today’s outline - november 07, 2012 - iitphys.iit.edu/~segre/phys405/12f/lecture_21.pdf ·...

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



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

2.

What is the total angularmomentum of the atom?


↑↑ ↑↓ ↓↑ ↓↓


~S ≡ ~S (1) + ~S (2)


Szχ1χ2 = (S(1)z + S

(2)z )χ1χ2 = (S

(1)z χ1)χ2 + χ1(S

(2)z χ2)

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







↑↑ ↑↓ ↓↑ ↓↓


~S ≡ ~S (1) + ~S (2)


Szχ1χ2 = (S(1)z + S

(2)z )χ1χ2 = (S

(1)z χ1)χ2 + χ1(S

(2)z χ2)

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







↑↑

↑↓ ↓↑ ↓↓


~S ≡ ~S (1) + ~S (2)


Szχ1χ2 = (S(1)z + S

(2)z )χ1χ2 = (S

(1)z χ1)χ2 + χ1(S

(2)z χ2)

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







↑↑ ↑↓

↓↑ ↓↓


~S ≡ ~S (1) + ~S (2)


Szχ1χ2 = (S(1)z + S

(2)z )χ1χ2 = (S

(1)z χ1)χ2 + χ1(S

(2)z χ2)

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







↑↑ ↑↓ ↓↑

↓↓


~S ≡ ~S (1) + ~S (2)


Szχ1χ2 = (S(1)z + S

(2)z )χ1χ2 = (S

(1)z χ1)χ2 + χ1(S

(2)z χ2)

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







↑↑ ↑↓ ↓↑ ↓↓


~S ≡ ~S (1) + ~S (2)


Szχ1χ2 = (S(1)z + S

(2)z )χ1χ2 = (S

(1)z χ1)χ2 + χ1(S

(2)z χ2)

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







↑↑ ↑↓ ↓↑ ↓↓


~S ≡ ~S (1) + ~S (2)


Szχ1χ2 = (S(1)z + S

(2)z )χ1χ2

= (S(1)z χ1)χ2 + χ1(S

(2)z χ2)

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







↑↑ ↑↓ ↓↑ ↓↓


~S ≡ ~S (1) + ~S (2)


Szχ1χ2 = (S(1)z + S

(2)z )χ1χ2 = (S

(1)z χ1)χ2 + χ1(S

(2)z χ2)

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







↑↑ ↑↓ ↓↑ ↓↓


~S ≡ ~S (1) + ~S (2)


Szχ1χ2 = (S(1)z + S

(2)z )χ1χ2 = (S

(1)z χ1)χ2 + χ1(S

(2)z χ2)

= (~m1χ1)χ2 + χ1(~m2χ2)

= ~(m1 + m2)χ1χ2







↑↑ ↑↓ ↓↑ ↓↓


~S ≡ ~S (1) + ~S (2)


Szχ1χ2 = (S(1)z + S

(2)z )χ1χ2 = (S

(1)z χ1)χ2 + χ1(S

(2)z χ2)

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


Triplet state

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

Let’s check this

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

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

(2)x + S

(1)y S

(2)y + S

(1)z S

(2)z )

S(1)x S

(2)x ↑↑ =

~2

(0 11 0

)(10

)~2

(0 11 0

)(10

)=

~2

4

(01

)(01

)=

~2

4↓↓

S(1)y S

(2)y ↑↑ =

~2

(0 −ii 0

)(10

)~2

(0 −ii 0

)(10

)= −~2

4

(01

)(01

)= −~2

4↓↓

S(1)z S

(2)z ↑↑ =

~2

(1 00 −1

)(10

)~2

(1 00 −1

)(10

)=

~2

4

(10

)(10

)=

~2

4↑↑

S2↑↑ = ~21

2

(3

2

)↑↑+ ~2

1

2

(3

2

)↑↑+ 2

(~2

4↓↓ − ~2

4↓↓+

~2

4↑↑)

= 2~2↑↑


Triplet state

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

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

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

(2)x + S

(1)y S

(2)y + S

(1)z S

(2)z )

S(1)x S

(2)x ↑↑ =

~2

(0 11 0

)(10

)~2

(0 11 0

)(10

)=

~2

4

(01

)(01

)=

~2

4↓↓

S(1)y S

(2)y ↑↑ =

~2

(0 −ii 0

)(10

)~2

(0 −ii 0

)(10

)= −~2

4

(01

)(01

)= −~2

4↓↓

S(1)z S

(2)z ↑↑ =

~2

(1 00 −1

)(10

)~2

(1 00 −1

)(10

)=

~2

4

(10

)(10

)=

~2

4↑↑

S2↑↑ = ~21

2

(3

2

)↑↑+ ~2

1

2

(3

2

)↑↑+ 2

(~2

4↓↓ − ~2

4↓↓+

~2

4↑↑)

= 2~2↑↑


Triplet state


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

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

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

(2)x + S

(1)y S

(2)y + S

(1)z S

(2)z )

S(1)x S

(2)x ↑↑ =

~2

(0 11 0

)(10

)~2

(0 11 0

)(10

)=

~2

4

(01

)(01

)=

~2

4↓↓

S(1)y S

(2)y ↑↑ =

~2

(0 −ii 0

)(10

)~2

(0 −ii 0

)(10

)= −~2

4

(01

)(01

)= −~2

4↓↓

S(1)z S

(2)z ↑↑ =

~2

(1 00 −1

)(10

)~2

(1 00 −1

)(10

)=

~2

4

(10

)(10

)=

~2

4↑↑

S2↑↑ = ~21

2

(3

2

)↑↑+ ~2

1

2

(3

2

)↑↑+ 2

(~2

4↓↓ − ~2

4↓↓+

~2

4↑↑)

= 2~2↑↑


Triplet state


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

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

(2)x + S

(1)y S

(2)y + S

(1)z S

(2)z )

S(1)x S

(2)x ↑↑ =

~2

(0 11 0

)(10

)~2

(0 11 0

)(10

)=

~2

4

(01

)(01

)=

~2

4↓↓

S(1)y S

(2)y ↑↑ =

~2

(0 −ii 0

)(10

)~2

(0 −ii 0

)(10

)= −~2

4

(01

)(01

)= −~2

4↓↓

S(1)z S

(2)z ↑↑ =

~2

(1 00 −1

)(10

)~2

(1 00 −1

)(10

)=

~2

4

(10

)(10

)=

~2

4↑↑

S2↑↑ = ~21

2

(3

2

)↑↑+ ~2

1

2

(3

2

)↑↑+ 2

(~2

4↓↓ − ~2

4↓↓+

~2

4↑↑)

= 2~2↑↑


Triplet state


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

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

(2)x + S

(1)y S

(2)y + S

(1)z S

(2)z )

S(1)x S

(2)x ↑↑

=~2

(0 11 0

)(10

)~2

(0 11 0

)(10

)=

~2

4

(01

)(01

)=

~2

4↓↓

S(1)y S

(2)y ↑↑ =

~2

(0 −ii 0

)(10

)~2

(0 −ii 0

)(10

)= −~2

4

(01

)(01

)= −~2

4↓↓

S(1)z S

(2)z ↑↑ =

~2

(1 00 −1

)(10

)~2

(1 00 −1

)(10

)=

~2

4

(10

)(10

)=

~2

4↑↑

S2↑↑ = ~21

2

(3

2

)↑↑+ ~2

1

2

(3

2

)↑↑+ 2

(~2

4↓↓ − ~2

4↓↓+

~2

4↑↑)

= 2~2↑↑


Triplet state


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

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

(2)x + S

(1)y S

(2)y + S

(1)z S

(2)z )

S(1)x S

(2)x ↑↑ =

~2

(0 11 0

)(10

)~2

(0 11 0

)(10

)

=~2

4

(01

)(01

)=

~2

4↓↓

S(1)y S

(2)y ↑↑ =

~2

(0 −ii 0

)(10

)~2

(0 −ii 0

)(10

)= −~2

4

(01

)(01

)= −~2

4↓↓

S(1)z S

(2)z ↑↑ =

~2

(1 00 −1

)(10

)~2

(1 00 −1

)(10

)=

~2

4

(10

)(10

)=

~2

4↑↑

S2↑↑ = ~21

2

(3

2

)↑↑+ ~2

1

2

(3

2

)↑↑+ 2

(~2

4↓↓ − ~2

4↓↓+

~2

4↑↑)

= 2~2↑↑


Triplet state


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

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

(2)x + S

(1)y S

(2)y + S

(1)z S

(2)z )

S(1)x S

(2)x ↑↑ =

~2

(0 11 0

)(10

)~2

(0 11 0

)(10

)=

~2

4

(01

)(01

)

=~2

4↓↓

S(1)y S

(2)y ↑↑ =

~2

(0 −ii 0

)(10

)~2

(0 −ii 0

)(10

)= −~2

4

(01

)(01

)= −~2

4↓↓

S(1)z S

(2)z ↑↑ =

~2

(1 00 −1

)(10

)~2

(1 00 −1

)(10

)=

~2

4

(10

)(10

)=

~2

4↑↑

S2↑↑ = ~21

2

(3

2

)↑↑+ ~2

1

2

(3

2

)↑↑+ 2

(~2

4↓↓ − ~2

4↓↓+

~2

4↑↑)

= 2~2↑↑


Triplet state


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

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

(2)x + S

(1)y S

(2)y + S

(1)z S

(2)z )

S(1)x S

(2)x ↑↑ =

~2

(0 11 0

)(10

)~2

(0 11 0

)(10

)=

~2

4

(01

)(01

)=

~2

4↓↓

S(1)y S

(2)y ↑↑ =

~2

(0 −ii 0

)(10

)~2

(0 −ii 0

)(10

)= −~2

4

(01

)(01

)= −~2

4↓↓

S(1)z S

(2)z ↑↑ =

~2

(1 00 −1

)(10

)~2

(1 00 −1

)(10

)=

~2

4

(10

)(10

)=

~2

4↑↑

S2↑↑ = ~21

2

(3

2

)↑↑+ ~2

1

2

(3

2

)↑↑+ 2

(~2

4↓↓ − ~2

4↓↓+

~2

4↑↑)

= 2~2↑↑


Triplet state


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

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

(2)x + S

(1)y S

(2)y + S

(1)z S

(2)z )

S(1)x S

(2)x ↑↑ =

~2

(0 11 0

)(10

)~2

(0 11 0

)(10

)=

~2

4

(01

)(01

)=

~2

4↓↓

S(1)y S

(2)y ↑↑

=~2

(0 −ii 0

)(10

)~2

(0 −ii 0

)(10

)= −~2

4

(01

)(01

)= −~2

4↓↓

S(1)z S

(2)z ↑↑ =

~2

(1 00 −1

)(10

)~2

(1 00 −1

)(10

)=

~2

4

(10

)(10

)=

~2

4↑↑

S2↑↑ = ~21

2

(3

2

)↑↑+ ~2

1

2

(3

2

)↑↑+ 2

(~2

4↓↓ − ~2

4↓↓+

~2

4↑↑)

= 2~2↑↑


Triplet state


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

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

(2)x + S

(1)y S

(2)y + S

(1)z S

(2)z )

S(1)x S

(2)x ↑↑ =

~2

(0 11 0

)(10

)~2

(0 11 0

)(10

)=

~2

4

(01

)(01

)=

~2

4↓↓

S(1)y S

(2)y ↑↑ =

~2

(0 −ii 0

)(10

)~2

(0 −ii 0

)(10

)

= −~2

4

(01

)(01

)= −~2

4↓↓

S(1)z S

(2)z ↑↑ =

~2

(1 00 −1

)(10

)~2

(1 00 −1

)(10

)=

~2

4

(10

)(10

)=

~2

4↑↑

S2↑↑ = ~21

2

(3

2

)↑↑+ ~2

1

2

(3

2

)↑↑+ 2

(~2

4↓↓ − ~2

4↓↓+

~2

4↑↑)

= 2~2↑↑


Triplet state


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

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

(2)x + S

(1)y S

(2)y + S

(1)z S

(2)z )

S(1)x S

(2)x ↑↑ =

~2

(0 11 0

)(10

)~2

(0 11 0

)(10

)=

~2

4

(01

)(01

)=

~2

4↓↓

S(1)y S

(2)y ↑↑ =

~2

(0 −ii 0

)(10

)~2

(0 −ii 0

)(10

)= −~2

4

(01

)(01

)

= −~2

4↓↓

S(1)z S

(2)z ↑↑ =

~2

(1 00 −1

)(10

)~2

(1 00 −1

)(10

)=

~2

4

(10

)(10

)=

~2

4↑↑

S2↑↑ = ~21

2

(3

2

)↑↑+ ~2

1

2

(3

2

)↑↑+ 2

(~2

4↓↓ − ~2

4↓↓+

~2

4↑↑)

= 2~2↑↑


Triplet state


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

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

(2)x + S

(1)y S

(2)y + S

(1)z S

(2)z )

S(1)x S

(2)x ↑↑ =

~2

(0 11 0

)(10

)~2

(0 11 0

)(10

)=

~2

4

(01

)(01

)=

~2

4↓↓

S(1)y S

(2)y ↑↑ =

~2

(0 −ii 0

)(10

)~2

(0 −ii 0

)(10

)= −~2

4

(01

)(01

)= −~2

4↓↓

S(1)z S

(2)z ↑↑ =

~2

(1 00 −1

)(10

)~2

(1 00 −1

)(10

)=

~2

4

(10

)(10

)=

~2

4↑↑

S2↑↑ = ~21

2

(3

2

)↑↑+ ~2

1

2

(3

2

)↑↑+ 2

(~2

4↓↓ − ~2

4↓↓+

~2

4↑↑)

= 2~2↑↑


Triplet state


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

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

(2)x + S

(1)y S

(2)y + S

(1)z S

(2)z )

S(1)x S

(2)x ↑↑ =

~2

(0 11 0

)(10

)~2

(0 11 0

)(10

)=

~2

4

(01

)(01

)=

~2

4↓↓

S(1)y S

(2)y ↑↑ =

~2

(0 −ii 0

)(10

)~2

(0 −ii 0

)(10

)= −~2

4

(01

)(01

)= −~2

4↓↓

S(1)z S

(2)z ↑↑

=~2

(1 00 −1

)(10

)~2

(1 00 −1

)(10

)=

~2

4

(10

)(10

)=

~2

4↑↑

S2↑↑ = ~21

2

(3

2

)↑↑+ ~2

1

2

(3

2

)↑↑+ 2

(~2

4↓↓ − ~2

4↓↓+

~2

4↑↑)

= 2~2↑↑


Triplet state


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

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

(2)x + S

(1)y S

(2)y + S

(1)z S

(2)z )

S(1)x S

(2)x ↑↑ =

~2

(0 11 0

)(10

)~2

(0 11 0

)(10

)=

~2

4

(01

)(01

)=

~2

4↓↓

S(1)y S

(2)y ↑↑ =

~2

(0 −ii 0

)(10

)~2

(0 −ii 0

)(10

)= −~2

4

(01

)(01

)= −~2

4↓↓

S(1)z S

(2)z ↑↑ =

~2

(1 00 −1

)(10

)~2

(1 00 −1

)(10

)

=~2

4

(10

)(10

)=

~2

4↑↑

S2↑↑ = ~21

2

(3

2

)↑↑+ ~2

1

2

(3

2

)↑↑+ 2

(~2

4↓↓ − ~2

4↓↓+

~2

4↑↑)

= 2~2↑↑


Triplet state


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

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

(2)x + S

(1)y S

(2)y + S

(1)z S

(2)z )

S(1)x S

(2)x ↑↑ =

~2

(0 11 0

)(10

)~2

(0 11 0

)(10

)=

~2

4

(01

)(01

)=

~2

4↓↓

S(1)y S

(2)y ↑↑ =

~2

(0 −ii 0

)(10

)~2

(0 −ii 0

)(10

)= −~2

4

(01

)(01

)= −~2

4↓↓

S(1)z S

(2)z ↑↑ =

~2

(1 00 −1

)(10

)~2

(1 00 −1

)(10

)=

~2

4

(10

)(10

)

=~2

4↑↑

S2↑↑ = ~21

2

(3

2

)↑↑+ ~2

1

2

(3

2

)↑↑+ 2

(~2

4↓↓ − ~2

4↓↓+

~2

4↑↑)

= 2~2↑↑


Triplet state


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

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

(2)x + S

(1)y S

(2)y + S

(1)z S

(2)z )

S(1)x S

(2)x ↑↑ =

~2

(0 11 0

)(10

)~2

(0 11 0

)(10

)=

~2

4

(01

)(01

)=

~2

4↓↓

S(1)y S

(2)y ↑↑ =

~2

(0 −ii 0

)(10

)~2

(0 −ii 0

)(10

)= −~2

4

(01

)(01

)= −~2

4↓↓

S(1)z S

(2)z ↑↑ =

~2

(1 00 −1

)(10

)~2

(1 00 −1

)(10

)=

~2

4

(10

)(10

)=

~2

4↑↑

S2↑↑ = ~21

2

(3

2

)↑↑+ ~2

1

2

(3

2

)↑↑+ 2

(~2

4↓↓ − ~2

4↓↓+

~2

4↑↑)

= 2~2↑↑


Triplet state


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

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

(2)x + S

(1)y S

(2)y + S

(1)z S

(2)z )

S(1)x S

(2)x ↑↑ =

~2

(0 11 0

)(10

)~2

(0 11 0

)(10

)=

~2

4

(01

)(01

)=

~2

4↓↓

S(1)y S

(2)y ↑↑ =

~2

(0 −ii 0

)(10

)~2

(0 −ii 0

)(10

)= −~2

4

(01

)(01

)= −~2

4↓↓

S(1)z S

(2)z ↑↑ =

~2

(1 00 −1

)(10

)~2

(1 00 −1

)(10

)=

~2

4

(10

)(10

)=

~2

4↑↑

S2↑↑ =

~21

2

(3

2

)↑↑+ ~2

1

2

(3

2

)↑↑+ 2

(~2

4↓↓ − ~2

4↓↓+

~2

4↑↑)

= 2~2↑↑


Triplet state


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

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

(2)x + S

(1)y S

(2)y + S

(1)z S

(2)z )

S(1)x S

(2)x ↑↑ =

~2

(0 11 0

)(10

)~2

(0 11 0

)(10

)=

~2

4

(01

)(01

)=

~2

4↓↓

S(1)y S

(2)y ↑↑ =

~2

(0 −ii 0

)(10

)~2

(0 −ii 0

)(10

)= −~2

4

(01

)(01

)= −~2

4↓↓

S(1)z S

(2)z ↑↑ =

~2

(1 00 −1

)(10

)~2

(1 00 −1

)(10

)=

~2

4

(10

)(10

)=

~2

4↑↑

S2↑↑ = ~21

2

(3

2

)↑↑

+ ~21

2

(3

2

)↑↑+ 2

(~2

4↓↓ − ~2

4↓↓+

~2

4↑↑)

= 2~2↑↑


Triplet state


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

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

(2)x + S

(1)y S

(2)y + S

(1)z S

(2)z )

S(1)x S

(2)x ↑↑ =

~2

(0 11 0

)(10

)~2

(0 11 0

)(10

)=

~2

4

(01

)(01

)=

~2

4↓↓

S(1)y S

(2)y ↑↑ =

~2

(0 −ii 0

)(10

)~2

(0 −ii 0

)(10

)= −~2

4

(01

)(01

)= −~2

4↓↓

S(1)z S

(2)z ↑↑ =

~2

(1 00 −1

)(10

)~2

(1 00 −1

)(10

)=

~2

4

(10

)(10

)=

~2

4↑↑

S2↑↑ = ~21

2

(3

2

)↑↑+ ~2

1

2

(3

2

)↑↑

+ 2

(~2

4↓↓ − ~2

4↓↓+

~2

4↑↑)

= 2~2↑↑


Triplet state


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

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

(2)x + S

(1)y S

(2)y + S

(1)z S

(2)z )

S(1)x S

(2)x ↑↑ =

~2

(0 11 0

)(10

)~2

(0 11 0

)(10

)=

~2

4

(01

)(01

)=

~2

4↓↓

S(1)y S

(2)y ↑↑ =

~2

(0 −ii 0

)(10

)~2

(0 −ii 0

)(10

)= −~2

4

(01

)(01

)= −~2

4↓↓

S(1)z S

(2)z ↑↑ =

~2

(1 00 −1

)(10

)~2

(1 00 −1

)(10

)=

~2

4

(10

)(10

)=

~2

4↑↑

S2↑↑ = ~21

2

(3

2

)↑↑+ ~2

1

2

(3

2

)↑↑+ 2

(~2

4↓↓ − ~2

4↓↓+

~2

4↑↑)

= 2~2↑↑


Triplet state (cont.)

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

S−↑↑ = (S(1)− + S

(2)− )↑↑ = (S

(1)− ↑)↑+ ↑(S (2)

− ↑)

= ~

√1

2

(3

2

)− 1

2

(−1

2

)(↓↑+ ↑↓) = ~(↓↑+ ↑↓) =

√2~

1√2

(↓↑+ ↑↓)

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

S−1√2

(↓↑+ ↑↓) = (S(1)− + S

(2)− )

1√2

(↓↑+ ↑↓)

=1√2

[S(1)− (↓↑+ ↑↓) + S

(2)− (↓↑+ ↑↓)

]=

1√2

[~↓↓+ ~↓↓]

=√

2~↓↓

The triplet state is thus

|1 1〉 = ↑↑, |1 0〉 =1√2

(↓↑+ ↑↓), |1−1〉 = ↓↓




S−↑↑ = (S(1)− + S

(2)− )↑↑

= (S(1)− ↑)↑+ ↑(S (2)

− ↑)

= ~

√1

2

(3

2

)− 1

2

(−1

2

)(↓↑+ ↑↓) = ~(↓↑+ ↑↓) =

√2~

1√2

(↓↑+ ↑↓)


S−1√2

(↓↑+ ↑↓) = (S(1)− + S

(2)− )

1√2

(↓↑+ ↑↓)

=1√2

[S(1)− (↓↑+ ↑↓) + S

(2)− (↓↑+ ↑↓)

]=

1√2

[~↓↓+ ~↓↓]

=√

2~↓↓


|1 1〉 = ↑↑, |1 0〉 =1√2

(↓↑+ ↑↓), |1−1〉 = ↓↓




S−↑↑ = (S(1)− + S

(2)− )↑↑ = (S

(1)− ↑)↑+ ↑(S (2)

− ↑)

= ~

√1

2

(3

2

)− 1

2

(−1

2

)(↓↑+ ↑↓) = ~(↓↑+ ↑↓) =

√2~

1√2

(↓↑+ ↑↓)


S−1√2

(↓↑+ ↑↓) = (S(1)− + S

(2)− )

1√2

(↓↑+ ↑↓)

=1√2

[S(1)− (↓↑+ ↑↓) + S

(2)− (↓↑+ ↑↓)

]=

1√2

[~↓↓+ ~↓↓]

=√

2~↓↓


|1 1〉 = ↑↑, |1 0〉 =1√2

(↓↑+ ↑↓), |1−1〉 = ↓↓




S−↑↑ = (S(1)− + S

(2)− )↑↑ = (S

(1)− ↑)↑+ ↑(S (2)

− ↑)

= ~

√1

2

(3

2

)− 1

2

(−1

2

)(↓↑+ ↑↓)

= ~(↓↑+ ↑↓) =√

2~1√2

(↓↑+ ↑↓)


S−1√2

(↓↑+ ↑↓) = (S(1)− + S

(2)− )

1√2

(↓↑+ ↑↓)

=1√2

[S(1)− (↓↑+ ↑↓) + S

(2)− (↓↑+ ↑↓)

]=

1√2

[~↓↓+ ~↓↓]

=√

2~↓↓


|1 1〉 = ↑↑, |1 0〉 =1√2

(↓↑+ ↑↓), |1−1〉 = ↓↓




S−↑↑ = (S(1)− + S

(2)− )↑↑ = (S

(1)− ↑)↑+ ↑(S (2)

− ↑)

= ~

√1

2

(3

2

)− 1

2

(−1

2

)(↓↑+ ↑↓) = ~(↓↑+ ↑↓)

=√

2~1√2

(↓↑+ ↑↓)


S−1√2

(↓↑+ ↑↓) = (S(1)− + S

(2)− )

1√2

(↓↑+ ↑↓)

=1√2

[S(1)− (↓↑+ ↑↓) + S

(2)− (↓↑+ ↑↓)

]=

1√2

[~↓↓+ ~↓↓]

=√

2~↓↓


|1 1〉 = ↑↑, |1 0〉 =1√2

(↓↑+ ↑↓), |1−1〉 = ↓↓




S−↑↑ = (S(1)− + S

(2)− )↑↑ = (S

(1)− ↑)↑+ ↑(S (2)

− ↑)

= ~

√1

2

(3

2

)− 1

2

(−1

2

)(↓↑+ ↑↓) = ~(↓↑+ ↑↓) =

√2~

1√2

(↓↑+ ↑↓)


S−1√2

(↓↑+ ↑↓) = (S(1)− + S

(2)− )

1√2

(↓↑+ ↑↓)

=1√2

[S(1)− (↓↑+ ↑↓) + S

(2)− (↓↑+ ↑↓)

]=

1√2

[~↓↓+ ~↓↓]

=√

2~↓↓


|1 1〉 = ↑↑, |1 0〉 =1√2

(↓↑+ ↑↓), |1−1〉 = ↓↓




S−↑↑ = (S(1)− + S

(2)− )↑↑ = (S

(1)− ↑)↑+ ↑(S (2)

− ↑)

= ~

√1

2

(3

2

)− 1

2

(−1

2

)(↓↑+ ↑↓) = ~(↓↑+ ↑↓) =

√2~

1√2

(↓↑+ ↑↓)


S−1√2

(↓↑+ ↑↓) = (S(1)− + S

(2)− )

1√2

(↓↑+ ↑↓)

=1√2

[S(1)− (↓↑+ ↑↓) + S

(2)− (↓↑+ ↑↓)

]

=1√2

[~↓↓+ ~↓↓]

=√

2~↓↓


|1 1〉 = ↑↑, |1 0〉 =1√2

(↓↑+ ↑↓), |1−1〉 = ↓↓




S−↑↑ = (S(1)− + S

(2)− )↑↑ = (S

(1)− ↑)↑+ ↑(S (2)

− ↑)

= ~

√1

2

(3

2

)− 1

2

(−1

2

)(↓↑+ ↑↓) = ~(↓↑+ ↑↓) =

√2~

1√2

(↓↑+ ↑↓)


S−1√2

(↓↑+ ↑↓) = (S(1)− + S

(2)− )

1√2

(↓↑+ ↑↓)

=1√2

[S(1)− (↓↑+ ↑↓) + S

(2)− (↓↑+ ↑↓)

]=

1√2

[~↓↓+ ~↓↓]

=√

2~↓↓


|1 1〉 = ↑↑, |1 0〉 =1√2

(↓↑+ ↑↓), |1−1〉 = ↓↓


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


Singlet state


|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


Singlet state


|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


Singlet state


|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


Singlet state


|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


Singlet state


|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


Singlet state


|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


Singlet state


|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


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~↓↓


Eigenvalues of Sz


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~↓↓


Eigenvalues of Sz


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~↓↓


Eigenvalues of Sz


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~↓↓


Eigenvalues of Sz


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~↓↓


Eigenvalues of Sz


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~↓↓


Eigenvalues of Sz


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~↓↓


“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



|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




|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



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


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 >




|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 >




|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 >




|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 >




|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 >




|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 >




|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 >




|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 >


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?


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



R21

(√1/3Y 0

1 χ+ +√

2/3Y 11 χ−

)L2


〈L2〉 = ~2l(l + 1)

= ~2[1(2)] = 2~2

Lz


〈Lz〉 = 0, P = 13

〈Lz〉 = ~, P = 23

S2



〈S2〉 = ~2s(s + 1) = ~2[ 12( 32)] = 3

4~2

Sz

One term has m = 12


2so we will have

〈Sz〉 = 12~, P = 1

3

〈Sz〉 = − 12~, P = 2

3



R21

(√1/3Y 0

1 χ+ +√

2/3Y 11 χ−

)L2


〈L2〉 = ~2l(l + 1) = ~2[1(2)] = 2~2

Lz


〈Lz〉 = 0, P = 13

〈Lz〉 = ~, P = 23

S2



〈S2〉 = ~2s(s + 1) = ~2[ 12( 32)] = 3

4~2

Sz

One term has m = 12


2so we will have

〈Sz〉 = 12~, P = 1

3

〈Sz〉 = − 12~, P = 2

3



R21

(√1/3Y 0

1 χ+ +√

2/3Y 11 χ−

)L2


〈L2〉 = ~2l(l + 1) = ~2[1(2)] = 2~2

Lz


〈Lz〉 = 0, P = 13

〈Lz〉 = ~, P = 23

S2



〈S2〉 = ~2s(s + 1)

= ~2[ 12( 32)] = 3

4~2

Sz

One term has m = 12


2so we will have

〈Sz〉 = 12~, P = 1

3

〈Sz〉 = − 12~, P = 2

3



R21

(√1/3Y 0

1 χ+ +√

2/3Y 11 χ−

)L2


〈L2〉 = ~2l(l + 1) = ~2[1(2)] = 2~2

Lz


〈Lz〉 = 0, P = 13

〈Lz〉 = ~, P = 23

S2



〈S2〉 = ~2s(s + 1) = ~2[ 12( 32)] = 3

4~2

Sz

One term has m = 12


2so we will have

〈Sz〉 = 12~, P = 1

3

〈Sz〉 = − 12~, P = 2

3



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~



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〉


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~




1/3Y 01 χ+ +

√2/3Y 1

1 χ−

)√

13|1 0〉 | 1

212〉+

√23|1 1〉 | 1

2−12〉


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~




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~




1/3Y 01 χ+ +

√2/3Y 1

1 χ−

)√

13|1 0〉 | 1

212〉+

√23|1 1〉 | 1

2−12〉


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~




1/3Y 01 χ+ +

√2/3Y 1

1 χ−

)√

13|1 0〉 | 1

212〉+

√23|1 1〉 | 1

2−12〉


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~




1/3Y 01 χ+ +

√2/3Y 1

1 χ−

)√

13|1 0〉 | 1

212〉+

√23|1 1〉 | 1

2−12〉


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~


today’s outline - november 07, 2012 - iitphys.iit.edu/~segre/phys405/12f/lecture_21.pdf ·...

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