operator to generate the state with total s = 1 and m = 0
S−↑↑ = (S(1)− + S
(2)− )↑↑ = (S
(1)− ↑)↑+ ↑(S (2)
− ↑)
= ~
√1
2
(3
2
)− 1
2
(−1
2
)(↓↑+ ↑↓)
= ~(↓↑+ ↑↓) =√
2~1√2
(↓↑+ ↑↓)
Apply the lowering operator again to get s = 1, m = −1
S−1√2
(↓↑+ ↑↓) = (S(1)− + S
(2)− )
1√2
(↓↑+ ↑↓)
=1√2
[S(1)− (↓↑+ ↑↓) + S
(2)− (↓↑+ ↑↓)
]=
1√2
[~↓↓+ ~↓↓]
=√
2~↓↓
The triplet state is thus
|1 1〉 = ↑↑, |1 0〉 =1√2
(↓↑+ ↑↓), |1−1〉 = ↓↓
C. Segre (IIT) PHYS 405 - Fall 2012 November 07, 2012 4 / 12
Triplet state (cont.)
Use the lowering operator to generate the state with total s = 1 and m = 0
S−↑↑ = (S(1)− + S
(2)− )↑↑ = (S
(1)− ↑)↑+ ↑(S (2)
− ↑)
= ~
√1
2
(3
2
)− 1
2
(−1
2
)(↓↑+ ↑↓) = ~(↓↑+ ↑↓)
=√
2~1√2
(↓↑+ ↑↓)
Apply the lowering operator again to get s = 1, m = −1
S−1√2
(↓↑+ ↑↓) = (S(1)− + S
(2)− )
1√2
(↓↑+ ↑↓)
=1√2
[S(1)− (↓↑+ ↑↓) + S
(2)− (↓↑+ ↑↓)
]=
1√2
[~↓↓+ ~↓↓]
=√
2~↓↓
The triplet state is thus
|1 1〉 = ↑↑, |1 0〉 =1√2
(↓↑+ ↑↓), |1−1〉 = ↓↓
C. Segre (IIT) PHYS 405 - Fall 2012 November 07, 2012 4 / 12
Triplet state (cont.)
Use the lowering operator to generate the state with total s = 1 and m = 0
S−↑↑ = (S(1)− + S
(2)− )↑↑ = (S
(1)− ↑)↑+ ↑(S (2)
− ↑)
= ~
√1
2
(3
2
)− 1
2
(−1
2
)(↓↑+ ↑↓) = ~(↓↑+ ↑↓) =
√2~
1√2
(↓↑+ ↑↓)
Apply the lowering operator again to get s = 1, m = −1
S−1√2
(↓↑+ ↑↓) = (S(1)− + S
(2)− )
1√2
(↓↑+ ↑↓)
=1√2
[S(1)− (↓↑+ ↑↓) + S
(2)− (↓↑+ ↑↓)
]=
1√2
[~↓↓+ ~↓↓]
=√
2~↓↓
The triplet state is thus
|1 1〉 = ↑↑, |1 0〉 =1√2
(↓↑+ ↑↓), |1−1〉 = ↓↓
C. Segre (IIT) PHYS 405 - Fall 2012 November 07, 2012 4 / 12
Triplet state (cont.)
Use the lowering operator to generate the state with total s = 1 and m = 0
S−↑↑ = (S(1)− + S
(2)− )↑↑ = (S
(1)− ↑)↑+ ↑(S (2)
− ↑)
= ~
√1
2
(3
2
)− 1
2
(−1
2
)(↓↑+ ↑↓) = ~(↓↑+ ↑↓) =
√2~
1√2
(↓↑+ ↑↓)
Apply the lowering operator again to get s = 1, m = −1
S−1√2
(↓↑+ ↑↓) = (S(1)− + S
(2)− )
1√2
(↓↑+ ↑↓)
=1√2
[S(1)− (↓↑+ ↑↓) + S
(2)− (↓↑+ ↑↓)
]=
1√2
[~↓↓+ ~↓↓]
=√
2~↓↓
The triplet state is thus
|1 1〉 = ↑↑, |1 0〉 =1√2
(↓↑+ ↑↓), |1−1〉 = ↓↓
C. Segre (IIT) PHYS 405 - Fall 2012 November 07, 2012 4 / 12
Triplet state (cont.)
Use the lowering operator to generate the state with total s = 1 and m = 0
S−↑↑ = (S(1)− + S
(2)− )↑↑ = (S
(1)− ↑)↑+ ↑(S (2)
− ↑)
= ~
√1
2
(3
2
)− 1
2
(−1
2
)(↓↑+ ↑↓) = ~(↓↑+ ↑↓) =
√2~
1√2
(↓↑+ ↑↓)
Apply the lowering operator again to get s = 1, m = −1
S−1√2
(↓↑+ ↑↓) = (S(1)− + S
(2)− )
1√2
(↓↑+ ↑↓)
=1√2
[S(1)− (↓↑+ ↑↓) + S
(2)− (↓↑+ ↑↓)
]=
1√2
[~↓↓+ ~↓↓]
=√
2~↓↓
The triplet state is thus
|1 1〉 = ↑↑, |1 0〉 =1√2
(↓↑+ ↑↓), |1−1〉 = ↓↓
C. Segre (IIT) PHYS 405 - Fall 2012 November 07, 2012 4 / 12
Triplet state (cont.)
Use the lowering operator to generate the state with total s = 1 and m = 0
S−↑↑ = (S(1)− + S
(2)− )↑↑ = (S
(1)− ↑)↑+ ↑(S (2)
− ↑)
= ~
√1
2
(3
2
)− 1
2
(−1
2
)(↓↑+ ↑↓) = ~(↓↑+ ↑↓) =
√2~
1√2
(↓↑+ ↑↓)
Apply the lowering operator again to get s = 1, m = −1
S−1√2
(↓↑+ ↑↓) = (S(1)− + S
(2)− )
1√2
(↓↑+ ↑↓)
=1√2
[S(1)− (↓↑+ ↑↓) + S
(2)− (↓↑+ ↑↓)
]
=1√2
[~↓↓+ ~↓↓]
=√
2~↓↓
The triplet state is thus
|1 1〉 = ↑↑, |1 0〉 =1√2
(↓↑+ ↑↓), |1−1〉 = ↓↓
C. Segre (IIT) PHYS 405 - Fall 2012 November 07, 2012 4 / 12
Triplet state (cont.)
Use the lowering operator to generate the state with total s = 1 and m = 0
S−↑↑ = (S(1)− + S
(2)− )↑↑ = (S
(1)− ↑)↑+ ↑(S (2)
− ↑)
= ~
√1
2
(3
2
)− 1
2
(−1
2
)(↓↑+ ↑↓) = ~(↓↑+ ↑↓) =
√2~
1√2
(↓↑+ ↑↓)
Apply the lowering operator again to get s = 1, m = −1
S−1√2
(↓↑+ ↑↓) = (S(1)− + S
(2)− )
1√2
(↓↑+ ↑↓)
=1√2
[S(1)− (↓↑+ ↑↓) + S
(2)− (↓↑+ ↑↓)
]=
1√2
[~↓↓+ ~↓↓]
=√
2~↓↓
The triplet state is thus
|1 1〉 = ↑↑, |1 0〉 =1√2
(↓↑+ ↑↓), |1−1〉 = ↓↓
C. Segre (IIT) PHYS 405 - Fall 2012 November 07, 2012 4 / 12
Triplet state (cont.)
Use the lowering operator to generate the state with total s = 1 and m = 0
S−↑↑ = (S(1)− + S
(2)− )↑↑ = (S
(1)− ↑)↑+ ↑(S (2)
− ↑)
= ~
√1
2
(3
2
)− 1
2
(−1
2
)(↓↑+ ↑↓) = ~(↓↑+ ↑↓) =
√2~
1√2
(↓↑+ ↑↓)
Apply the lowering operator again to get s = 1, m = −1
S−1√2
(↓↑+ ↑↓) = (S(1)− + S
(2)− )
1√2
(↓↑+ ↑↓)
=1√2
[S(1)− (↓↑+ ↑↓) + S
(2)− (↓↑+ ↑↓)
]=
1√2
[~↓↓+ ~↓↓]
=√
2~↓↓
The triplet state is thus
|1 1〉 = ↑↑, |1 0〉 =1√2
(↓↑+ ↑↓), |1−1〉 = ↓↓
C. Segre (IIT) PHYS 405 - Fall 2012 November 07, 2012 4 / 12
Triplet state (cont.)
Use the lowering operator to generate the state with total s = 1 and m = 0
S−↑↑ = (S(1)− + S
(2)− )↑↑ = (S
(1)− ↑)↑+ ↑(S (2)
− ↑)
= ~
√1
2
(3
2
)− 1
2
(−1
2
)(↓↑+ ↑↓) = ~(↓↑+ ↑↓) =
√2~
1√2
(↓↑+ ↑↓)
Apply the lowering operator again to get s = 1, m = −1
S−1√2
(↓↑+ ↑↓) = (S(1)− + S
(2)− )
1√2
(↓↑+ ↑↓)
=1√2
[S(1)− (↓↑+ ↑↓) + S
(2)− (↓↑+ ↑↓)
]=
1√2
[~↓↓+ ~↓↓]
=√
2~↓↓
The triplet state is thus
|1 1〉 = ↑↑, |1 0〉 =1√2
(↓↑+ ↑↓), |1−1〉 = ↓↓
C. Segre (IIT) PHYS 405 - Fall 2012 November 07, 2012 4 / 12
Triplet state (cont.)
Use the lowering operator to generate the state with total s = 1 and m = 0
S−↑↑ = (S(1)− + S
(2)− )↑↑ = (S
(1)− ↑)↑+ ↑(S (2)
− ↑)
= ~
√1
2
(3
2
)− 1
2
(−1
2
)(↓↑+ ↑↓) = ~(↓↑+ ↑↓) =
√2~
1√2
(↓↑+ ↑↓)
Apply the lowering operator again to get s = 1, m = −1
S−1√2
(↓↑+ ↑↓) = (S(1)− + S
(2)− )
1√2
(↓↑+ ↑↓)
=1√2
[S(1)− (↓↑+ ↑↓) + S
(2)− (↓↑+ ↑↓)
]=
1√2
[~↓↓+ ~↓↓]
=√
2~↓↓
The triplet state is thus
|1 1〉 = ↑↑, |1 0〉 =1√2
(↓↑+ ↑↓), |1−1〉 = ↓↓
C. Segre (IIT) PHYS 405 - Fall 2012 November 07, 2012 4 / 12
Singlet state
Recall that we had four possible combinations of the two spins, but wehave used all four to generate only two states so there is one combinationremaining and it must be
|0 0〉 =1√2
(↑↓ − ↓↑)
S2 1√2
(↑↓ − ↓↑) = (S (1))21√2
(↑↓ − ↓↑) + (S (2))21√2
(↑↓ − ↓↑)
+ 2(S(1)x S
(2)x + S
(1)y S
(2)y + S
(1)z S
(2)z )
1√2
(↑↓ − ↓↑)
S(1)x S
(2)x ↑↓ =
1
4~2↓↑, S
(1)y S
(2)y ↑↓ = −i2 1
4~2↓↑, S
(1)z S
(2)z ↑↓ = −1
4~2↑↓
S2 1√2
(↑↓ − ↓↑) =
[
3
4~2 +
3
4~2 − 2
3
4~2
]1√2
(↑↓ − ↓↑) = 0
C. Segre (IIT) PHYS 405 - Fall 2012 November 07, 2012 5 / 12
Singlet state
Recall that we had four possible combinations of the two spins, but wehave used all four to generate only two states so there is one combinationremaining and it must be
|0 0〉 =1√2
(↑↓ − ↓↑)
S2 1√2
(↑↓ − ↓↑) = (S (1))21√2
(↑↓ − ↓↑) + (S (2))21√2
(↑↓ − ↓↑)
+ 2(S(1)x S
(2)x + S
(1)y S
(2)y + S
(1)z S
(2)z )
1√2
(↑↓ − ↓↑)
S(1)x S
(2)x ↑↓ =
1
4~2↓↑, S
(1)y S
(2)y ↑↓ = −i2 1
4~2↓↑, S
(1)z S
(2)z ↑↓ = −1
4~2↑↓
S2 1√2
(↑↓ − ↓↑) =
[
3
4~2 +
3
4~2 − 2
3
4~2
]1√2
(↑↓ − ↓↑) = 0
C. Segre (IIT) PHYS 405 - Fall 2012 November 07, 2012 5 / 12
Singlet state
Recall that we had four possible combinations of the two spins, but wehave used all four to generate only two states so there is one combinationremaining and it must be
|0 0〉 =1√2
(↑↓ − ↓↑)
S2 1√2
(↑↓ − ↓↑)
= (S (1))21√2
(↑↓ − ↓↑) + (S (2))21√2
(↑↓ − ↓↑)
+ 2(S(1)x S
(2)x + S
(1)y S
(2)y + S
(1)z S
(2)z )
1√2
(↑↓ − ↓↑)
S(1)x S
(2)x ↑↓ =
1
4~2↓↑, S
(1)y S
(2)y ↑↓ = −i2 1
4~2↓↑, S
(1)z S
(2)z ↑↓ = −1
4~2↑↓
S2 1√2
(↑↓ − ↓↑) =
[
3
4~2 +
3
4~2 − 2
3
4~2
]1√2
(↑↓ − ↓↑) = 0
C. Segre (IIT) PHYS 405 - Fall 2012 November 07, 2012 5 / 12
Singlet state
Recall that we had four possible combinations of the two spins, but wehave used all four to generate only two states so there is one combinationremaining and it must be
|0 0〉 =1√2
(↑↓ − ↓↑)
S2 1√2
(↑↓ − ↓↑) = (S (1))21√2
(↑↓ − ↓↑) + (S (2))21√2
(↑↓ − ↓↑)
+ 2(S(1)x S
(2)x + S
(1)y S
(2)y + S
(1)z S
(2)z )
1√2
(↑↓ − ↓↑)
S(1)x S
(2)x ↑↓ =
1
4~2↓↑, S
(1)y S
(2)y ↑↓ = −i2 1
4~2↓↑, S
(1)z S
(2)z ↑↓ = −1
4~2↑↓
S2 1√2
(↑↓ − ↓↑) =
[
3
4~2 +
3
4~2 − 2
3
4~2
]1√2
(↑↓ − ↓↑) = 0
C. Segre (IIT) PHYS 405 - Fall 2012 November 07, 2012 5 / 12
Singlet state
Recall that we had four possible combinations of the two spins, but wehave used all four to generate only two states so there is one combinationremaining and it must be
|0 0〉 =1√2
(↑↓ − ↓↑)
S2 1√2
(↑↓ − ↓↑) = (S (1))21√2
(↑↓ − ↓↑) + (S (2))21√2
(↑↓ − ↓↑)
+ 2(S(1)x S
(2)x + S
(1)y S
(2)y + S
(1)z S
(2)z )
1√2
(↑↓ − ↓↑)
S(1)x S
(2)x ↑↓ =
1
4~2↓↑, S
(1)y S
(2)y ↑↓ = −i2 1
4~2↓↑, S
(1)z S
(2)z ↑↓ = −1
4~2↑↓
S2 1√2
(↑↓ − ↓↑) =
[
3
4~2 +
3
4~2 − 2
3
4~2
]1√2
(↑↓ − ↓↑) = 0
C. Segre (IIT) PHYS 405 - Fall 2012 November 07, 2012 5 / 12
Singlet state
Recall that we had four possible combinations of the two spins, but wehave used all four to generate only two states so there is one combinationremaining and it must be
|0 0〉 =1√2
(↑↓ − ↓↑)
S2 1√2
(↑↓ − ↓↑) = (S (1))21√2
(↑↓ − ↓↑) + (S (2))21√2
(↑↓ − ↓↑)
+ 2(S(1)x S
(2)x + S
(1)y S
(2)y + S
(1)z S
(2)z )
1√2
(↑↓ − ↓↑)
S(1)x S
(2)x ↑↓ =
1
4~2↓↑, S
(1)y S
(2)y ↑↓ = −i2 1
4~2↓↑, S
(1)z S
(2)z ↑↓ = −1
4~2↑↓
S2 1√2
(↑↓ − ↓↑) =
[
3
4~2 +
3
4~2 − 2
3
4~2
]1√2
(↑↓ − ↓↑)
= 0
C. Segre (IIT) PHYS 405 - Fall 2012 November 07, 2012 5 / 12
Singlet state
Recall that we had four possible combinations of the two spins, but wehave used all four to generate only two states so there is one combinationremaining and it must be
|0 0〉 =1√2
(↑↓ − ↓↑)
S2 1√2
(↑↓ − ↓↑) = (S (1))21√2
(↑↓ − ↓↑) + (S (2))21√2
(↑↓ − ↓↑)
+ 2(S(1)x S
(2)x + S
(1)y S
(2)y + S
(1)z S
(2)z )
1√2
(↑↓ − ↓↑)
S(1)x S
(2)x ↑↓ =
1
4~2↓↑, S
(1)y S
(2)y ↑↓ = −i2 1
4~2↓↑, S
(1)z S
(2)z ↑↓ = −1
4~2↑↓
S2 1√2
(↑↓ − ↓↑) =
[3
4~2
+3
4~2 − 2
3
4~2
]1√2
(↑↓ − ↓↑)
= 0
C. Segre (IIT) PHYS 405 - Fall 2012 November 07, 2012 5 / 12
Singlet state
Recall that we had four possible combinations of the two spins, but wehave used all four to generate only two states so there is one combinationremaining and it must be
|0 0〉 =1√2
(↑↓ − ↓↑)
S2 1√2
(↑↓ − ↓↑) = (S (1))21√2
(↑↓ − ↓↑) + (S (2))21√2
(↑↓ − ↓↑)
+ 2(S(1)x S
(2)x + S
(1)y S
(2)y + S
(1)z S
(2)z )
1√2
(↑↓ − ↓↑)
S(1)x S
(2)x ↑↓ =
1
4~2↓↑, S
(1)y S
(2)y ↑↓ = −i2 1
4~2↓↑, S
(1)z S
(2)z ↑↓ = −1
4~2↑↓
S2 1√2
(↑↓ − ↓↑) =
[3
4~2 +
3
4~2
− 23
4~2
]1√2
(↑↓ − ↓↑)
= 0
C. Segre (IIT) PHYS 405 - Fall 2012 November 07, 2012 5 / 12
Singlet state
Recall that we had four possible combinations of the two spins, but wehave used all four to generate only two states so there is one combinationremaining and it must be
|0 0〉 =1√2
(↑↓ − ↓↑)
S2 1√2
(↑↓ − ↓↑) = (S (1))21√2
(↑↓ − ↓↑) + (S (2))21√2
(↑↓ − ↓↑)
+ 2(S(1)x S
(2)x + S
(1)y S
(2)y + S
(1)z S
(2)z )
1√2
(↑↓ − ↓↑)
S(1)x S
(2)x ↑↓ =
1
4~2↓↑, S
(1)y S
(2)y ↑↓ = −i2 1
4~2↓↑, S
(1)z S
(2)z ↑↓ = −1
4~2↑↓
S2 1√2
(↑↓ − ↓↑) =
[3
4~2 +
3
4~2 − 2
3
4~2]
1√2
(↑↓ − ↓↑)
= 0
C. Segre (IIT) PHYS 405 - Fall 2012 November 07, 2012 5 / 12
Singlet state
Recall that we had four possible combinations of the two spins, but wehave used all four to generate only two states so there is one combinationremaining and it must be
|0 0〉 =1√2
(↑↓ − ↓↑)
S2 1√2
(↑↓ − ↓↑) = (S (1))21√2
(↑↓ − ↓↑) + (S (2))21√2
(↑↓ − ↓↑)
+ 2(S(1)x S
(2)x + S
(1)y S
(2)y + S
(1)z S
(2)z )
1√2
(↑↓ − ↓↑)
S(1)x S
(2)x ↑↓ =
1
4~2↓↑, S
(1)y S
(2)y ↑↓ = −i2 1
4~2↓↑, S
(1)z S
(2)z ↑↓ = −1
4~2↑↓
S2 1√2
(↑↓ − ↓↑) =
[3
4~2 +
3
4~2 − 2
3
4~2]
1√2
(↑↓ − ↓↑) = 0
C. Segre (IIT) PHYS 405 - Fall 2012 November 07, 2012 5 / 12
Eigenvalues of Sz
We can also check that these are eigenfunctions of the Sz operator
Sz |0 0〉 = (S(1)z + S
(2)z )
1√2
(↑↓ − ↓↑)
=1√2
(S(1)z ↑↓+ S
(2)z ↑↓ − S
(1)z ↓↑ − S
(2)z ↓↑)
=1√2~(
1
2↑↓ − 1
2↑↓+
1
2↓↑ − 1
2↓↑)
= 0
Sz |1 1〉 = (S(1)z + S
(2)z )↑↑ = ~
(1
2+
1
2
)↑↑ = +1~↑↑
Sz |1 0〉 = (S(1)z + S
(2)z )
1√2
(↑↓+ ↓↑)
=1√2~(
1
2↑↓ − 1
2↑↓ − 1
2↓↑+
1
2↓↑)
= 0
Sz |1−1〉 = (S(1)z + S
(2)z )↓↓ = ~
(−1
2− 1
2
)↓↓ = −1~↓↓
C. Segre (IIT) PHYS 405 - Fall 2012 November 07, 2012 6 / 12
Eigenvalues of Sz
We can also check that these are eigenfunctions of the Sz operator
Sz |0 0〉 = (S(1)z + S
(2)z )
1√2
(↑↓ − ↓↑)
=1√2
(S(1)z ↑↓+ S
(2)z ↑↓ − S
(1)z ↓↑ − S
(2)z ↓↑)
=1√2~(
1
2↑↓ − 1
2↑↓+
1
2↓↑ − 1
2↓↑)
= 0
Sz |1 1〉 = (S(1)z + S
(2)z )↑↑ = ~
(1
2+
1
2
)↑↑ = +1~↑↑
Sz |1 0〉 = (S(1)z + S
(2)z )
1√2
(↑↓+ ↓↑)
=1√2~(
1
2↑↓ − 1
2↑↓ − 1
2↓↑+
1
2↓↑)
= 0
Sz |1−1〉 = (S(1)z + S
(2)z )↓↓ = ~
(−1
2− 1
2
)↓↓ = −1~↓↓
C. Segre (IIT) PHYS 405 - Fall 2012 November 07, 2012 6 / 12
Eigenvalues of Sz
We can also check that these are eigenfunctions of the Sz operator
Sz |0 0〉 = (S(1)z + S
(2)z )
1√2
(↑↓ − ↓↑)
=1√2
(S(1)z ↑↓+ S
(2)z ↑↓ − S
(1)z ↓↑ − S
(2)z ↓↑)
=1√2~(
1
2↑↓ − 1
2↑↓+
1
2↓↑ − 1
2↓↑)
= 0
Sz |1 1〉 = (S(1)z + S
(2)z )↑↑ = ~
(1
2+
1
2
)↑↑ = +1~↑↑
Sz |1 0〉 = (S(1)z + S
(2)z )
1√2
(↑↓+ ↓↑)
=1√2~(
1
2↑↓ − 1
2↑↓ − 1
2↓↑+
1
2↓↑)
= 0
Sz |1−1〉 = (S(1)z + S
(2)z )↓↓ = ~
(−1
2− 1
2
)↓↓ = −1~↓↓
C. Segre (IIT) PHYS 405 - Fall 2012 November 07, 2012 6 / 12
Eigenvalues of Sz
We can also check that these are eigenfunctions of the Sz operator
Sz |0 0〉 = (S(1)z + S
(2)z )
1√2
(↑↓ − ↓↑)
=1√2
(S(1)z ↑↓+ S
(2)z ↑↓ − S
(1)z ↓↑ − S
(2)z ↓↑)
=1√2~(
1
2↑↓ − 1
2↑↓+
1
2↓↑ − 1
2↓↑)
= 0
Sz |1 1〉 = (S(1)z + S
(2)z )↑↑ = ~
(1
2+
1
2
)↑↑ = +1~↑↑
Sz |1 0〉 = (S(1)z + S
(2)z )
1√2
(↑↓+ ↓↑)
=1√2~(
1
2↑↓ − 1
2↑↓ − 1
2↓↑+
1
2↓↑)
= 0
Sz |1−1〉 = (S(1)z + S
(2)z )↓↓ = ~
(−1
2− 1
2
)↓↓ = −1~↓↓
C. Segre (IIT) PHYS 405 - Fall 2012 November 07, 2012 6 / 12
Eigenvalues of Sz
We can also check that these are eigenfunctions of the Sz operator
Sz |0 0〉 = (S(1)z + S
(2)z )
1√2
(↑↓ − ↓↑)
=1√2
(S(1)z ↑↓+ S
(2)z ↑↓ − S
(1)z ↓↑ − S
(2)z ↓↑)
=1√2~(
1
2↑↓ − 1
2↑↓+
1
2↓↑ − 1
2↓↑)
= 0
Sz |1 1〉 = (S(1)z + S
(2)z )↑↑ = ~
(1
2+
1
2
)↑↑ = +1~↑↑
Sz |1 0〉 = (S(1)z + S
(2)z )
1√2
(↑↓+ ↓↑)
=1√2~(
1
2↑↓ − 1
2↑↓ − 1
2↓↑+
1
2↓↑)
= 0
Sz |1−1〉 = (S(1)z + S
(2)z )↓↓ = ~
(−1
2− 1
2
)↓↓ = −1~↓↓
C. Segre (IIT) PHYS 405 - Fall 2012 November 07, 2012 6 / 12
Eigenvalues of Sz
We can also check that these are eigenfunctions of the Sz operator
Sz |0 0〉 = (S(1)z + S
(2)z )
1√2
(↑↓ − ↓↑)
=1√2
(S(1)z ↑↓+ S
(2)z ↑↓ − S
(1)z ↓↑ − S
(2)z ↓↑)
=1√2~(
1
2↑↓ − 1
2↑↓+
1
2↓↑ − 1
2↓↑)
= 0
Sz |1 1〉 = (S(1)z + S
(2)z )↑↑
= ~(
1
2+
1
2
)↑↑ = +1~↑↑
Sz |1 0〉 = (S(1)z + S
(2)z )
1√2
(↑↓+ ↓↑)
=1√2~(
1
2↑↓ − 1
2↑↓ − 1
2↓↑+
1
2↓↑)
= 0
Sz |1−1〉 = (S(1)z + S
(2)z )↓↓ = ~
(−1
2− 1
2
)↓↓ = −1~↓↓
C. Segre (IIT) PHYS 405 - Fall 2012 November 07, 2012 6 / 12
Eigenvalues of Sz
We can also check that these are eigenfunctions of the Sz operator
Sz |0 0〉 = (S(1)z + S
(2)z )
1√2
(↑↓ − ↓↑)
=1√2
(S(1)z ↑↓+ S
(2)z ↑↓ − S
(1)z ↓↑ − S
(2)z ↓↑)
=1√2~(
1
2↑↓ − 1
2↑↓+
1
2↓↑ − 1
2↓↑)
= 0
Sz |1 1〉 = (S(1)z + S
(2)z )↑↑ = ~
(1
2+
1
2
)↑↑
= +1~↑↑
Sz |1 0〉 = (S(1)z + S
(2)z )
1√2
(↑↓+ ↓↑)
=1√2~(
1
2↑↓ − 1
2↑↓ − 1
2↓↑+
1
2↓↑)
= 0
Sz |1−1〉 = (S(1)z + S
(2)z )↓↓ = ~
(−1
2− 1
2
)↓↓ = −1~↓↓
C. Segre (IIT) PHYS 405 - Fall 2012 November 07, 2012 6 / 12
Eigenvalues of Sz
We can also check that these are eigenfunctions of the Sz operator
Sz |0 0〉 = (S(1)z + S
(2)z )
1√2
(↑↓ − ↓↑)
=1√2
(S(1)z ↑↓+ S
(2)z ↑↓ − S
(1)z ↓↑ − S
(2)z ↓↑)
=1√2~(
1
2↑↓ − 1
2↑↓+
1
2↓↑ − 1
2↓↑)
= 0
Sz |1 1〉 = (S(1)z + S
(2)z )↑↑ = ~
(1
2+
1
2
)↑↑ = +1~↑↑
Sz |1 0〉 = (S(1)z + S
(2)z )
1√2
(↑↓+ ↓↑)
=1√2~(
1
2↑↓ − 1
2↑↓ − 1
2↓↑+
1
2↓↑)
= 0
Sz |1−1〉 = (S(1)z + S
(2)z )↓↓ = ~
(−1
2− 1
2
)↓↓ = −1~↓↓
C. Segre (IIT) PHYS 405 - Fall 2012 November 07, 2012 6 / 12
Eigenvalues of Sz
We can also check that these are eigenfunctions of the Sz operator
Sz |0 0〉 = (S(1)z + S
(2)z )
1√2
(↑↓ − ↓↑)
=1√2
(S(1)z ↑↓+ S
(2)z ↑↓ − S
(1)z ↓↑ − S
(2)z ↓↑)
=1√2~(
1
2↑↓ − 1
2↑↓+
1
2↓↑ − 1
2↓↑)
= 0
Sz |1 1〉 = (S(1)z + S
(2)z )↑↑ = ~
(1
2+
1
2
)↑↑ = +1~↑↑
Sz |1 0〉 = (S(1)z + S
(2)z )
1√2
(↑↓+ ↓↑)
=1√2~(
1
2↑↓ − 1
2↑↓ − 1
2↓↑+
1
2↓↑)
= 0
Sz |1−1〉 = (S(1)z + S
(2)z )↓↓ = ~
(−1
2− 1
2
)↓↓ = −1~↓↓
C. Segre (IIT) PHYS 405 - Fall 2012 November 07, 2012 6 / 12
Eigenvalues of Sz
We can also check that these are eigenfunctions of the Sz operator
Sz |0 0〉 = (S(1)z + S
(2)z )
1√2
(↑↓ − ↓↑)
=1√2
(S(1)z ↑↓+ S
(2)z ↑↓ − S
(1)z ↓↑ − S
(2)z ↓↑)
=1√2~(
1
2↑↓ − 1
2↑↓+
1
2↓↑ − 1
2↓↑)
= 0
Sz |1 1〉 = (S(1)z + S
(2)z )↑↑ = ~
(1
2+
1
2
)↑↑ = +1~↑↑
Sz |1 0〉 = (S(1)z + S
(2)z )
1√2
(↑↓+ ↓↑)
=1√2~(
1
2↑↓ − 1
2↑↓ − 1
2↓↑+
1
2↓↑)
= 0
Sz |1−1〉 = (S(1)z + S
(2)z )↓↓ = ~
(−1
2− 1
2
)↓↓ = −1~↓↓
C. Segre (IIT) PHYS 405 - Fall 2012 November 07, 2012 6 / 12
Eigenvalues of Sz
We can also check that these are eigenfunctions of the Sz operator
Sz |0 0〉 = (S(1)z + S
(2)z )
1√2
(↑↓ − ↓↑)
=1√2
(S(1)z ↑↓+ S
(2)z ↑↓ − S
(1)z ↓↑ − S
(2)z ↓↑)
=1√2~(
1
2↑↓ − 1
2↑↓+
1
2↓↑ − 1
2↓↑)
= 0
Sz |1 1〉 = (S(1)z + S
(2)z )↑↑ = ~
(1
2+
1
2
)↑↑ = +1~↑↑
Sz |1 0〉 = (S(1)z + S
(2)z )
1√2
(↑↓+ ↓↑)
=1√2~(
1
2↑↓ − 1
2↑↓ − 1
2↓↑+
1
2↓↑)
= 0
Sz |1−1〉 = (S(1)z + S
(2)z )↓↓ = ~
(−1
2− 1
2
)↓↓ = −1~↓↓
C. Segre (IIT) PHYS 405 - Fall 2012 November 07, 2012 6 / 12
Eigenvalues of Sz
We can also check that these are eigenfunctions of the Sz operator
Sz |0 0〉 = (S(1)z + S
(2)z )
1√2
(↑↓ − ↓↑)
=1√2
(S(1)z ↑↓+ S
(2)z ↑↓ − S
(1)z ↓↑ − S
(2)z ↓↑)
=1√2~(
1
2↑↓ − 1
2↑↓+
1
2↓↑ − 1
2↓↑)
= 0
Sz |1 1〉 = (S(1)z + S
(2)z )↑↑ = ~
(1
2+
1
2
)↑↑ = +1~↑↑
Sz |1 0〉 = (S(1)z + S
(2)z )
1√2
(↑↓+ ↓↑)
=1√2~(
1
2↑↓ − 1
2↑↓ − 1
2↓↑+
1
2↓↑)
= 0
Sz |1−1〉 = (S(1)z + S
(2)z )↓↓
= ~(−1
2− 1
2
)↓↓ = −1~↓↓
C. Segre (IIT) PHYS 405 - Fall 2012 November 07, 2012 6 / 12
Eigenvalues of Sz
We can also check that these are eigenfunctions of the Sz operator
Sz |0 0〉 = (S(1)z + S
(2)z )
1√2
(↑↓ − ↓↑)
=1√2
(S(1)z ↑↓+ S
(2)z ↑↓ − S
(1)z ↓↑ − S
(2)z ↓↑)
=1√2~(
1
2↑↓ − 1
2↑↓+
1
2↓↑ − 1
2↓↑)
= 0
Sz |1 1〉 = (S(1)z + S
(2)z )↑↑ = ~
(1
2+
1
2
)↑↑ = +1~↑↑
Sz |1 0〉 = (S(1)z + S
(2)z )
1√2
(↑↓+ ↓↑)
=1√2~(
1
2↑↓ − 1
2↑↓ − 1
2↓↑+
1
2↓↑)
= 0
Sz |1−1〉 = (S(1)z + S
(2)z )↓↓ = ~
(−1
2− 1
2
)↓↓
= −1~↓↓
C. Segre (IIT) PHYS 405 - Fall 2012 November 07, 2012 6 / 12
Eigenvalues of Sz
We can also check that these are eigenfunctions of the Sz operator
Sz |0 0〉 = (S(1)z + S
(2)z )
1√2
(↑↓ − ↓↑)
=1√2
(S(1)z ↑↓+ S
(2)z ↑↓ − S
(1)z ↓↑ − S
(2)z ↓↑)
=1√2~(
1
2↑↓ − 1
2↑↓+
1
2↓↑ − 1
2↓↑)
= 0
Sz |1 1〉 = (S(1)z + S
(2)z )↑↑ = ~
(1
2+
1
2
)↑↑ = +1~↑↑
Sz |1 0〉 = (S(1)z + S
(2)z )
1√2
(↑↓+ ↓↑)
=1√2~(
1
2↑↓ − 1
2↑↓ − 1
2↓↑+
1
2↓↑)
= 0
Sz |1−1〉 = (S(1)z + S
(2)z )↓↓ = ~
(−1
2− 1
2
)↓↓ = −1~↓↓
C. Segre (IIT) PHYS 405 - Fall 2012 November 07, 2012 6 / 12
“Good” quantum numbers
|0 0〉 =1√2
(↑↓ − ↓↑)
|1 1〉 = ↑↑, |1 0〉 =1√2
(↓↑+ ↑↓), |1−1〉 = ↓↓
Inspecting the four solutions, we see that these states are still eigenstatesof the individual spin operators, S2 and S2 but not of their z-components,
S(1)z and S
(2)z
Thus the valid quantum numbers for these states are s, m, s, s – a generalresult
C. Segre (IIT) PHYS 405 - Fall 2012 November 07, 2012 7 / 12
“Good” quantum numbers
|0 0〉 =1√2
(↑↓ − ↓↑)
|1 1〉 = ↑↑, |1 0〉 =1√2
(↓↑+ ↑↓), |1−1〉 = ↓↓
Inspecting the four solutions, we see that these states are still eigenstatesof the individual spin operators, S2 and S2
but not of their z-components,
S(1)z and S
(2)z
Thus the valid quantum numbers for these states are s, m, s, s – a generalresult
C. Segre (IIT) PHYS 405 - Fall 2012 November 07, 2012 7 / 12
“Good” quantum numbers
|0 0〉 =1√2
(↑↓ − ↓↑)
|1 1〉 = ↑↑, |1 0〉 =1√2
(↓↑+ ↑↓), |1−1〉 = ↓↓
Inspecting the four solutions, we see that these states are still eigenstatesof the individual spin operators, S2 and S2 but not of their z-components,
S(1)z and S
(2)z
Thus the valid quantum numbers for these states are s, m, s, s – a generalresult
C. Segre (IIT) PHYS 405 - Fall 2012 November 07, 2012 7 / 12
“Good” quantum numbers
|0 0〉 =1√2
(↑↓ − ↓↑)
|1 1〉 = ↑↑, |1 0〉 =1√2
(↓↑+ ↑↓), |1−1〉 = ↓↓
Inspecting the four solutions, we see that these states are still eigenstatesof the individual spin operators, S2 and S2 but not of their z-components,
S(1)z and S
(2)z
Thus the valid quantum numbers for these states are s, m, s, s – a generalresult
C. Segre (IIT) PHYS 405 - Fall 2012 November 07, 2012 7 / 12
Adding angular momentum
Addition of spin for hydrogen is the simplest case of a more general case,which can be to add any form of angular momentum from multiple sources
For two spins s1 and s2, their“sum” gives all possible values of the spinfrom (s1 + s2) down to |s1 − s2| in integer steps
s = (s1 + s2), (s1 + s2 − 1), (s1 + s2 − 2), . . . , |s1 − s2|
the combined eigenstate of the total spin, |s m〉 is given by a linearcombination of the composite states
|s m〉 =∑
m1+m2=m
C s1s2m1m2m
|s1 m1〉|s2 m2〉
The probabilities are given by the Clebsch-Gordan coefficients, C s1s2m1m2m
C. Segre (IIT) PHYS 405 - Fall 2012 November 07, 2012 8 / 12
Adding angular momentum
Addition of spin for hydrogen is the simplest case of a more general case,which can be to add any form of angular momentum from multiple sources
For two spins s1 and s2, their“sum” gives all possible values of the spinfrom (s1 + s2) down to |s1 − s2| in integer steps
s = (s1 + s2), (s1 + s2 − 1), (s1 + s2 − 2), . . . , |s1 − s2|
the combined eigenstate of the total spin, |s m〉 is given by a linearcombination of the composite states
|s m〉 =∑
m1+m2=m
C s1s2m1m2m
|s1 m1〉|s2 m2〉
The probabilities are given by the Clebsch-Gordan coefficients, C s1s2m1m2m
C. Segre (IIT) PHYS 405 - Fall 2012 November 07, 2012 8 / 12
Adding angular momentum
Addition of spin for hydrogen is the simplest case of a more general case,which can be to add any form of angular momentum from multiple sources
For two spins s1 and s2, their“sum” gives all possible values of the spinfrom (s1 + s2) down to |s1 − s2| in integer steps
s = (s1 + s2), (s1 + s2 − 1), (s1 + s2 − 2), . . . , |s1 − s2|
the combined eigenstate of the total spin, |s m〉 is given by a linearcombination of the composite states
|s m〉 =∑
m1+m2=m
C s1s2m1m2m
|s1 m1〉|s2 m2〉
The probabilities are given by the Clebsch-Gordan coefficients, C s1s2m1m2m
C. Segre (IIT) PHYS 405 - Fall 2012 November 07, 2012 8 / 12
Adding angular momentum
Addition of spin for hydrogen is the simplest case of a more general case,which can be to add any form of angular momentum from multiple sources
For two spins s1 and s2, their“sum” gives all possible values of the spinfrom (s1 + s2) down to |s1 − s2| in integer steps
s = (s1 + s2), (s1 + s2 − 1), (s1 + s2 − 2), . . . , |s1 − s2|
the combined eigenstate of the total spin, |s m〉 is given by a linearcombination of the composite states
|s m〉 =∑
m1+m2=m
C s1s2m1m2m
|s1 m1〉|s2 m2〉
The probabilities are given by the Clebsch-Gordan coefficients, C s1s2m1m2m
C. Segre (IIT) PHYS 405 - Fall 2012 November 07, 2012 8 / 12
Adding angular momentum
Addition of spin for hydrogen is the simplest case of a more general case,which can be to add any form of angular momentum from multiple sources
For two spins s1 and s2, their“sum” gives all possible values of the spinfrom (s1 + s2) down to |s1 − s2| in integer steps
s = (s1 + s2), (s1 + s2 − 1), (s1 + s2 − 2), . . . , |s1 − s2|
the combined eigenstate of the total spin, |s m〉 is given by a linearcombination of the composite states
|s m〉 =∑
m1+m2=m
C s1s2m1m2m
|s1 m1〉|s2 m2〉
The probabilities are given by the Clebsch-Gordan coefficients, C s1s2m1m2m
C. Segre (IIT) PHYS 405 - Fall 2012 November 07, 2012 8 / 12
Adding angular momentum
Addition of spin for hydrogen is the simplest case of a more general case,which can be to add any form of angular momentum from multiple sources
For two spins s1 and s2, their“sum” gives all possible values of the spinfrom (s1 + s2) down to |s1 − s2| in integer steps
s = (s1 + s2), (s1 + s2 − 1), (s1 + s2 − 2), . . . , |s1 − s2|
the combined eigenstate of the total spin, |s m〉 is given by a linearcombination of the composite states
|s m〉 =∑
m1+m2=m
C s1s2m1m2m
|s1 m1〉|s2 m2〉
The probabilities are given by the Clebsch-Gordan coefficients, C s1s2m1m2m
C. Segre (IIT) PHYS 405 - Fall 2012 November 07, 2012 8 / 12
Clebsch-Gordan coefficients
consider the 2× 1 table of Clebsch-Gordan coefficients
|3 0 > =
√1
5|2 1 > |1−1 > +
√3
5|2 0 > |1 0 > +
√1
5|2−1 > |1 1 >
|2 0 > |1 0 > =
√3
5|3 0 > −
√2
5|1 0 >
C. Segre (IIT) PHYS 405 - Fall 2012 November 07, 2012 9 / 12
Clebsch-Gordan coefficients
consider the 2× 1 table of Clebsch-Gordan coefficients
|3 0 > =
√1
5|2 1 > |1−1 > +
√3
5|2 0 > |1 0 > +
√1
5|2−1 > |1 1 >
|2 0 > |1 0 > =
√3
5|3 0 > −
√2
5|1 0 >
C. Segre (IIT) PHYS 405 - Fall 2012 November 07, 2012 9 / 12
Clebsch-Gordan coefficients
consider the 2× 1 table of Clebsch-Gordan coefficients
|3 0 >
=
√1
5|2 1 > |1−1 > +
√3
5|2 0 > |1 0 > +
√1
5|2−1 > |1 1 >
|2 0 > |1 0 > =
√3
5|3 0 > −
√2
5|1 0 >
C. Segre (IIT) PHYS 405 - Fall 2012 November 07, 2012 9 / 12
Clebsch-Gordan coefficients
consider the 2× 1 table of Clebsch-Gordan coefficients
|3 0 > =
√1
5|2 1 > |1−1 >
+
√3
5|2 0 > |1 0 > +
√1
5|2−1 > |1 1 >
|2 0 > |1 0 > =
√3
5|3 0 > −
√2
5|1 0 >
C. Segre (IIT) PHYS 405 - Fall 2012 November 07, 2012 9 / 12
Clebsch-Gordan coefficients
consider the 2× 1 table of Clebsch-Gordan coefficients
|3 0 > =
√1
5|2 1 > |1−1 > +
√3
5|2 0 > |1 0 >
+
√1
5|2−1 > |1 1 >
|2 0 > |1 0 > =
√3
5|3 0 > −
√2
5|1 0 >
C. Segre (IIT) PHYS 405 - Fall 2012 November 07, 2012 9 / 12
Clebsch-Gordan coefficients
consider the 2× 1 table of Clebsch-Gordan coefficients
|3 0 > =
√1
5|2 1 > |1−1 > +
√3
5|2 0 > |1 0 > +
√1
5|2−1 > |1 1 >
|2 0 > |1 0 > =
√3
5|3 0 > −
√2
5|1 0 >
C. Segre (IIT) PHYS 405 - Fall 2012 November 07, 2012 9 / 12
Clebsch-Gordan coefficients
consider the 2× 1 table of Clebsch-Gordan coefficients
|3 0 > =
√1
5|2 1 > |1−1 > +
√3
5|2 0 > |1 0 > +
√1
5|2−1 > |1 1 >
|2 0 > |1 0 >
=
√3
5|3 0 > −
√2
5|1 0 >
C. Segre (IIT) PHYS 405 - Fall 2012 November 07, 2012 9 / 12
Clebsch-Gordan coefficients
consider the 2× 1 table of Clebsch-Gordan coefficients
|3 0 > =
√1
5|2 1 > |1−1 > +
√3
5|2 0 > |1 0 > +
√1
5|2−1 > |1 1 >
|2 0 > |1 0 > =
√3
5|3 0 >
−√
2
5|1 0 >
C. Segre (IIT) PHYS 405 - Fall 2012 November 07, 2012 9 / 12
Clebsch-Gordan coefficients
consider the 2× 1 table of Clebsch-Gordan coefficients
|3 0 > =
√1
5|2 1 > |1−1 > +
√3
5|2 0 > |1 0 > +
√1
5|2−1 > |1 1 >
|2 0 > |1 0 > =
√3
5|3 0 > −
√2
5|1 0 >
C. Segre (IIT) PHYS 405 - Fall 2012 November 07, 2012 9 / 12
Problem 4.55
The electron in a hydrogen atom occupies the combined spin and positionstate
R21
(√1/3Y 0
1 χ+ +√
2/3Y 11 χ−
)and let ~J ≡ ~L + ~S
(a) If you measured L2, what values might you get, and what is theprobability of each?
Repeat for (b) Lz , (c) S2, (d) Sz , (e) J2, (f) Jz .
(g) If you measured the position of the particle, what is the probabilitydensity for finding it at r , θ, φ?
(h) If you measured both the Z component of the spin and the distancefrom the origin, what is teh probability density for finding the particle withspin up and at a radius r?
C. Segre (IIT) PHYS 405 - Fall 2012 November 07, 2012 10 / 12
Problem 4.55
The electron in a hydrogen atom occupies the combined spin and positionstate
R21
(√1/3Y 0
1 χ+ +√
2/3Y 11 χ−
)and let ~J ≡ ~L + ~S
(a) If you measured L2, what values might you get, and what is theprobability of each?
Repeat for (b) Lz , (c) S2, (d) Sz , (e) J2, (f) Jz .
(g) If you measured the position of the particle, what is the probabilitydensity for finding it at r , θ, φ?
(h) If you measured both the Z component of the spin and the distancefrom the origin, what is teh probability density for finding the particle withspin up and at a radius r?
C. Segre (IIT) PHYS 405 - Fall 2012 November 07, 2012 10 / 12
Problem 4.55
The electron in a hydrogen atom occupies the combined spin and positionstate
R21
(√1/3Y 0
1 χ+ +√
2/3Y 11 χ−
)and let ~J ≡ ~L + ~S
(a) If you measured L2, what values might you get, and what is theprobability of each?
Repeat for (b) Lz , (c) S2, (d) Sz , (e) J2, (f) Jz .
(g) If you measured the position of the particle, what is the probabilitydensity for finding it at r , θ, φ?
(h) If you measured both the Z component of the spin and the distancefrom the origin, what is teh probability density for finding the particle withspin up and at a radius r?
C. Segre (IIT) PHYS 405 - Fall 2012 November 07, 2012 10 / 12
Problem 4.55
The electron in a hydrogen atom occupies the combined spin and positionstate
R21
(√1/3Y 0
1 χ+ +√
2/3Y 11 χ−
)and let ~J ≡ ~L + ~S
(a) If you measured L2, what values might you get, and what is theprobability of each?
Repeat for (b) Lz , (c) S2, (d) Sz , (e) J2, (f) Jz .
(g) If you measured the position of the particle, what is the probabilitydensity for finding it at r , θ, φ?
(h) If you measured both the Z component of the spin and the distancefrom the origin, what is teh probability density for finding the particle withspin up and at a radius r?
C. Segre (IIT) PHYS 405 - Fall 2012 November 07, 2012 10 / 12
Problem 4.55
The electron in a hydrogen atom occupies the combined spin and positionstate
R21
(√1/3Y 0
1 χ+ +√
2/3Y 11 χ−
)and let ~J ≡ ~L + ~S
(a) If you measured L2, what values might you get, and what is theprobability of each?
Repeat for (b) Lz , (c) S2, (d) Sz , (e) J2, (f) Jz .
(g) If you measured the position of the particle, what is the probabilitydensity for finding it at r , θ, φ?
(h) If you measured both the Z component of the spin and the distancefrom the origin, what is teh probability density for finding the particle withspin up and at a radius r?
C. Segre (IIT) PHYS 405 - Fall 2012 November 07, 2012 10 / 12
Problem 4.55
The electron in a hydrogen atom occupies the combined spin and positionstate
R21
(√1/3Y 0
1 χ+ +√
2/3Y 11 χ−
)and let ~J ≡ ~L + ~S
(a) If you measured L2, what values might you get, and what is theprobability of each?
Repeat for (b) Lz , (c) S2, (d) Sz , (e) J2, (f) Jz .
(g) If you measured the position of the particle, what is the probabilitydensity for finding it at r , θ, φ?
(h) If you measured both the Z component of the spin and the distancefrom the origin, what is teh probability density for finding the particle withspin up and at a radius r?
C. Segre (IIT) PHYS 405 - Fall 2012 November 07, 2012 10 / 12
Problem 4.55 (cont.)
R21
(√1/3Y 0
1 χ+ +√
2/3Y 11 χ−
)L2
Both terms have l = 1 so the mea-sured value of L2 is, with P = 1
〈L2〉 = ~2l(l + 1) = ~2[1(2)] = 2~2
Lz
One term has m = 0 and the otherhas m = 1 so we will have
〈Lz〉 = 0, P = 13
〈Lz〉 = ~, P = 23
S2
Both terms have s = 12
so the mea-sured value of S2 is, with P = 1
〈S2〉 = ~2s(s + 1) = ~2[ 12( 32)] = 3
4~2
Sz
One term has m = 12
and the otherhas m = − 1
2so we will have
〈Sz〉 = 12~, P = 1
3
〈Sz〉 = − 12~, P = 2
3
C. Segre (IIT) PHYS 405 - Fall 2012 November 07, 2012 11 / 12
Problem 4.55 (cont.)
R21
(√1/3Y 0
1 χ+ +√
2/3Y 11 χ−
)L2
Both terms have l = 1 so the mea-sured value of L2 is, with P = 1
〈L2〉 = ~2l(l + 1)
= ~2[1(2)] = 2~2
Lz
One term has m = 0 and the otherhas m = 1 so we will have
〈Lz〉 = 0, P = 13
〈Lz〉 = ~, P = 23
S2
Both terms have s = 12
so the mea-sured value of S2 is, with P = 1
〈S2〉 = ~2s(s + 1) = ~2[ 12( 32)] = 3
4~2
Sz
One term has m = 12
and the otherhas m = − 1
2so we will have
〈Sz〉 = 12~, P = 1
3
〈Sz〉 = − 12~, P = 2
3
C. Segre (IIT) PHYS 405 - Fall 2012 November 07, 2012 11 / 12
Problem 4.55 (cont.)
R21
(√1/3Y 0
1 χ+ +√
2/3Y 11 χ−
)L2
Both terms have l = 1 so the mea-sured value of L2 is, with P = 1
〈L2〉 = ~2l(l + 1) = ~2[1(2)] = 2~2
Lz
One term has m = 0 and the otherhas m = 1 so we will have
〈Lz〉 = 0, P = 13
〈Lz〉 = ~, P = 23
S2
Both terms have s = 12
so the mea-sured value of S2 is, with P = 1
〈S2〉 = ~2s(s + 1) = ~2[ 12( 32)] = 3
4~2
Sz
One term has m = 12
and the otherhas m = − 1
2so we will have
〈Sz〉 = 12~, P = 1
3
〈Sz〉 = − 12~, P = 2
3
C. Segre (IIT) PHYS 405 - Fall 2012 November 07, 2012 11 / 12
Problem 4.55 (cont.)
R21
(√1/3Y 0
1 χ+ +√
2/3Y 11 χ−
)L2
Both terms have l = 1 so the mea-sured value of L2 is, with P = 1
〈L2〉 = ~2l(l + 1) = ~2[1(2)] = 2~2
Lz
One term has m = 0 and the otherhas m = 1 so we will have
〈Lz〉 = 0, P = 13
〈Lz〉 = ~, P = 23
S2
Both terms have s = 12
so the mea-sured value of S2 is, with P = 1
〈S2〉 = ~2s(s + 1) = ~2[ 12( 32)] = 3
4~2
Sz
One term has m = 12
and the otherhas m = − 1
2so we will have
〈Sz〉 = 12~, P = 1
3
〈Sz〉 = − 12~, P = 2
3
C. Segre (IIT) PHYS 405 - Fall 2012 November 07, 2012 11 / 12
Problem 4.55 (cont.)
R21
(√1/3Y 0
1 χ+ +√
2/3Y 11 χ−
)L2
Both terms have l = 1 so the mea-sured value of L2 is, with P = 1
〈L2〉 = ~2l(l + 1) = ~2[1(2)] = 2~2
Lz
One term has m = 0 and the otherhas m = 1 so we will have
〈Lz〉 = 0, P = 13
〈Lz〉 = ~, P = 23
S2
Both terms have s = 12
so the mea-sured value of S2 is, with P = 1
〈S2〉 = ~2s(s + 1) = ~2[ 12( 32)] = 3
4~2
Sz
One term has m = 12
and the otherhas m = − 1
2so we will have
〈Sz〉 = 12~, P = 1
3
〈Sz〉 = − 12~, P = 2
3
C. Segre (IIT) PHYS 405 - Fall 2012 November 07, 2012 11 / 12
Problem 4.55 (cont.)
R21
(√1/3Y 0
1 χ+ +√
2/3Y 11 χ−
)L2
Both terms have l = 1 so the mea-sured value of L2 is, with P = 1
〈L2〉 = ~2l(l + 1) = ~2[1(2)] = 2~2
Lz
One term has m = 0 and the otherhas m = 1 so we will have
〈Lz〉 = 0, P = 13
〈Lz〉 = ~, P = 23
S2
Both terms have s = 12
so the mea-sured value of S2 is, with P = 1
〈S2〉 = ~2s(s + 1) = ~2[ 12( 32)] = 3
4~2
Sz
One term has m = 12
and the otherhas m = − 1
2so we will have
〈Sz〉 = 12~, P = 1
3
〈Sz〉 = − 12~, P = 2
3
C. Segre (IIT) PHYS 405 - Fall 2012 November 07, 2012 11 / 12
Problem 4.55 (cont.)
R21
(√1/3Y 0
1 χ+ +√
2/3Y 11 χ−
)L2
Both terms have l = 1 so the mea-sured value of L2 is, with P = 1
〈L2〉 = ~2l(l + 1) = ~2[1(2)] = 2~2
Lz
One term has m = 0 and the otherhas m = 1 so we will have
〈Lz〉 = 0, P = 13
〈Lz〉 = ~, P = 23
S2
Both terms have s = 12
so the mea-sured value of S2 is, with P = 1
〈S2〉 = ~2s(s + 1) = ~2[ 12( 32)] = 3
4~2
Sz
One term has m = 12
and the otherhas m = − 1
2so we will have
〈Sz〉 = 12~, P = 1
3
〈Sz〉 = − 12~, P = 2
3
C. Segre (IIT) PHYS 405 - Fall 2012 November 07, 2012 11 / 12
Problem 4.55 (cont.)
R21
(√1/3Y 0
1 χ+ +√
2/3Y 11 χ−
)L2
Both terms have l = 1 so the mea-sured value of L2 is, with P = 1
〈L2〉 = ~2l(l + 1) = ~2[1(2)] = 2~2
Lz
One term has m = 0 and the otherhas m = 1 so we will have
〈Lz〉 = 0, P = 13
〈Lz〉 = ~, P = 23
S2
Both terms have s = 12
so the mea-sured value of S2 is, with P = 1
〈S2〉 = ~2s(s + 1)
= ~2[ 12( 32)] = 3
4~2
Sz
One term has m = 12
and the otherhas m = − 1
2so we will have
〈Sz〉 = 12~, P = 1
3
〈Sz〉 = − 12~, P = 2
3
C. Segre (IIT) PHYS 405 - Fall 2012 November 07, 2012 11 / 12
Problem 4.55 (cont.)
R21
(√1/3Y 0
1 χ+ +√
2/3Y 11 χ−
)L2
Both terms have l = 1 so the mea-sured value of L2 is, with P = 1
〈L2〉 = ~2l(l + 1) = ~2[1(2)] = 2~2
Lz
One term has m = 0 and the otherhas m = 1 so we will have
〈Lz〉 = 0, P = 13
〈Lz〉 = ~, P = 23
S2
Both terms have s = 12
so the mea-sured value of S2 is, with P = 1
〈S2〉 = ~2s(s + 1) = ~2[ 12( 32)] = 3
4~2
Sz
One term has m = 12
and the otherhas m = − 1
2so we will have
〈Sz〉 = 12~, P = 1
3
〈Sz〉 = − 12~, P = 2
3
C. Segre (IIT) PHYS 405 - Fall 2012 November 07, 2012 11 / 12
Problem 4.55 (cont.)
R21
(√1/3Y 0
1 χ+ +√
2/3Y 11 χ−
)L2
Both terms have l = 1 so the mea-sured value of L2 is, with P = 1
〈L2〉 = ~2l(l + 1) = ~2[1(2)] = 2~2
Lz
One term has m = 0 and the otherhas m = 1 so we will have
〈Lz〉 = 0, P = 13
〈Lz〉 = ~, P = 23
S2
Both terms have s = 12
so the mea-sured value of S2 is, with P = 1
〈S2〉 = ~2s(s + 1) = ~2[ 12( 32)] = 3
4~2
Sz
One term has m = 12
and the otherhas m = − 1
2so we will have
〈Sz〉 = 12~, P = 1
3
〈Sz〉 = − 12~, P = 2
3
C. Segre (IIT) PHYS 405 - Fall 2012 November 07, 2012 11 / 12
Problem 4.55 (cont.)
R21
(√1/3Y 0
1 χ+ +√
2/3Y 11 χ−
)L2
Both terms have l = 1 so the mea-sured value of L2 is, with P = 1
〈L2〉 = ~2l(l + 1) = ~2[1(2)] = 2~2
Lz
One term has m = 0 and the otherhas m = 1 so we will have
〈Lz〉 = 0, P = 13
〈Lz〉 = ~, P = 23
S2
Both terms have s = 12
so the mea-sured value of S2 is, with P = 1
〈S2〉 = ~2s(s + 1) = ~2[ 12( 32)] = 3
4~2
Sz
One term has m = 12
and the otherhas m = − 1
2so we will have
〈Sz〉 = 12~, P = 1
3
〈Sz〉 = − 12~, P = 2
3
C. Segre (IIT) PHYS 405 - Fall 2012 November 07, 2012 11 / 12
Problem 4.55 (cont.)
R21
(√1/3Y 0
1 χ+ +√
2/3Y 11 χ−
)L2
Both terms have l = 1 so the mea-sured value of L2 is, with P = 1
〈L2〉 = ~2l(l + 1) = ~2[1(2)] = 2~2
Lz
One term has m = 0 and the otherhas m = 1 so we will have
〈Lz〉 = 0, P = 13
〈Lz〉 = ~, P = 23
S2
Both terms have s = 12
so the mea-sured value of S2 is, with P = 1
〈S2〉 = ~2s(s + 1) = ~2[ 12( 32)] = 3
4~2
Sz
One term has m = 12
and the otherhas m = − 1
2so we will have
〈Sz〉 = 12~, P = 1
3
〈Sz〉 = − 12~, P = 2
3
C. Segre (IIT) PHYS 405 - Fall 2012 November 07, 2012 11 / 12
Problem 4.55 (cont.)
Rewriting the angular and spin portion of the wave function with the firstbra-ket corresponding to the angular momentum
(√1/3Y 0
1 χ+ +√
2/3Y 11 χ−
)√
13|1 0〉 | 1
212〉+
√23|1 1〉 | 1
2−12〉
convert to eigenfunctions of J2 andJz using the 1×1/2 Clebsch-Gordantable √
13
[√23| 32
12〉 −
√13| 12
12〉]
+√
23
[√13| 32
12〉 −
√23| 12
12〉]
2√
23| 32
12〉+ 1
3| 12
12〉
for j = 32 , P = 8
9 , 〈J2〉 = 154 ~
2 ; for j = 12 , P = 1
9 , 〈J2〉 = 34~
2
for m = 12 , P = 1, 〈Jz〉 = 1
2~
C. Segre (IIT) PHYS 405 - Fall 2012 November 07, 2012 12 / 12
Problem 4.55 (cont.)
Rewriting the angular and spin portion of the wave function with the firstbra-ket corresponding to the angular momentum(√
1/3Y 01 χ+ +
√2/3Y 1
1 χ−
)
√13|1 0〉 | 1
212〉+
√23|1 1〉 | 1
2−12〉
convert to eigenfunctions of J2 andJz using the 1×1/2 Clebsch-Gordantable √
13
[√23| 32
12〉 −
√13| 12
12〉]
+√
23
[√13| 32
12〉 −
√23| 12
12〉]
2√
23| 32
12〉+ 1
3| 12
12〉
for j = 32 , P = 8
9 , 〈J2〉 = 154 ~
2 ; for j = 12 , P = 1
9 , 〈J2〉 = 34~
2
for m = 12 , P = 1, 〈Jz〉 = 1
2~
C. Segre (IIT) PHYS 405 - Fall 2012 November 07, 2012 12 / 12
Problem 4.55 (cont.)
Rewriting the angular and spin portion of the wave function with the firstbra-ket corresponding to the angular momentum(√
1/3Y 01 χ+ +
√2/3Y 1
1 χ−
)√
13|1 0〉 | 1
212〉+
√23|1 1〉 | 1
2−12〉
convert to eigenfunctions of J2 andJz using the 1×1/2 Clebsch-Gordantable √
13
[√23| 32
12〉 −
√13| 12
12〉]
+√
23
[√13| 32
12〉 −
√23| 12
12〉]
2√
23| 32
12〉+ 1
3| 12
12〉
for j = 32 , P = 8
9 , 〈J2〉 = 154 ~
2 ; for j = 12 , P = 1
9 , 〈J2〉 = 34~
2
for m = 12 , P = 1, 〈Jz〉 = 1
2~
C. Segre (IIT) PHYS 405 - Fall 2012 November 07, 2012 12 / 12
Problem 4.55 (cont.)
Rewriting the angular and spin portion of the wave function with the firstbra-ket corresponding to the angular momentum(√
1/3Y 01 χ+ +
√2/3Y 1
1 χ−
)√
13|1 0〉 | 1
212〉+
√23|1 1〉 | 1
2−12〉
convert to eigenfunctions of J2 andJz using the 1×1/2 Clebsch-Gordantable
√13
[√23| 32
12〉 −
√13| 12
12〉]
+√
23
[√13| 32
12〉 −
√23| 12
12〉]
2√
23| 32
12〉+ 1
3| 12
12〉
for j = 32 , P = 8
9 , 〈J2〉 = 154 ~
2 ; for j = 12 , P = 1
9 , 〈J2〉 = 34~
2
for m = 12 , P = 1, 〈Jz〉 = 1
2~
C. Segre (IIT) PHYS 405 - Fall 2012 November 07, 2012 12 / 12
Problem 4.55 (cont.)
Rewriting the angular and spin portion of the wave function with the firstbra-ket corresponding to the angular momentum(√
1/3Y 01 χ+ +
√2/3Y 1
1 χ−
)√
13|1 0〉 | 1
212〉+
√23|1 1〉 | 1
2−12〉
convert to eigenfunctions of J2 andJz using the 1×1/2 Clebsch-Gordantable
√13
[√23| 32
12〉 −
√13| 12
12〉]
+√
23
[√13| 32
12〉 −
√23| 12
12〉]
2√
23| 32
12〉+ 1
3| 12
12〉
for j = 32 , P = 8
9 , 〈J2〉 = 154 ~
2 ; for j = 12 , P = 1
9 , 〈J2〉 = 34~
2
for m = 12 , P = 1, 〈Jz〉 = 1
2~
C. Segre (IIT) PHYS 405 - Fall 2012 November 07, 2012 12 / 12
Problem 4.55 (cont.)
Rewriting the angular and spin portion of the wave function with the firstbra-ket corresponding to the angular momentum(√
1/3Y 01 χ+ +
√2/3Y 1
1 χ−
)√
13|1 0〉 | 1
212〉+
√23|1 1〉 | 1
2−12〉
convert to eigenfunctions of J2 andJz using the 1×1/2 Clebsch-Gordantable √
13
[√23| 32
12〉 −
√13| 12
12〉]
+√
23
[√13| 32
12〉 −
√23| 12
12〉]
2√
23| 32
12〉+ 1
3| 12
12〉
for j = 32 , P = 8
9 , 〈J2〉 = 154 ~
2 ; for j = 12 , P = 1
9 , 〈J2〉 = 34~
2
for m = 12 , P = 1, 〈Jz〉 = 1
2~
C. Segre (IIT) PHYS 405 - Fall 2012 November 07, 2012 12 / 12
Problem 4.55 (cont.)
Rewriting the angular and spin portion of the wave function with the firstbra-ket corresponding to the angular momentum(√
1/3Y 01 χ+ +
√2/3Y 1
1 χ−
)√
13|1 0〉 | 1
212〉+
√23|1 1〉 | 1
2−12〉
convert to eigenfunctions of J2 andJz using the 1×1/2 Clebsch-Gordantable √
13
[√23| 32
12〉 −
√13| 12
12〉]
+√
23
[√13| 32
12〉 −
√23| 12
12〉]
2√
23| 32
12〉+ 1
3| 12
12〉
for j = 32 , P = 8
9 , 〈J2〉 = 154 ~
2 ; for j = 12 , P = 1
9 , 〈J2〉 = 34~
2
for m = 12 , P = 1, 〈Jz〉 = 1
2~
C. Segre (IIT) PHYS 405 - Fall 2012 November 07, 2012 12 / 12
Problem 4.55 (cont.)
Rewriting the angular and spin portion of the wave function with the firstbra-ket corresponding to the angular momentum(√
1/3Y 01 χ+ +
√2/3Y 1
1 χ−
)√
13|1 0〉 | 1
212〉+
√23|1 1〉 | 1
2−12〉
convert to eigenfunctions of J2 andJz using the 1×1/2 Clebsch-Gordantable √
13
[√23| 32
12〉 −
√13| 12
12〉]
+√
23
[√13| 32
12〉 −
√23| 12
12〉]
2√
23| 32
12〉+ 1
3| 12
12〉
for j = 32 , P = 8
9 , 〈J2〉 = 154 ~
2 ; for j = 12 , P = 1
9 , 〈J2〉 = 34~
2
for m = 12 , P = 1, 〈Jz〉 = 1
2~
C. Segre (IIT) PHYS 405 - Fall 2012 November 07, 2012 12 / 12
Problem 4.55 (cont.)
Rewriting the angular and spin portion of the wave function with the firstbra-ket corresponding to the angular momentum(√
1/3Y 01 χ+ +
√2/3Y 1
1 χ−
)√
13|1 0〉 | 1
212〉+
√23|1 1〉 | 1
2−12〉
convert to eigenfunctions of J2 andJz using the 1×1/2 Clebsch-Gordantable √
13
[√23| 32
12〉 −
√13| 12
12〉]
+√
23
[√13| 32
12〉 −
√23| 12
12〉]
2√
23| 32
12〉+ 1
3| 12
12〉
for j = 32 , P = 8
9 , 〈J2〉 = 154 ~
2
; for j = 12 , P = 1
9 , 〈J2〉 = 34~
2
for m = 12 , P = 1, 〈Jz〉 = 1
2~
C. Segre (IIT) PHYS 405 - Fall 2012 November 07, 2012 12 / 12
Problem 4.55 (cont.)
Rewriting the angular and spin portion of the wave function with the firstbra-ket corresponding to the angular momentum(√
1/3Y 01 χ+ +
√2/3Y 1
1 χ−
)√
13|1 0〉 | 1
212〉+
√23|1 1〉 | 1
2−12〉
convert to eigenfunctions of J2 andJz using the 1×1/2 Clebsch-Gordantable √
13
[√23| 32
12〉 −
√13| 12
12〉]
+√
23
[√13| 32
12〉 −
√23| 12
12〉]
2√
23| 32
12〉+ 1
3| 12
12〉
for j = 32 , P = 8
9 , 〈J2〉 = 154 ~
2 ; for j = 12 , P = 1
9 , 〈J2〉 = 34~
2
for m = 12 , P = 1, 〈Jz〉 = 1
2~
C. Segre (IIT) PHYS 405 - Fall 2012 November 07, 2012 12 / 12
Problem 4.55 (cont.)
Rewriting the angular and spin portion of the wave function with the firstbra-ket corresponding to the angular momentum(√
1/3Y 01 χ+ +
√2/3Y 1
1 χ−
)√
13|1 0〉 | 1
212〉+
√23|1 1〉 | 1
2−12〉
convert to eigenfunctions of J2 andJz using the 1×1/2 Clebsch-Gordantable √
13
[√23| 32
12〉 −
√13| 12
12〉]
+√
23
[√13| 32
12〉 −
√23| 12
12〉]
2√
23| 32
12〉+ 1
3| 12
12〉
for j = 32 , P = 8
9 , 〈J2〉 = 154 ~
2 ; for j = 12 , P = 1
9 , 〈J2〉 = 34~
2
for m = 12 , P = 1, 〈Jz〉 = 1
2~
C. Segre (IIT) PHYS 405 - Fall 2012 November 07, 2012 12 / 12
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