csrri.iit.educsrri.iit.edu/~segre/phys405/16F/lecture_18.pdf · Computing the Clebsch-Gordan coe...

Today’s Outline - October 31, 2016

• Generating the Clebsch-Gordan coefficients

• Chapter 4 problems

Midterm Exam 2: Wednesday, November 9, 2016covers through Chapter 4.4.2 (at least!)

Homework Assignment #08:Chapter 4: 10,13,14,15,16,38due Wednesday, November 02, 2016

C. Segre (IIT) PHYS 405 - Fall 2016 October 31, 2016 1 / 18

Computing the Clebsch-Gordan coefficients

Deriving the Clebsch-Gordan coefficients is non-trivial and there are anumber of ways to do it in the literature which give a general formula forthe coefficients.

We will focus on generating the coefficients, starting witha general procedure, then applying it to a specific case s1 = 1, s2 = 1

The goal is to find an expression for each of the total spin states |s m〉 interms of the states of the individual particle spins.

starting with the initial basis states

each of these are eigenstates of thetotal spin in the z direction,Sz = S1z + S2z

the state with highest m = s1 + s2must be a singlet and haves = s1 + s2 so we can extract thefirst Clebsch-Gordan coefficient

|s1m1 s2m2〉 ≡ |m1 m2〉

Sz |m1 m2〉 = ~(m1 + m2)|m1 m2〉

|(s1+s2) (s1+s2)〉 = |s1 s2〉

〈(s1+s2) (s1+s2)|s1 s2〉 = 1

Deriving the Clebsch-Gordan coefficients is non-trivial and there are anumber of ways to do it in the literature which give a general formula forthe coefficients. We will focus on generating the coefficients, starting witha general procedure, then applying it to a specific case s1 = 1, s2 = 1

|s1m1 s2m2〉 ≡ |m1 m2〉

Sz |m1 m2〉 = ~(m1 + m2)|m1 m2〉

|(s1+s2) (s1+s2)〉 = |s1 s2〉

〈(s1+s2) (s1+s2)|s1 s2〉 = 1

|s1m1 s2m2〉 ≡ |m1 m2〉

Sz |m1 m2〉 = ~(m1 + m2)|m1 m2〉

|(s1+s2) (s1+s2)〉 = |s1 s2〉

〈(s1+s2) (s1+s2)|s1 s2〉 = 1

|s1m1 s2m2〉 ≡ |m1 m2〉

Sz |m1 m2〉 = ~(m1 + m2)|m1 m2〉

|(s1+s2) (s1+s2)〉 = |s1 s2〉

〈(s1+s2) (s1+s2)|s1 s2〉 = 1

|s1m1 s2m2〉

≡ |m1 m2〉

Sz |m1 m2〉 = ~(m1 + m2)|m1 m2〉

|(s1+s2) (s1+s2)〉 = |s1 s2〉

〈(s1+s2) (s1+s2)|s1 s2〉 = 1

|s1m1 s2m2〉 ≡ |m1 m2〉

Sz |m1 m2〉 = ~(m1 + m2)|m1 m2〉

|(s1+s2) (s1+s2)〉 = |s1 s2〉

〈(s1+s2) (s1+s2)|s1 s2〉 = 1

|s1m1 s2m2〉 ≡ |m1 m2〉

Sz |m1 m2〉 = ~(m1 + m2)|m1 m2〉

|(s1+s2) (s1+s2)〉 = |s1 s2〉

〈(s1+s2) (s1+s2)|s1 s2〉 = 1

|s1m1 s2m2〉 ≡ |m1 m2〉

Sz |m1 m2〉 = ~(m1 + m2)|m1 m2〉

|(s1+s2) (s1+s2)〉 = |s1 s2〉

〈(s1+s2) (s1+s2)|s1 s2〉 = 1

the state with highest m = s1 + s2must be a singlet and haves = s1 + s2

so we can extract thefirst Clebsch-Gordan coefficient

|s1m1 s2m2〉 ≡ |m1 m2〉

Sz |m1 m2〉 = ~(m1 + m2)|m1 m2〉

|(s1+s2) (s1+s2)〉 = |s1 s2〉

〈(s1+s2) (s1+s2)|s1 s2〉 = 1

the state with highest m = s1 + s2must be a singlet and haves = s1 + s2

so we can extract thefirst Clebsch-Gordan coefficient

|s1m1 s2m2〉 ≡ |m1 m2〉

Sz |m1 m2〉 = ~(m1 + m2)|m1 m2〉

|(s1+s2) (s1+s2)〉 = |s1 s2〉

〈(s1+s2) (s1+s2)|s1 s2〉 = 1

|s1m1 s2m2〉 ≡ |m1 m2〉

Sz |m1 m2〉 = ~(m1 + m2)|m1 m2〉

|(s1+s2) (s1+s2)〉 = |s1 s2〉

〈(s1+s2) (s1+s2)|s1 s2〉 = 1

|s1m1 s2m2〉 ≡ |m1 m2〉

Sz |m1 m2〉 = ~(m1 + m2)|m1 m2〉

|(s1+s2) (s1+s2)〉 = |s1 s2〉

〈(s1+s2) (s1+s2)|s1 s2〉

|s1m1 s2m2〉 ≡ |m1 m2〉

Sz |m1 m2〉 = ~(m1 + m2)|m1 m2〉

|(s1+s2) (s1+s2)〉 = |s1 s2〉

〈(s1+s2) (s1+s2)|s1 s2〉 = 1

Using the lowering operator

The next states will by necessityhave a value of m lower than themaximum value by 1

there are two possible initial stateswhich satisfy this condition

m = s1+s2 − 1

|(s1−1) s2〉, |s1 (s2−1)〉

One combination must be the state which is obtained by applying thelowering operator S− to the first state.

S−|(s1+s2) (s1+s2)〉

= (S1− + S2−)|s1 s2〉

this is repeated to generate all the states with the same value of s anddifferent values of m

states with the same value of m but different values of s are generatedusing the unused combinations of the original basis functions, then lowered

this is best understood with an example

m = s1+s2 − 1

|(s1−1) s2〉, |s1 (s2−1)〉

S−|(s1+s2) (s1+s2)〉

= (S1− + S2−)|s1 s2〉

m = s1+s2 − 1

|(s1−1) s2〉, |s1 (s2−1)〉

S−|(s1+s2) (s1+s2)〉

= (S1− + S2−)|s1 s2〉

m = s1+s2 − 1

|(s1−1) s2〉

, |s1 (s2−1)〉

S−|(s1+s2) (s1+s2)〉

= (S1− + S2−)|s1 s2〉

m = s1+s2 − 1

|(s1−1) s2〉, |s1 (s2−1)〉

S−|(s1+s2) (s1+s2)〉

= (S1− + S2−)|s1 s2〉

m = s1+s2 − 1

|(s1−1) s2〉, |s1 (s2−1)〉

S−|(s1+s2) (s1+s2)〉

= (S1− + S2−)|s1 s2〉

m = s1+s2 − 1

|(s1−1) s2〉, |s1 (s2−1)〉

S−|(s1+s2) (s1+s2)〉

= (S1− + S2−)|s1 s2〉

m = s1+s2 − 1

|(s1−1) s2〉, |s1 (s2−1)〉

S−|(s1+s2) (s1+s2)〉 = (S1− + S2−)|s1 s2〉

m = s1+s2 − 1

|(s1−1) s2〉, |s1 (s2−1)〉

S−|(s1+s2) (s1+s2)〉 = (S1− + S2−)|s1 s2〉

m = s1+s2 − 1

|(s1−1) s2〉, |s1 (s2−1)〉

S−|(s1+s2) (s1+s2)〉 = (S1− + S2−)|s1 s2〉

m = s1+s2 − 1

|(s1−1) s2〉, |s1 (s2−1)〉

S−|(s1+s2) (s1+s2)〉 = (S1− + S2−)|s1 s2〉

Example Clebsch-Gordan computation

Suppose s1 = s2 = 1, the “top”state thus has s = 2, m = 2 and isexpressed as

the next lower state m = 1 isobtained (dropping the ~ on bothsides)

|2 2〉 = |1 1〉

S−|2 2〉 = (S1− + S2−)|1 1〉

√(2)(2 + 1)− (2)(2− 1)|2 1〉

(1)(1 + 1)− (1)(1− 1)|0 1〉

(1)(1 + 1)− (1)(1− 1)|1 0〉

Solving for the |2 1〉 eigenfunction

The Clebsch-Gordan coefficients

are thus both√

√4|2 1〉

2|0 1〉+√

2|1 0〉

|2 1〉 =√

12 |0 1〉+

√12 |1 0〉

|2 2〉 = |1 1〉

S−|2 2〉 = (S1− + S2−)|1 1〉

√(2)(2 + 1)− (2)(2− 1)|2 1〉

(1)(1 + 1)− (1)(1− 1)|0 1〉

(1)(1 + 1)− (1)(1− 1)|1 0〉

are thus both√

√4|2 1〉

2|0 1〉+√

2|1 0〉

|2 1〉 =√

12 |0 1〉+

√12 |1 0〉

|2 2〉 = |1 1〉

S−|2 2〉 = (S1− + S2−)|1 1〉

√(2)(2 + 1)− (2)(2− 1)|2 1〉

(1)(1 + 1)− (1)(1− 1)|0 1〉

(1)(1 + 1)− (1)(1− 1)|1 0〉

are thus both√

√4|2 1〉

2|0 1〉+√

2|1 0〉

|2 1〉 =√

12 |0 1〉+

√12 |1 0〉

|2 2〉 = |1 1〉

S−|2 2〉 = (S1− + S2−)|1 1〉

√(2)(2 + 1)− (2)(2− 1)|2 1〉

(1)(1 + 1)− (1)(1− 1)|0 1〉

(1)(1 + 1)− (1)(1− 1)|1 0〉

are thus both√

√4|2 1〉

2|0 1〉+√

2|1 0〉

|2 1〉 =√

12 |0 1〉+

√12 |1 0〉

|2 2〉 = |1 1〉

S−|2 2〉 = (S1− + S2−)|1 1〉

√(2)(2 + 1)− (2)(2− 1)|2 1〉

(1)(1 + 1)− (1)(1− 1)|0 1〉

(1)(1 + 1)− (1)(1− 1)|1 0〉

are thus both√

√4|2 1〉

2|0 1〉+√

2|1 0〉

|2 1〉 =√

12 |0 1〉+

√12 |1 0〉

|2 2〉 = |1 1〉

S−|2 2〉 = (S1− + S2−)|1 1〉

√(2)(2 + 1)− (2)(2− 1)|2 1〉 =

√(1)(1 + 1)− (1)(1− 1)|0 1〉

(1)(1 + 1)− (1)(1− 1)|1 0〉

are thus both√

√4|2 1〉

2|0 1〉+√

2|1 0〉

|2 1〉 =√

12 |0 1〉+

√12 |1 0〉

|2 2〉 = |1 1〉

S−|2 2〉 = (S1− + S2−)|1 1〉

√(2)(2 + 1)− (2)(2− 1)|2 1〉 =

√(1)(1 + 1)− (1)(1− 1)|0 1〉

(1)(1 + 1)− (1)(1− 1)|1 0〉

are thus both√

√4|2 1〉

2|0 1〉+√

2|1 0〉

|2 1〉 =√

12 |0 1〉+

√12 |1 0〉

|2 2〉 = |1 1〉

S−|2 2〉 = (S1− + S2−)|1 1〉

√(2)(2 + 1)− (2)(2− 1)|2 1〉 =

√(1)(1 + 1)− (1)(1− 1)|0 1〉

(1)(1 + 1)− (1)(1− 1)|1 0〉

are thus both√

√4|2 1〉

2|0 1〉+√

2|1 0〉

|2 1〉 =√

12 |0 1〉+

√12 |1 0〉

|2 2〉 = |1 1〉

S−|2 2〉 = (S1− + S2−)|1 1〉

√(2)(2 + 1)− (2)(2− 1)|2 1〉 =

√(1)(1 + 1)− (1)(1− 1)|0 1〉

(1)(1 + 1)− (1)(1− 1)|1 0〉

are thus both√

√4|2 1〉

2|0 1〉+√

2|1 0〉

|2 1〉 =√

12 |0 1〉+

√12 |1 0〉

|2 2〉 = |1 1〉

S−|2 2〉 = (S1− + S2−)|1 1〉

√(2)(2 + 1)− (2)(2− 1)|2 1〉 =

√(1)(1 + 1)− (1)(1− 1)|0 1〉

(1)(1 + 1)− (1)(1− 1)|1 0〉

are thus both√

√4|2 1〉 =

√2|0 1〉

2|1 0〉

|2 1〉 =√

12 |0 1〉+

√12 |1 0〉

|2 2〉 = |1 1〉

S−|2 2〉 = (S1− + S2−)|1 1〉

√(2)(2 + 1)− (2)(2− 1)|2 1〉 =

√(1)(1 + 1)− (1)(1− 1)|0 1〉

(1)(1 + 1)− (1)(1− 1)|1 0〉

are thus both√

√4|2 1〉 =

√2|0 1〉+

√2|1 0〉

|2 1〉 =√

12 |0 1〉+

√12 |1 0〉

|2 2〉 = |1 1〉

S−|2 2〉 = (S1− + S2−)|1 1〉

√(2)(2 + 1)− (2)(2− 1)|2 1〉 =

√(1)(1 + 1)− (1)(1− 1)|0 1〉

(1)(1 + 1)− (1)(1− 1)|1 0〉

are thus both√

√4|2 1〉 =

√2|0 1〉+

√2|1 0〉

|2 1〉 =√

12 |0 1〉+

√12 |1 0〉

|2 2〉 = |1 1〉

S−|2 2〉 = (S1− + S2−)|1 1〉

√(2)(2 + 1)− (2)(2− 1)|2 1〉 =

√(1)(1 + 1)− (1)(1− 1)|0 1〉

(1)(1 + 1)− (1)(1− 1)|1 0〉

are thus both√

√4|2 1〉 =

√2|0 1〉+

√2|1 0〉

|2 1〉 =√

12 |0 1〉+

√12 |1 0〉

Keep lowering. . .

The |2 0〉 state is similarly obtained

S−|2 1〉 = (S1− + S2−)√

[|0 1〉+ |1 0〉

(2)(2 + 1)− (1)(1− 1)|2 0〉

[√(1)(1 + 1)− (0)(0− 1)|−1 1〉

(1)(1 + 1)− (1)(1− 1)|0 0〉

(1)(1 + 1)− (0)(0− 1)|1 −1〉]

√6|2 0〉 =

√2|−1 1〉+ 2

√2|0 0〉+

√2|1 −1〉

]|2 0〉 =

√16 |−1 1〉+

√23 |0 0〉+

√16 |1 −1〉

the other two eigenvectors with s = 2 are generated in the same way

|2 −1〉 =√

12 |0 −1〉+

√12 |−1 0〉,

|2 −2〉 = |−1 −1〉

Keep lowering. . .

S−|2 1〉 = (S1− + S2−)√

[|0 1〉+ |1 0〉

√(2)(2 + 1)− (1)(1− 1)|2 0〉

[√(1)(1 + 1)− (0)(0− 1)|−1 1〉

(1)(1 + 1)− (1)(1− 1)|0 0〉

(1)(1 + 1)− (0)(0− 1)|1 −1〉]

√6|2 0〉 =

√2|−1 1〉+ 2

√2|0 0〉+

√2|1 −1〉

]|2 0〉 =

√16 |−1 1〉+

√23 |0 0〉+

√16 |1 −1〉

|2 −1〉 =√

12 |0 −1〉+

√12 |−1 0〉,

|2 −2〉 = |−1 −1〉

Keep lowering. . .

S−|2 1〉 = (S1− + S2−)√

[|0 1〉+ |1 0〉

(2)(2 + 1)− (1)(1− 1)|2 0〉

[√(1)(1 + 1)− (0)(0− 1)|−1 1〉

(1)(1 + 1)− (1)(1− 1)|0 0〉

(1)(1 + 1)− (0)(0− 1)|1 −1〉]

√6|2 0〉 =

√2|−1 1〉+ 2

√2|0 0〉+

√2|1 −1〉

]|2 0〉 =

√16 |−1 1〉+

√23 |0 0〉+

√16 |1 −1〉

|2 −1〉 =√

12 |0 −1〉+

√12 |−1 0〉,

|2 −2〉 = |−1 −1〉

Keep lowering. . .

S−|2 1〉 = (S1− + S2−)√

[|0 1〉+ |1 0〉

(2)(2 + 1)− (1)(1− 1)|2 0〉 =√

[√(1)(1 + 1)− (0)(0− 1)|−1 1〉

(1)(1 + 1)− (1)(1− 1)|0 0〉

(1)(1 + 1)− (0)(0− 1)|1 −1〉

√6|2 0〉 =

√2|−1 1〉+ 2

√2|0 0〉+

√2|1 −1〉

]|2 0〉 =

√16 |−1 1〉+

√23 |0 0〉+

√16 |1 −1〉

|2 −1〉 =√

12 |0 −1〉+

√12 |−1 0〉,

|2 −2〉 = |−1 −1〉

Keep lowering. . .

S−|2 1〉 = (S1− + S2−)√

[|0 1〉+ |1 0〉

(2)(2 + 1)− (1)(1− 1)|2 0〉 =√

[√(1)(1 + 1)− (0)(0− 1)|−1 1〉

(1)(1 + 1)− (1)(1− 1)|0 0〉

(1)(1 + 1)− (0)(0− 1)|1 −1〉

√6|2 0〉 =

√2|−1 1〉+ 2

√2|0 0〉+

√2|1 −1〉

]|2 0〉 =

√16 |−1 1〉+

√23 |0 0〉+

√16 |1 −1〉

|2 −1〉 =√

12 |0 −1〉+

√12 |−1 0〉,

|2 −2〉 = |−1 −1〉

Keep lowering. . .

S−|2 1〉 = (S1− + S2−)√

[|0 1〉+ |1 0〉

(2)(2 + 1)− (1)(1− 1)|2 0〉 =√

[√(1)(1 + 1)− (0)(0− 1)|−1 1〉

(1)(1 + 1)− (1)(1− 1)|0 0〉

(1)(1 + 1)− (0)(0− 1)|1 −1〉

√6|2 0〉 =

√2|−1 1〉+ 2

√2|0 0〉+

√2|1 −1〉

]|2 0〉 =

√16 |−1 1〉+

√23 |0 0〉+

√16 |1 −1〉

|2 −1〉 =√

12 |0 −1〉+

√12 |−1 0〉,

|2 −2〉 = |−1 −1〉

Keep lowering. . .

S−|2 1〉 = (S1− + S2−)√

[|0 1〉+ |1 0〉

(2)(2 + 1)− (1)(1− 1)|2 0〉 =√

[√(1)(1 + 1)− (0)(0− 1)|−1 1〉

(1)(1 + 1)− (1)(1− 1)|0 0〉

(1)(1 + 1)− (0)(0− 1)|1 −1〉]

√6|2 0〉 =

√2|−1 1〉+ 2

√2|0 0〉+

√2|1 −1〉

]|2 0〉 =

√16 |−1 1〉+

√23 |0 0〉+

√16 |1 −1〉

|2 −1〉 =√

12 |0 −1〉+

√12 |−1 0〉,

|2 −2〉 = |−1 −1〉

Keep lowering. . .

S−|2 1〉 = (S1− + S2−)√

[|0 1〉+ |1 0〉

(2)(2 + 1)− (1)(1− 1)|2 0〉 =√

[√(1)(1 + 1)− (0)(0− 1)|−1 1〉

(1)(1 + 1)− (1)(1− 1)|0 0〉

(1)(1 + 1)− (0)(0− 1)|1 −1〉]

√6|2 0〉 =

√2|−1 1〉+ 2

√2|0 0〉+

√2|1 −1〉

|2 0〉 =√

16 |−1 1〉+

√23 |0 0〉+

√16 |1 −1〉

|2 −1〉 =√

12 |0 −1〉+

√12 |−1 0〉,

|2 −2〉 = |−1 −1〉

Keep lowering. . .

S−|2 1〉 = (S1− + S2−)√

[|0 1〉+ |1 0〉

(2)(2 + 1)− (1)(1− 1)|2 0〉 =√

[√(1)(1 + 1)− (0)(0− 1)|−1 1〉

(1)(1 + 1)− (1)(1− 1)|0 0〉

(1)(1 + 1)− (0)(0− 1)|1 −1〉]

√6|2 0〉 =

[√2|−1 1〉

+ 2√

2|0 0〉+√

2|1 −1〉

|2 0〉 =√

16 |−1 1〉+

√23 |0 0〉+

√16 |1 −1〉

|2 −1〉 =√

12 |0 −1〉+

√12 |−1 0〉,

|2 −2〉 = |−1 −1〉

Keep lowering. . .

S−|2 1〉 = (S1− + S2−)√

[|0 1〉+ |1 0〉

(2)(2 + 1)− (1)(1− 1)|2 0〉 =√

[√(1)(1 + 1)− (0)(0− 1)|−1 1〉

(1)(1 + 1)− (1)(1− 1)|0 0〉

(1)(1 + 1)− (0)(0− 1)|1 −1〉]

√6|2 0〉 =

[√2|−1 1〉+ 2

√2|0 0〉

2|1 −1〉

|2 0〉 =√

16 |−1 1〉+

√23 |0 0〉+

√16 |1 −1〉

|2 −1〉 =√

12 |0 −1〉+

√12 |−1 0〉,

|2 −2〉 = |−1 −1〉

Keep lowering. . .

S−|2 1〉 = (S1− + S2−)√

[|0 1〉+ |1 0〉

(2)(2 + 1)− (1)(1− 1)|2 0〉 =√

[√(1)(1 + 1)− (0)(0− 1)|−1 1〉

(1)(1 + 1)− (1)(1− 1)|0 0〉

(1)(1 + 1)− (0)(0− 1)|1 −1〉]

√6|2 0〉 =

[√2|−1 1〉+ 2

√2|0 0〉+

√2|1 −1〉

|2 0〉 =√

16 |−1 1〉+

√23 |0 0〉+

√16 |1 −1〉

|2 −1〉 =√

12 |0 −1〉+

√12 |−1 0〉,

|2 −2〉 = |−1 −1〉

Keep lowering. . .

S−|2 1〉 = (S1− + S2−)√

[|0 1〉+ |1 0〉

(2)(2 + 1)− (1)(1− 1)|2 0〉 =√

[√(1)(1 + 1)− (0)(0− 1)|−1 1〉

(1)(1 + 1)− (1)(1− 1)|0 0〉

(1)(1 + 1)− (0)(0− 1)|1 −1〉]

√6|2 0〉 =

[√2|−1 1〉+ 2

√2|0 0〉+

√2|1 −1〉

]|2 0〉 =

√16 |−1 1〉+

√23 |0 0〉+

√16 |1 −1〉

|2 −1〉 =√

12 |0 −1〉+

√12 |−1 0〉,

|2 −2〉 = |−1 −1〉

Keep lowering. . .

S−|2 1〉 = (S1− + S2−)√

[|0 1〉+ |1 0〉

(2)(2 + 1)− (1)(1− 1)|2 0〉 =√

[√(1)(1 + 1)− (0)(0− 1)|−1 1〉

(1)(1 + 1)− (1)(1− 1)|0 0〉

(1)(1 + 1)− (0)(0− 1)|1 −1〉]

√6|2 0〉 =

[√2|−1 1〉+ 2

√2|0 0〉+

√2|1 −1〉

]|2 0〉 =

√16 |−1 1〉+

√23 |0 0〉+

√16 |1 −1〉

|2 −1〉 =√

12 |0 −1〉+

√12 |−1 0〉,

|2 −2〉 = |−1 −1〉

Keep lowering. . .

S−|2 1〉 = (S1− + S2−)√

[|0 1〉+ |1 0〉

(2)(2 + 1)− (1)(1− 1)|2 0〉 =√

[√(1)(1 + 1)− (0)(0− 1)|−1 1〉

(1)(1 + 1)− (1)(1− 1)|0 0〉

(1)(1 + 1)− (0)(0− 1)|1 −1〉]

√6|2 0〉 =

[√2|−1 1〉+ 2

√2|0 0〉+

√2|1 −1〉

]|2 0〉 =

√16 |−1 1〉+

√23 |0 0〉+

√16 |1 −1〉

|2 −1〉 =√

12 |0 −1〉+

√12 |−1 0〉,

|2 −2〉 = |−1 −1〉

Keep lowering. . .

S−|2 1〉 = (S1− + S2−)√

[|0 1〉+ |1 0〉

(2)(2 + 1)− (1)(1− 1)|2 0〉 =√

[√(1)(1 + 1)− (0)(0− 1)|−1 1〉

(1)(1 + 1)− (1)(1− 1)|0 0〉

(1)(1 + 1)− (0)(0− 1)|1 −1〉]

√6|2 0〉 =

[√2|−1 1〉+ 2

√2|0 0〉+

√2|1 −1〉

]|2 0〉 =

√16 |−1 1〉+

√23 |0 0〉+

√16 |1 −1〉

|2 −1〉 =√

12 |0 −1〉+

√12 |−1 0〉, |2 −2〉 = |−1 −1〉

What about the other combinations?

The remaining eigenvectors must be orthogonal combinations of the onesalready generated so to start, for |1 1〉

|1 1〉 =√

12 |0 1〉 −

√12 |1 0〉

and applying the lowering operator as before

S−|1 1〉 = (S1− + S2−)√

[|0 1〉 − |1 0〉

(1)(1 + 1)− (1)(1− 1)|1 0〉

[√(1)(1 + 1)− (0)(0− 1)|−1 1〉

−√

(1)(1 + 1)− (1)(1− 1)|0 0〉

−√

(1)(1 + 1)− (0)(0− 1)|1 −1〉

√2|1 0〉 =

√2|−1 1〉 −

√2|1 −1〉

]|1 0〉 =

√12 |−1 1〉 −

√12 |1 −1〉,

|1 −1〉 =√

12 |−1 0〉 −

√12 |0 −1〉

|1 1〉 =√

12 |0 1〉 −

√12 |1 0〉

S−|1 1〉 = (S1− + S2−)√

[|0 1〉 − |1 0〉

(1)(1 + 1)− (1)(1− 1)|1 0〉

[√(1)(1 + 1)− (0)(0− 1)|−1 1〉

−√

(1)(1 + 1)− (1)(1− 1)|0 0〉

−√

(1)(1 + 1)− (0)(0− 1)|1 −1〉

√2|1 0〉 =

√2|−1 1〉 −

√2|1 −1〉

]|1 0〉 =

√12 |−1 1〉 −

√12 |1 −1〉,

|1 −1〉 =√

12 |−1 0〉 −

√12 |0 −1〉

|1 1〉 =√

12 |0 1〉 −

√12 |1 0〉

S−|1 1〉 = (S1− + S2−)√

[|0 1〉 − |1 0〉

(1)(1 + 1)− (1)(1− 1)|1 0〉

[√(1)(1 + 1)− (0)(0− 1)|−1 1〉

−√

(1)(1 + 1)− (1)(1− 1)|0 0〉

−√

(1)(1 + 1)− (0)(0− 1)|1 −1〉

√2|1 0〉 =

√2|−1 1〉 −

√2|1 −1〉

]|1 0〉 =

√12 |−1 1〉 −

√12 |1 −1〉,

|1 −1〉 =√

12 |−1 0〉 −

√12 |0 −1〉

|1 1〉 =√

12 |0 1〉 −

√12 |1 0〉

S−|1 1〉 = (S1− + S2−)√

[|0 1〉 − |1 0〉

√(1)(1 + 1)− (1)(1− 1)|1 0〉

[√(1)(1 + 1)− (0)(0− 1)|−1 1〉

−√

(1)(1 + 1)− (1)(1− 1)|0 0〉

−√

(1)(1 + 1)− (0)(0− 1)|1 −1〉

√2|1 0〉 =

√2|−1 1〉 −

√2|1 −1〉

]|1 0〉 =

√12 |−1 1〉 −

√12 |1 −1〉,

|1 −1〉 =√

12 |−1 0〉 −

√12 |0 −1〉

|1 1〉 =√

12 |0 1〉 −

√12 |1 0〉

S−|1 1〉 = (S1− + S2−)√

[|0 1〉 − |1 0〉

(1)(1 + 1)− (1)(1− 1)|1 0〉

[√(1)(1 + 1)− (0)(0− 1)|−1 1〉

−√

(1)(1 + 1)− (1)(1− 1)|0 0〉

−√

(1)(1 + 1)− (0)(0− 1)|1 −1〉√

2|1 0〉 =√

√2|−1 1〉 −

√2|1 −1〉

]|1 0〉 =

√12 |−1 1〉 −

√12 |1 −1〉,

|1 −1〉 =√

12 |−1 0〉 −

√12 |0 −1〉

|1 1〉 =√

12 |0 1〉 −

√12 |1 0〉

S−|1 1〉 = (S1− + S2−)√

[|0 1〉 − |1 0〉

(1)(1 + 1)− (1)(1− 1)|1 0〉 =√

[√(1)(1 + 1)− (0)(0− 1)|−1 1〉

−√

(1)(1 + 1)− (1)(1− 1)|0 0〉

−√

(1)(1 + 1)− (0)(0− 1)|1 −1〉√

2|1 0〉 =√

√2|−1 1〉 −

√2|1 −1〉

]|1 0〉 =

√12 |−1 1〉 −

√12 |1 −1〉,

|1 −1〉 =√

12 |−1 0〉 −

√12 |0 −1〉

|1 1〉 =√

12 |0 1〉 −

√12 |1 0〉

S−|1 1〉 = (S1− + S2−)√

[|0 1〉 − |1 0〉

(1)(1 + 1)− (1)(1− 1)|1 0〉 =√

[√(1)(1 + 1)− (0)(0− 1)|−1 1〉

−√

(1)(1 + 1)− (1)(1− 1)|0 0〉

−√

(1)(1 + 1)− (0)(0− 1)|1 −1〉√

2|1 0〉 =√

√2|−1 1〉 −

√2|1 −1〉

]|1 0〉 =

√12 |−1 1〉 −

√12 |1 −1〉,

|1 −1〉 =√

12 |−1 0〉 −

√12 |0 −1〉

|1 1〉 =√

12 |0 1〉 −

√12 |1 0〉

S−|1 1〉 = (S1− + S2−)√

[|0 1〉 − |1 0〉

(1)(1 + 1)− (1)(1− 1)|1 0〉 =√

[√(1)(1 + 1)− (0)(0− 1)|−1 1〉

−√

(1)(1 + 1)− (1)(1− 1)|0 0〉

−√

(1)(1 + 1)− (0)(0− 1)|1 −1〉√

2|1 0〉 =√

√2|−1 1〉 −

√2|1 −1〉

]|1 0〉 =

√12 |−1 1〉 −

√12 |1 −1〉,

|1 −1〉 =√

12 |−1 0〉 −

√12 |0 −1〉

|1 1〉 =√

12 |0 1〉 −

√12 |1 0〉

S−|1 1〉 = (S1− + S2−)√

[|0 1〉 − |1 0〉

(1)(1 + 1)− (1)(1− 1)|1 0〉 =√

[√(1)(1 + 1)− (0)(0− 1)|−1 1〉

−√

(1)(1 + 1)− (1)(1− 1)|0 0〉

−√

(1)(1 + 1)− (0)(0− 1)|1 −1〉

√2|1 0〉 =

√2|−1 1〉 −

√2|1 −1〉

]|1 0〉 =

√12 |−1 1〉 −

√12 |1 −1〉,

|1 −1〉 =√

12 |−1 0〉 −

√12 |0 −1〉

|1 1〉 =√

12 |0 1〉 −

√12 |1 0〉

S−|1 1〉 = (S1− + S2−)√

[|0 1〉 − |1 0〉

(1)(1 + 1)− (1)(1− 1)|1 0〉 =√

[√(1)(1 + 1)− (0)(0− 1)|−1 1〉

−√

(1)(1 + 1)− (1)(1− 1)|0 0〉

−√

(1)(1 + 1)− (0)(0− 1)|1 −1〉√

2|1 0〉 =√

√2|−1 1〉 −

√2|1 −1〉

|1 0〉 =√

12 |−1 1〉 −

√12 |1 −1〉,

|1 −1〉 =√

12 |−1 0〉 −

√12 |0 −1〉

|1 1〉 =√

12 |0 1〉 −

√12 |1 0〉

S−|1 1〉 = (S1− + S2−)√

[|0 1〉 − |1 0〉

(1)(1 + 1)− (1)(1− 1)|1 0〉 =√

[√(1)(1 + 1)− (0)(0− 1)|−1 1〉

−√

(1)(1 + 1)− (1)(1− 1)|0 0〉

−√

(1)(1 + 1)− (0)(0− 1)|1 −1〉√

2|1 0〉 =√

[√2|−1 1〉

−√

2|1 −1〉

|1 0〉 =√

12 |−1 1〉 −

√12 |1 −1〉,

|1 −1〉 =√

12 |−1 0〉 −

√12 |0 −1〉

|1 1〉 =√

12 |0 1〉 −

√12 |1 0〉

S−|1 1〉 = (S1− + S2−)√

[|0 1〉 − |1 0〉

(1)(1 + 1)− (1)(1− 1)|1 0〉 =√

[√(1)(1 + 1)− (0)(0− 1)|−1 1〉

−√

(1)(1 + 1)− (1)(1− 1)|0 0〉

−√

(1)(1 + 1)− (0)(0− 1)|1 −1〉√

2|1 0〉 =√

[√2|−1 1〉 −

√2|1 −1〉

|1 0〉 =√

12 |−1 1〉 −

√12 |1 −1〉,

|1 −1〉 =√

12 |−1 0〉 −

√12 |0 −1〉

|1 1〉 =√

12 |0 1〉 −

√12 |1 0〉

S−|1 1〉 = (S1− + S2−)√

[|0 1〉 − |1 0〉

(1)(1 + 1)− (1)(1− 1)|1 0〉 =√

[√(1)(1 + 1)− (0)(0− 1)|−1 1〉

−√

(1)(1 + 1)− (1)(1− 1)|0 0〉

−√

(1)(1 + 1)− (0)(0− 1)|1 −1〉√

2|1 0〉 =√

[√2|−1 1〉 −

√2|1 −1〉

]|1 0〉 =

√12 |−1 1〉 −

√12 |1 −1〉,

|1 −1〉 =√

12 |−1 0〉 −

√12 |0 −1〉

|1 1〉 =√

12 |0 1〉 −

√12 |1 0〉

S−|1 1〉 = (S1− + S2−)√

[|0 1〉 − |1 0〉

(1)(1 + 1)− (1)(1− 1)|1 0〉 =√

[√(1)(1 + 1)− (0)(0− 1)|−1 1〉

−√

(1)(1 + 1)− (1)(1− 1)|0 0〉

−√

(1)(1 + 1)− (0)(0− 1)|1 −1〉√

2|1 0〉 =√

[√2|−1 1〉 −

√2|1 −1〉

]|1 0〉 =

√12 |−1 1〉 −

√12 |1 −1〉, |1 −1〉 =

√12 |−1 0〉 −

√12 |0 −1〉

Just one left!

So, we started with 9 |m1 m2〉 states and have generated 8 |s m〉 states,the final one must be

|0 0〉 = a|−1 1〉+ b|0 0〉+ c |1 −1〉

where a, b, and c are chosen to ensure that |0 0〉 is orthogonal to theother states with m = 0

|2 0〉 =√

16 |−1 1〉+

√23 |0 0〉+

√16 |1 −1〉

|1 0〉 =√

12 |−1 1〉 −

√12 |1 −1〉

〈2 0|0 0〉 = 0 =√

√46b +

√16c

= a + 2b + c

〈1 0|0 0〉 = 0 =√

12a−

√12c

= a− c −→ a = −b = c

|0 0〉 =√

13 |−1 1〉 −

√13 |0 0〉+

√13 |1 −1〉

Just one left!

|0 0〉 = a|−1 1〉+ b|0 0〉+ c |1 −1〉

|2 0〉 =√

16 |−1 1〉+

√23 |0 0〉+

√16 |1 −1〉

|1 0〉 =√

12 |−1 1〉 −

√12 |1 −1〉

〈2 0|0 0〉 = 0 =√

√46b +

√16c

= a + 2b + c

〈1 0|0 0〉 = 0 =√

12a−

√12c

= a− c −→ a = −b = c

|0 0〉 =√

13 |−1 1〉 −

√13 |0 0〉+

√13 |1 −1〉

Just one left!

|0 0〉 = a|−1 1〉+ b|0 0〉+ c |1 −1〉

|2 0〉 =√

16 |−1 1〉+

√23 |0 0〉+

√16 |1 −1〉

|1 0〉 =√

12 |−1 1〉 −

√12 |1 −1〉

〈2 0|0 0〉 = 0 =√

√46b +

√16c

= a + 2b + c

〈1 0|0 0〉 = 0 =√

12a−

√12c

= a− c −→ a = −b = c

|0 0〉 =√

13 |−1 1〉 −

√13 |0 0〉+

√13 |1 −1〉

Just one left!

|0 0〉 = a|−1 1〉+ b|0 0〉+ c |1 −1〉

|2 0〉 =√

16 |−1 1〉+

√23 |0 0〉+

√16 |1 −1〉

|1 0〉 =√

12 |−1 1〉 −

√12 |1 −1〉

〈2 0|0 0〉 = 0 =√

√46b +

√16c

= a + 2b + c

〈1 0|0 0〉 = 0 =√

12a−

√12c

= a− c −→ a = −b = c

|0 0〉 =√

13 |−1 1〉 −

√13 |0 0〉+

√13 |1 −1〉

Just one left!

|0 0〉 = a|−1 1〉+ b|0 0〉+ c |1 −1〉

|2 0〉 =√

16 |−1 1〉+

√23 |0 0〉+

√16 |1 −1〉

|1 0〉 =√

12 |−1 1〉 −

√12 |1 −1〉

〈2 0|0 0〉 = 0 =√

√46b +

√16c

= a + 2b + c

〈1 0|0 0〉 = 0 =√

12a−

√12c

= a− c −→ a = −b = c

|0 0〉 =√

13 |−1 1〉 −

√13 |0 0〉+

√13 |1 −1〉

Just one left!

|0 0〉 = a|−1 1〉+ b|0 0〉+ c |1 −1〉

|2 0〉 =√

16 |−1 1〉+

√23 |0 0〉+

√16 |1 −1〉

|1 0〉 =√

12 |−1 1〉 −

√12 |1 −1〉

〈2 0|0 0〉 = 0 =√

√46b +

√16c

= a + 2b + c

〈1 0|0 0〉 = 0 =√

12a−

√12c

= a− c −→ a = −b = c

|0 0〉 =√

13 |−1 1〉 −

√13 |0 0〉+

√13 |1 −1〉

Just one left!

|0 0〉 = a|−1 1〉+ b|0 0〉+ c |1 −1〉

|2 0〉 =√

16 |−1 1〉+

√23 |0 0〉+

√16 |1 −1〉

|1 0〉 =√

12 |−1 1〉 −

√12 |1 −1〉

〈2 0|0 0〉 = 0 =√

√46b +

√16c

= a + 2b + c

〈1 0|0 0〉 = 0 =√

12a−

√12c

= a− c −→ a = −b = c

|0 0〉 =√

13 |−1 1〉 −

√13 |0 0〉+

√13 |1 −1〉

Just one left!

|0 0〉 = a|−1 1〉+ b|0 0〉+ c |1 −1〉

|2 0〉 =√

16 |−1 1〉+

√23 |0 0〉+

√16 |1 −1〉

|1 0〉 =√

12 |−1 1〉 −

√12 |1 −1〉

〈2 0|0 0〉 = 0 =√

√46b +

√16c = a + 2b + c

〈1 0|0 0〉 = 0 =√

12a−

√12c

= a− c −→ a = −b = c

|0 0〉 =√

13 |−1 1〉 −

√13 |0 0〉+

√13 |1 −1〉

Just one left!

|0 0〉 = a|−1 1〉+ b|0 0〉+ c |1 −1〉

|2 0〉 =√

16 |−1 1〉+

√23 |0 0〉+

√16 |1 −1〉

|1 0〉 =√

12 |−1 1〉 −

√12 |1 −1〉

〈2 0|0 0〉 = 0 =√

√46b +

√16c = a + 2b + c

〈1 0|0 0〉 = 0 =√

12a−

√12c = a− c

−→ a = −b = c

|0 0〉 =√

13 |−1 1〉 −

√13 |0 0〉+

√13 |1 −1〉

Just one left!

|0 0〉 = a|−1 1〉+ b|0 0〉+ c |1 −1〉

|2 0〉 =√

16 |−1 1〉+

√23 |0 0〉+

√16 |1 −1〉

|1 0〉 =√

12 |−1 1〉 −

√12 |1 −1〉

〈2 0|0 0〉 = 0 =√

√46b +

√16c = a + 2b + c

〈1 0|0 0〉 = 0 =√

12a−

√12c = a− c −→ a = −b = c

|0 0〉 =√

13 |−1 1〉 −

√13 |0 0〉+

√13 |1 −1〉

Just one left!

|0 0〉 = a|−1 1〉+ b|0 0〉+ c |1 −1〉

|2 0〉 =√

16 |−1 1〉+

√23 |0 0〉+

√16 |1 −1〉

|1 0〉 =√

12 |−1 1〉 −

√12 |1 −1〉

〈2 0|0 0〉 = 0 =√

√46b +

√16c = a + 2b + c

〈1 0|0 0〉 = 0 =√

12a−

√12c = a− c −→ a = −b = c

|0 0〉 =√

13 |−1 1〉 −

√13 |0 0〉+

√13 |1 −1〉

The 1×1 table

Problem 4.35

Quarks carry spin 1/2. Three quarks bind together to make abaryon (such as the proton or neutron); two quarks (or moreprecisely a quark and an antiquark) bind together to make ameson (such as the pion of the kaon). assume the quarks are inthe ground state (zero orbital angular momentum).

(a) What spins are possible for baryons?

(b) What spins are possible for mesons?

Problem 4.35 (cont.)

(a) Add the first two spins together

Then add in the third spin to eachof the two combinations

There are three possible spin states

12 1 0 1

Note that there are 23 = 8 ways to combine the three spins and these 8states are present in the resulting total spin states

(b) For mesons, one needs to combine two spins only

This gives the usual 4 states divided into a spin 0 singlet and a spin 1triplet.

12 1 0 1

Problem 4.37

Determine the commutator of S2 with S(1)z (where ~S ≡ ~S (1) + ~S (2)).

Generalize the result to show that

[S2, ~S (1)] = 2i~(~S (1) × ~S (2))

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

[S2,S(1)z ] =

��[(S (1))2,S

(1)z ] +��

[(S (2))2,S(1)z ] + 2[~S (1) · ~S (2),S

(1)z ]

[S(1)x ,S

(1)z ]S

(2)x + [S

(1)y ,S

(1)z ]S

(2)y +��

[S(1)z ,S

(1)z ]S

(−i~S (1)

y S(2)x + i~S (1)

x S(2)y

)= 2i~(~S (1) × ~S (2))z

Similarly we can obtain

[S2,S(1)x ] = 2i~(~S (1) × ~S (2))x

and [S2,S(1)y ] = 2i~(~S (1) × ~S (2))y

proving the general relation: [S2, ~S (1)] = 2i~(~S (1) × ~S (2))

Problem 4.37

[S2, ~S (1)] = 2i~(~S (1) × ~S (2))

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

[S2,S(1)z ] =

��[(S (1))2,S

(1)z ] +��

[(S (2))2,S(1)z ] + 2[~S (1) · ~S (2),S

(1)z ]

[S(1)x ,S

(1)z ]S

(2)x + [S

(1)y ,S

(1)z ]S

(2)y +��

[S(1)z ,S

(1)z ]S

(−i~S (1)

y S(2)x + i~S (1)

x S(2)y

)= 2i~(~S (1) × ~S (2))z

[S2,S(1)x ] = 2i~(~S (1) × ~S (2))x

and [S2,S(1)y ] = 2i~(~S (1) × ~S (2))y

Problem 4.37

[S2, ~S (1)] = 2i~(~S (1) × ~S (2))

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

[S2,S(1)z ] =

��[(S (1))2, S

(1)z ] +��

[(S (2))2, S(1)z ] + 2[~S (1) · ~S (2), S

(1)z ]

[S(1)x ,S

(1)z ]S

(2)x + [S

(1)y ,S

(1)z ]S

(2)y +��

[S(1)z ,S

(1)z ]S

(−i~S (1)

y S(2)x + i~S (1)

x S(2)y

)= 2i~(~S (1) × ~S (2))z

[S2,S(1)x ] = 2i~(~S (1) × ~S (2))x

and [S2,S(1)y ] = 2i~(~S (1) × ~S (2))y

Problem 4.37

[S2, ~S (1)] = 2i~(~S (1) × ~S (2))

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

[S2,S(1)z ] = [(S (1))2, S

(1)z ]

+��[(S (2))2, S

(1)z ] + 2[~S (1) · ~S (2), S

(1)z ]

[S(1)x ,S

(1)z ]S

(2)x + [S

(1)y ,S

(1)z ]S

(2)y +��

[S(1)z ,S

(1)z ]S

(−i~S (1)

y S(2)x + i~S (1)

x S(2)y

)= 2i~(~S (1) × ~S (2))z

[S2,S(1)x ] = 2i~(~S (1) × ~S (2))x

and [S2,S(1)y ] = 2i~(~S (1) × ~S (2))y

Problem 4.37

[S2, ~S (1)] = 2i~(~S (1) × ~S (2))

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

[S2,S(1)z ] = [(S (1))2, S

(1)z ] + [(S (2))2, S

(1)z ]

+ 2[~S (1) · ~S (2), S(1)z ]

[S(1)x ,S

(1)z ]S

(2)x + [S

(1)y ,S

(1)z ]S

(2)y +��

[S(1)z ,S

(1)z ]S

(−i~S (1)

y S(2)x + i~S (1)

x S(2)y

)= 2i~(~S (1) × ~S (2))z

[S2,S(1)x ] = 2i~(~S (1) × ~S (2))x

and [S2,S(1)y ] = 2i~(~S (1) × ~S (2))y

Problem 4.37

[S2, ~S (1)] = 2i~(~S (1) × ~S (2))

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

[S2,S(1)z ] = [(S (1))2, S

(1)z ] + [(S (2))2, S

(1)z ] + 2[~S (1) · ~S (2), S

(1)z ]

[S(1)x ,S

(1)z ]S

(2)x + [S

(1)y ,S

(1)z ]S

(2)y +��

[S(1)z ,S

(1)z ]S

(−i~S (1)

y S(2)x + i~S (1)

x S(2)y

)= 2i~(~S (1) × ~S (2))z

[S2,S(1)x ] = 2i~(~S (1) × ~S (2))x

and [S2,S(1)y ] = 2i~(~S (1) × ~S (2))y

Problem 4.37

[S2, ~S (1)] = 2i~(~S (1) × ~S (2))

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

[S2,S(1)z ] = ��

[(S (1))2, S(1)z ] + [(S (2))2, S

(1)z ] + 2[~S (1) · ~S (2), S

(1)z ]

[S(1)x ,S

(1)z ]S

(2)x + [S

(1)y ,S

(1)z ]S

(2)y +��

[S(1)z ,S

(1)z ]S

(−i~S (1)

y S(2)x + i~S (1)

x S(2)y

)= 2i~(~S (1) × ~S (2))z

[S2,S(1)x ] = 2i~(~S (1) × ~S (2))x

and [S2,S(1)y ] = 2i~(~S (1) × ~S (2))y

Problem 4.37

[S2, ~S (1)] = 2i~(~S (1) × ~S (2))

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

[S2,S(1)z ] = ��

[(S (1))2, S(1)z ] +��

[(S (2))2, S(1)z ] + 2[~S (1) · ~S (2), S

(1)z ]

[S(1)x ,S

(1)z ]S

(2)x + [S

(1)y , S

(1)z ]S

(2)y +��

[S(1)z ,S

(1)z ]S

= 2(−i~S (1)

y S(2)x + i~S (1)

x S(2)y

)= 2i~(~S (1) × ~S (2))z

[S2,S(1)x ] = 2i~(~S (1) × ~S (2))x

and [S2,S(1)y ] = 2i~(~S (1) × ~S (2))y

Problem 4.37

[S2, ~S (1)] = 2i~(~S (1) × ~S (2))

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

[S2,S(1)z ] = ��

[(S (1))2, S(1)z ] +��

[(S (2))2, S(1)z ] + 2[~S (1) · ~S (2), S

(1)z ]

[S(1)x ,S

(1)z ]S

+ [S(1)y , S

(1)z ]S

(2)y +��

[S(1)z ,S

(1)z ]S

= 2(−i~S (1)

y S(2)x + i~S (1)

x S(2)y

)= 2i~(~S (1) × ~S (2))z

[S2,S(1)x ] = 2i~(~S (1) × ~S (2))x

and [S2,S(1)y ] = 2i~(~S (1) × ~S (2))y

Problem 4.37

[S2, ~S (1)] = 2i~(~S (1) × ~S (2))

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

[S2,S(1)z ] = ��

[(S (1))2, S(1)z ] +��

[(S (2))2, S(1)z ] + 2[~S (1) · ~S (2), S

(1)z ]

[S(1)x ,S

(1)z ]S

(2)x + [S

(1)y , S

(1)z ]S

+��[S

(1)z ,S

(1)z ]S

= 2(−i~S (1)

y S(2)x + i~S (1)

x S(2)y

)= 2i~(~S (1) × ~S (2))z

[S2,S(1)x ] = 2i~(~S (1) × ~S (2))x

and [S2,S(1)y ] = 2i~(~S (1) × ~S (2))y

Problem 4.37

[S2, ~S (1)] = 2i~(~S (1) × ~S (2))

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

[S2,S(1)z ] = ��

[(S (1))2, S(1)z ] +��

[(S (2))2, S(1)z ] + 2[~S (1) · ~S (2), S

(1)z ]

[S(1)x ,S

(1)z ]S

(2)x + [S

(1)y , S

(1)z ]S

(2)y + [S

(1)z ,S

(1)z ]S

= 2(−i~S (1)

y S(2)x + i~S (1)

x S(2)y

)= 2i~(~S (1) × ~S (2))z

[S2,S(1)x ] = 2i~(~S (1) × ~S (2))x

and [S2,S(1)y ] = 2i~(~S (1) × ~S (2))y

Problem 4.37

[S2, ~S (1)] = 2i~(~S (1) × ~S (2))

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

[S2,S(1)z ] = ��

[(S (1))2, S(1)z ] +��

[(S (2))2, S(1)z ] + 2[~S (1) · ~S (2), S

(1)z ]

[S(1)x ,S

(1)z ]S

(2)x + [S

(1)y , S

(1)z ]S

(2)y +��

[S(1)z ,S

(1)z ]S

= 2(−i~S (1)

y S(2)x + i~S (1)

x S(2)y

)= 2i~(~S (1) × ~S (2))z

[S2,S(1)x ] = 2i~(~S (1) × ~S (2))x

and [S2,S(1)y ] = 2i~(~S (1) × ~S (2))y

Problem 4.37

[S2, ~S (1)] = 2i~(~S (1) × ~S (2))

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

[S2,S(1)z ] = ��

[(S (1))2, S(1)z ] +��

[(S (2))2, S(1)z ] + 2[~S (1) · ~S (2), S

(1)z ]

[S(1)x ,S

(1)z ]S

(2)x + [S

(1)y , S

(1)z ]S

(2)y +��

[S(1)z ,S

(1)z ]S

(−i~S (1)

y S(2)x + i~S (1)

x S(2)y

= 2i~(~S (1) × ~S (2))z

[S2,S(1)x ] = 2i~(~S (1) × ~S (2))x

and [S2,S(1)y ] = 2i~(~S (1) × ~S (2))y

Problem 4.37

[S2, ~S (1)] = 2i~(~S (1) × ~S (2))

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

[S2,S(1)z ] = ��

[(S (1))2, S(1)z ] +��

[(S (2))2, S(1)z ] + 2[~S (1) · ~S (2), S

(1)z ]

[S(1)x ,S

(1)z ]S

(2)x + [S

(1)y , S

(1)z ]S

(2)y +��

[S(1)z ,S

(1)z ]S

(−i~S (1)

y S(2)x + i~S (1)

x S(2)y

)= 2i~(~S (1) × ~S (2))z

[S2,S(1)x ] = 2i~(~S (1) × ~S (2))x

and [S2,S(1)y ] = 2i~(~S (1) × ~S (2))y

Problem 4.37

[S2, ~S (1)] = 2i~(~S (1) × ~S (2))

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

[S2,S(1)z ] = ��

[(S (1))2, S(1)z ] +��

[(S (2))2, S(1)z ] + 2[~S (1) · ~S (2), S

(1)z ]

[S(1)x ,S

(1)z ]S

(2)x + [S

(1)y , S

(1)z ]S

(2)y +��

[S(1)z ,S

(1)z ]S

(−i~S (1)

y S(2)x + i~S (1)

x S(2)y

)= 2i~(~S (1) × ~S (2))z

[S2,S(1)x ] = 2i~(~S (1) × ~S (2))x

and [S2,S(1)y ] = 2i~(~S (1) × ~S (2))y

Problem 4.37

[S2, ~S (1)] = 2i~(~S (1) × ~S (2))

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

[S2,S(1)z ] = ��

[(S (1))2, S(1)z ] +��

[(S (2))2, S(1)z ] + 2[~S (1) · ~S (2), S

(1)z ]

[S(1)x ,S

(1)z ]S

(2)x + [S

(1)y , S

(1)z ]S

(2)y +��

[S(1)z ,S

(1)z ]S

(−i~S (1)

y S(2)x + i~S (1)

x S(2)y

)= 2i~(~S (1) × ~S (2))z

[S2, S(1)x ] = 2i~(~S (1) × ~S (2))x

and [S2, S(1)y ] = 2i~(~S (1) × ~S (2))y

Problem 4.37

[S2, ~S (1)] = 2i~(~S (1) × ~S (2))

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

[S2,S(1)z ] = ��

[(S (1))2, S(1)z ] +��

[(S (2))2, S(1)z ] + 2[~S (1) · ~S (2), S

(1)z ]

[S(1)x ,S

(1)z ]S

(2)x + [S

(1)y , S

(1)z ]S

(2)y +��

[S(1)z ,S

(1)z ]S

(−i~S (1)

y S(2)x + i~S (1)

x S(2)y

)= 2i~(~S (1) × ~S (2))z

[S2, S(1)x ] = 2i~(~S (1) × ~S (2))x and [S2, S

(1)y ] = 2i~(~S (1) × ~S (2))y

Problem 4.37

[S2, ~S (1)] = 2i~(~S (1) × ~S (2))

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

[S2,S(1)z ] = ��

[(S (1))2, S(1)z ] +��

[(S (2))2, S(1)z ] + 2[~S (1) · ~S (2), S

(1)z ]

[S(1)x ,S

(1)z ]S

(2)x + [S

(1)y , S

(1)z ]S

(2)y +��

[S(1)z ,S

(1)z ]S

(−i~S (1)

y S(2)x + i~S (1)

x S(2)y

)= 2i~(~S (1) × ~S (2))z

[S2, S(1)x ] = 2i~(~S (1) × ~S (2))x and [S2, S

(1)y ] = 2i~(~S (1) × ~S (2))y

proving the general relation: [S2, ~S (1)] = 2i~(~S (1) × ~S (2))C. Segre (IIT) PHYS 405 - Fall 2016 October 31, 2016 11 / 18

Problem 4.52

Find the matrix representing Sx for a particle of spin 3/2 (using the basisof eigenstates of Sz). Solve the characteristic equation to determine theeigenvalues of Sx .

First we write down the eigenstates of Sz in the S = 3/2 system.

∣∣32

∣∣32 −

Problem 4.52

∣∣32

∣∣32 −

Problem 4.52

∣∣32

∣∣32 −

Problem 4.52

∣∣32

∣∣32 −

Problem 4.52

∣∣32

∣∣32 −

Problem 4.52

∣∣32

∣∣32 −

The raising and lowering operators are

S±|s m〉 = ~√s(s + 1)−m(m ± 1) |s (m ± 1)〉

we build the matrix for S+ by applying the raising operator to each of theeigenstates

S+ = ~

3 0 00 0 2 0

0 0 0√

30 0 0 0

S+∣∣32

⟩= 0

S+∣∣32

⟩=√

)− 1

)~∣∣32

⟩=√

3~∣∣32

⟩S+∣∣32 −

⟩=√

)~∣∣32

⟩= 2~

∣∣32

⟩S+∣∣32 −

⟩=√

)~∣∣32 −

⟩=√

3~∣∣32 −

S±|s m〉 = ~√s(s + 1)−m(m ± 1) |s (m ± 1)〉

S+ = ~

3 0 00 0 2 0

0 0 0√

30 0 0 0

S+∣∣32

⟩= 0

S+∣∣32

⟩=√

)− 1

)~∣∣32

⟩=√

3~∣∣32

⟩S+∣∣32 −

⟩=√

)~∣∣32

⟩= 2~

∣∣32

⟩S+∣∣32 −

⟩=√

)~∣∣32 −

⟩=√

3~∣∣32 −

S±|s m〉 = ~√s(s + 1)−m(m ± 1) |s (m ± 1)〉

S+ = ~

3 0 00 0 2 0

0 0 0√

30 0 0 0

S+∣∣32

⟩= 0

S+∣∣32

⟩=√

)− 1

)~∣∣32

⟩=√

3~∣∣32

⟩S+∣∣32 −

⟩=√

)~∣∣32

⟩= 2~

∣∣32

⟩S+∣∣32 −

⟩=√

)~∣∣32 −

⟩=√

3~∣∣32 −

S±|s m〉 = ~√s(s + 1)−m(m ± 1) |s (m ± 1)〉

S+ = ~

3 0 00 0 2 0

0 0 0√

30 0 0 0

S+∣∣32

⟩= 0

S+∣∣32

⟩=√

)− 1

)~∣∣32

⟩=√

3~∣∣32

⟩S+∣∣32 −

⟩=√

)~∣∣32

⟩= 2~

∣∣32

⟩S+∣∣32 −

⟩=√

)~∣∣32 −

⟩=√

3~∣∣32 −

S±|s m〉 = ~√s(s + 1)−m(m ± 1) |s (m ± 1)〉

S+ = ~

√3 0 0

0 0√

S+∣∣32

⟩= 0

S+∣∣32

⟩=√

)− 1

)~∣∣32

⟩=√

3~∣∣32

⟩S+∣∣32 −

⟩=√

)~∣∣32

⟩= 2~

∣∣32

⟩S+∣∣32 −

⟩=√

)~∣∣32 −

⟩=√

3~∣∣32 −

S±|s m〉 = ~√s(s + 1)−m(m ± 1) |s (m ± 1)〉

S+ = ~

√3 0 0

0 0√

S+∣∣32

⟩= 0

S+∣∣32

⟩=√

)− 1

)~∣∣32

3~∣∣32

⟩S+∣∣32 −

⟩=√

)~∣∣32

⟩= 2~

∣∣32

⟩S+∣∣32 −

⟩=√

)~∣∣32 −

⟩=√

3~∣∣32 −

S±|s m〉 = ~√s(s + 1)−m(m ± 1) |s (m ± 1)〉

S+ = ~

√3 0 0

0 0√

S+∣∣32

⟩= 0

S+∣∣32

⟩=√

)− 1

)~∣∣32

⟩=√

3~∣∣32

S+∣∣32 −

⟩=√

)~∣∣32

⟩= 2~

∣∣32

⟩S+∣∣32 −

⟩=√

)~∣∣32 −

⟩=√

3~∣∣32 −

S±|s m〉 = ~√s(s + 1)−m(m ± 1) |s (m ± 1)〉

S+ = ~

S+∣∣32

⟩= 0

S+∣∣32

⟩=√

)− 1

)~∣∣32

⟩=√

3~∣∣32

S+∣∣32 −

⟩=√

)~∣∣32

⟩= 2~

∣∣32

⟩S+∣∣32 −

⟩=√

)~∣∣32 −

⟩=√

3~∣∣32 −

S±|s m〉 = ~√s(s + 1)−m(m ± 1) |s (m ± 1)〉

S+ = ~

S+∣∣32

⟩= 0

S+∣∣32

⟩=√

)− 1

)~∣∣32

⟩=√

3~∣∣32

⟩S+∣∣32 −

⟩=√

)~∣∣32

= 2~∣∣32

⟩S+∣∣32 −

⟩=√

)~∣∣32 −

⟩=√

3~∣∣32 −

S±|s m〉 = ~√s(s + 1)−m(m ± 1) |s (m ± 1)〉

S+ = ~

S+∣∣32

⟩= 0

S+∣∣32

⟩=√

)− 1

)~∣∣32

⟩=√

3~∣∣32

⟩S+∣∣32 −

⟩=√

)~∣∣32

⟩= 2~

∣∣32

S+∣∣32 −

⟩=√

)~∣∣32 −

⟩=√

3~∣∣32 −

S±|s m〉 = ~√s(s + 1)−m(m ± 1) |s (m ± 1)〉

S+ = ~

S+∣∣32

⟩= 0

S+∣∣32

⟩=√

)− 1

)~∣∣32

⟩=√

3~∣∣32

⟩S+∣∣32 −

⟩=√

)~∣∣32

⟩= 2~

∣∣32

S+∣∣32 −

⟩=√

)~∣∣32 −

⟩=√

3~∣∣32 −

S±|s m〉 = ~√s(s + 1)−m(m ± 1) |s (m ± 1)〉

S+ = ~

S+∣∣32

⟩= 0

S+∣∣32

⟩=√

)− 1

)~∣∣32

⟩=√

3~∣∣32

⟩S+∣∣32 −

⟩=√

)~∣∣32

⟩= 2~

∣∣32

⟩S+∣∣32 −

⟩=√

)~∣∣32 −

3~∣∣32 −

S±|s m〉 = ~√s(s + 1)−m(m ± 1) |s (m ± 1)〉

S+ = ~

S+∣∣32

⟩= 0

S+∣∣32

⟩=√

)− 1

)~∣∣32

⟩=√

3~∣∣32

⟩S+∣∣32 −

⟩=√

)~∣∣32

⟩= 2~

∣∣32

⟩S+∣∣32 −

⟩=√

)~∣∣32 −

⟩=√

3~∣∣32 −

⟩C. Segre (IIT) PHYS 405 - Fall 2016 October 31, 2016 13 / 18

S±|s m〉 = ~√s(s + 1)−m(m ± 1) |s (m ± 1)〉

S+ = ~

3 0 00 0 2 0

0 0 0√

30 0 0 0

S+∣∣32

⟩= 0

S+∣∣32

⟩=√

)− 1

)~∣∣32

⟩=√

3~∣∣32

⟩S+∣∣32 −

⟩=√

)~∣∣32

⟩= 2~

∣∣32

⟩S+∣∣32 −

⟩=√

)~∣∣32 −

⟩=√

3~∣∣32 −

⟩C. Segre (IIT) PHYS 405 - Fall 2016 October 31, 2016 13 / 18

Similarly, we build the matrix for S− by applying the lowering operator toeach of the eigenstates

S− = ~

0 0 0 0√3 0 0 0

0 2 0 0

0 0√

S−∣∣32

⟩=√

)− 3

)~∣∣32

⟩=√

3~∣∣32

⟩S−∣∣32

⟩=√

)− 1

)~∣∣32

⟩= 2~

∣∣32 −

⟩S−∣∣32 −

⟩=√

)~∣∣32

⟩=√

3~∣∣32 −

⟩S−∣∣32 −

⟩= 0

S− = ~

0 0 0 0√3 0 0 0

0 2 0 0

0 0√

S−∣∣32

⟩=√

)− 3

)~∣∣32

⟩=√

3~∣∣32

⟩S−∣∣32

⟩=√

)− 1

)~∣∣32

⟩= 2~

∣∣32 −

⟩S−∣∣32 −

⟩=√

)~∣∣32

⟩=√

3~∣∣32 −

⟩S−∣∣32 −

⟩= 0

S− = ~

0 0 0 0√3 0 0 0

0 2 0 0

0 0√

S−∣∣32

⟩=√

)− 3

)~∣∣32

3~∣∣32

⟩S−∣∣32

⟩=√

)− 1

)~∣∣32

⟩= 2~

∣∣32 −

⟩S−∣∣32 −

⟩=√

)~∣∣32

⟩=√

3~∣∣32 −

⟩S−∣∣32 −

⟩= 0

S− = ~

0 0 0 0√3 0 0 0

0 2 0 0

0 0√

S−∣∣32

⟩=√

)− 3

)~∣∣32

⟩=√

3~∣∣32

S−∣∣32

⟩=√

)− 1

)~∣∣32

⟩= 2~

∣∣32 −

⟩S−∣∣32 −

⟩=√

)~∣∣32

⟩=√

3~∣∣32 −

⟩S−∣∣32 −

⟩= 0

S− = ~

S−∣∣32

⟩=√

)− 3

)~∣∣32

⟩=√

3~∣∣32

S−∣∣32

⟩=√

)− 1

)~∣∣32

⟩= 2~

∣∣32 −

⟩S−∣∣32 −

⟩=√

)~∣∣32

⟩=√

3~∣∣32 −

⟩S−∣∣32 −

⟩= 0

S− = ~

S−∣∣32

⟩=√

)− 3

)~∣∣32

⟩=√

3~∣∣32

⟩S−∣∣32

⟩=√

)− 1

)~∣∣32

= 2~∣∣32 −

⟩S−∣∣32 −

⟩=√

)~∣∣32

⟩=√

3~∣∣32 −

⟩S−∣∣32 −

⟩= 0

S− = ~

S−∣∣32

⟩=√

)− 3

)~∣∣32

⟩=√

3~∣∣32

⟩S−∣∣32

⟩=√

)− 1

)~∣∣32

⟩= 2~

∣∣32 −

S−∣∣32 −

⟩=√

)~∣∣32

⟩=√

3~∣∣32 −

⟩S−∣∣32 −

⟩= 0

S− = ~

√3 0

S−∣∣32

⟩=√

)− 3

)~∣∣32

⟩=√

3~∣∣32

⟩S−∣∣32

⟩=√

)− 1

)~∣∣32

⟩= 2~

∣∣32 −

S−∣∣32 −

⟩=√

)~∣∣32

⟩=√

3~∣∣32 −

⟩S−∣∣32 −

⟩= 0

S− = ~

√3 0

S−∣∣32

⟩=√

)− 3

)~∣∣32

⟩=√

3~∣∣32

⟩S−∣∣32

⟩=√

)− 1

)~∣∣32

⟩= 2~

∣∣32 −

⟩S−∣∣32 −

⟩=√

)~∣∣32

3~∣∣32 −

⟩S−∣∣32 −

⟩= 0

S− = ~

√3 0

S−∣∣32

⟩=√

)− 3

)~∣∣32

⟩=√

3~∣∣32

⟩S−∣∣32

⟩=√

)− 1

)~∣∣32

⟩= 2~

∣∣32 −

⟩S−∣∣32 −

⟩=√

)~∣∣32

⟩=√

3~∣∣32 −

S−∣∣32 −

⟩= 0

S− = ~

√3 0 0

0 0√

S−∣∣32

⟩=√

)− 3

)~∣∣32

⟩=√

3~∣∣32

⟩S−∣∣32

⟩=√

)− 1

)~∣∣32

⟩= 2~

∣∣32 −

⟩S−∣∣32 −

⟩=√

)~∣∣32

⟩=√

3~∣∣32 −

S−∣∣32 −

⟩= 0

S− = ~

√3 0 0

0 0√

S−∣∣32

⟩=√

)− 3

)~∣∣32

⟩=√

3~∣∣32

⟩S−∣∣32

⟩=√

)− 1

)~∣∣32

⟩= 2~

∣∣32 −

⟩S−∣∣32 −

⟩=√

)~∣∣32

⟩=√

3~∣∣32 −

⟩S−∣∣32 −

⟩= 0

S− = ~

0 0 0 0√3 0 0 0

0 2 0 0

0 0√

S−∣∣32

⟩=√

)− 3

)~∣∣32

⟩=√

3~∣∣32

⟩S−∣∣32

⟩=√

)− 1

)~∣∣32

⟩= 2~

∣∣32 −

⟩S−∣∣32 −

⟩=√

)~∣∣32

⟩=√

3~∣∣32 −

⟩S−∣∣32 −

⟩= 0

Sx = 12(S+ + S−)

3 0 0√3 0 2 0

0 2 0√

0 0√

The characteristic equation for eigenvalues is (Note that λ is really theeigenvalue divided by ~/2 for ease of notation)

∣∣∣∣∣∣∣∣−λ

√3 0 0√

3 −λ 2 0

0 2 −λ√

0 0√

3 −λ

∣∣∣∣∣∣∣∣ = −λ

∣∣∣∣∣∣−λ 2 0

2 −λ√

3 −λ

∣∣∣∣∣∣−√3

∣∣∣∣∣∣√

0 −λ√

3 −λ

∣∣∣∣∣∣= −λ[−λ(λ2 − 3)− 2(−2λ)]−

√3[√

3(λ2 − 3)]

= −λ[−λ3 + 7λ]−[3λ2 − 9] = λ4 − 10λ2 + 9

0 = λ4 − 10λ2 + 9 = (λ2 − 9)(λ2 − 1) = (λ+ 3)(λ− 3)(λ+ 1)(λ− 1)

the eigenvalues of Sx are : 32~,

12~,−

Sx = 12(S+ + S−) =

3 0 0√3 0 2 0

0 2 0√

0 0√

∣∣∣∣∣∣∣∣−λ

√3 0 0√

3 −λ 2 0

0 2 −λ√

0 0√

3 −λ

∣∣∣∣∣∣∣∣ = −λ

∣∣∣∣∣∣−λ 2 0

2 −λ√

3 −λ

∣∣∣∣∣∣−√3

∣∣∣∣∣∣√

0 −λ√

3 −λ

∣∣∣∣∣∣= −λ[−λ(λ2 − 3)− 2(−2λ)]−

√3[√

3(λ2 − 3)]

= −λ[−λ3 + 7λ]−[3λ2 − 9] = λ4 − 10λ2 + 9

0 = λ4 − 10λ2 + 9 = (λ2 − 9)(λ2 − 1) = (λ+ 3)(λ− 3)(λ+ 1)(λ− 1)

12~,−

Sx = 12(S+ + S−) =

3 0 0√3 0 2 0

0 2 0√

0 0√

∣∣∣∣∣∣∣∣−λ

√3 0 0√

3 −λ 2 0

0 2 −λ√

0 0√

3 −λ

∣∣∣∣∣∣∣∣ = −λ

∣∣∣∣∣∣−λ 2 0

2 −λ√

3 −λ

∣∣∣∣∣∣−√3

∣∣∣∣∣∣√

0 −λ√

3 −λ

∣∣∣∣∣∣= −λ[−λ(λ2 − 3)− 2(−2λ)]−

√3[√

3(λ2 − 3)]

= −λ[−λ3 + 7λ]−[3λ2 − 9] = λ4 − 10λ2 + 9

0 = λ4 − 10λ2 + 9 = (λ2 − 9)(λ2 − 1) = (λ+ 3)(λ− 3)(λ+ 1)(λ− 1)

12~,−

Sx = 12(S+ + S−) =

3 0 0√3 0 2 0

0 2 0√

0 0√

∣∣∣∣∣∣∣∣−λ

√3 0 0√

3 −λ 2 0

0 2 −λ√

0 0√

3 −λ

∣∣∣∣∣∣∣∣

= −λ

∣∣∣∣∣∣−λ 2 0

2 −λ√

3 −λ

∣∣∣∣∣∣−√3

∣∣∣∣∣∣√

0 −λ√

3 −λ

∣∣∣∣∣∣= −λ[−λ(λ2 − 3)− 2(−2λ)]−

√3[√

3(λ2 − 3)]

= −λ[−λ3 + 7λ]−[3λ2 − 9] = λ4 − 10λ2 + 9

0 = λ4 − 10λ2 + 9 = (λ2 − 9)(λ2 − 1) = (λ+ 3)(λ− 3)(λ+ 1)(λ− 1)

12~,−

Sx = 12(S+ + S−) =

3 0 0√3 0 2 0

0 2 0√

0 0√

∣∣∣∣∣∣∣∣−λ

√3 0 0√

3 −λ 2 0

0 2 −λ√

0 0√

3 −λ

∣∣∣∣∣∣∣∣ = −λ

∣∣∣∣∣∣−λ 2 0

2 −λ√

3 −λ

∣∣∣∣∣∣

−√

∣∣∣∣∣∣√

0 −λ√

3 −λ

∣∣∣∣∣∣= −λ[−λ(λ2 − 3)− 2(−2λ)]−

√3[√

3(λ2 − 3)]

= −λ[−λ3 + 7λ]−[3λ2 − 9] = λ4 − 10λ2 + 9

0 = λ4 − 10λ2 + 9 = (λ2 − 9)(λ2 − 1) = (λ+ 3)(λ− 3)(λ+ 1)(λ− 1)

12~,−

Sx = 12(S+ + S−) =

3 0 0√3 0 2 0

0 2 0√

0 0√

∣∣∣∣∣∣∣∣−λ

√3 0 0√

3 −λ 2 0

0 2 −λ√

0 0√

3 −λ

∣∣∣∣∣∣∣∣ = −λ

∣∣∣∣∣∣−λ 2 0

2 −λ√

3 −λ

∣∣∣∣∣∣−√3

∣∣∣∣∣∣√

0 −λ√

3 −λ

∣∣∣∣∣∣

= −λ[−λ(λ2 − 3)− 2(−2λ)]−√

3(λ2 − 3)]

= −λ[−λ3 + 7λ]−[3λ2 − 9] = λ4 − 10λ2 + 9

0 = λ4 − 10λ2 + 9 = (λ2 − 9)(λ2 − 1) = (λ+ 3)(λ− 3)(λ+ 1)(λ− 1)

12~,−

Sx = 12(S+ + S−) =

3 0 0√3 0 2 0

0 2 0√

0 0√

∣∣∣∣∣∣∣∣−λ

√3 0 0√

3 −λ 2 0

0 2 −λ√

0 0√

3 −λ

∣∣∣∣∣∣∣∣ = −λ

∣∣∣∣∣∣−λ 2 0

2 −λ√

3 −λ

∣∣∣∣∣∣−√3

∣∣∣∣∣∣√

0 −λ√

3 −λ

∣∣∣∣∣∣= −λ[−λ(λ2 − 3)− 2(−2λ)]−

√3[√

3(λ2 − 3)]

= −λ[−λ3 + 7λ]−[3λ2 − 9] = λ4 − 10λ2 + 9

0 = λ4 − 10λ2 + 9 = (λ2 − 9)(λ2 − 1) = (λ+ 3)(λ− 3)(λ+ 1)(λ− 1)

12~,−

Sx = 12(S+ + S−) =

3 0 0√3 0 2 0

0 2 0√

0 0√

∣∣∣∣∣∣∣∣−λ

√3 0 0√

3 −λ 2 0

0 2 −λ√

0 0√

3 −λ

∣∣∣∣∣∣∣∣ = −λ

∣∣∣∣∣∣−λ 2 0

2 −λ√

3 −λ

∣∣∣∣∣∣−√3

∣∣∣∣∣∣√

0 −λ√

3 −λ

∣∣∣∣∣∣= −λ[−λ(λ2 − 3)− 2(−2λ)]−

√3[√

3(λ2 − 3)]

= −λ[−λ3 + 7λ]−[3λ2 − 9]

= λ4 − 10λ2 + 9

0 = λ4 − 10λ2 + 9 = (λ2 − 9)(λ2 − 1) = (λ+ 3)(λ− 3)(λ+ 1)(λ− 1)

12~,−

Sx = 12(S+ + S−) =

3 0 0√3 0 2 0

0 2 0√

0 0√

∣∣∣∣∣∣∣∣−λ

√3 0 0√

3 −λ 2 0

0 2 −λ√

0 0√

3 −λ

∣∣∣∣∣∣∣∣ = −λ

∣∣∣∣∣∣−λ 2 0

2 −λ√

3 −λ

∣∣∣∣∣∣−√3

∣∣∣∣∣∣√

0 −λ√

3 −λ

∣∣∣∣∣∣= −λ[−λ(λ2 − 3)− 2(−2λ)]−

√3[√

3(λ2 − 3)]

= −λ[−λ3 + 7λ]−[3λ2 − 9] = λ4 − 10λ2 + 9

0 = λ4 − 10λ2 + 9 = (λ2 − 9)(λ2 − 1) = (λ+ 3)(λ− 3)(λ+ 1)(λ− 1)

12~,−

Sx = 12(S+ + S−) =

3 0 0√3 0 2 0

0 2 0√

0 0√

∣∣∣∣∣∣∣∣−λ

√3 0 0√

3 −λ 2 0

0 2 −λ√

0 0√

3 −λ

∣∣∣∣∣∣∣∣ = −λ

∣∣∣∣∣∣−λ 2 0

2 −λ√

3 −λ

∣∣∣∣∣∣−√3

∣∣∣∣∣∣√

0 −λ√

3 −λ

∣∣∣∣∣∣= −λ[−λ(λ2 − 3)− 2(−2λ)]−

√3[√

3(λ2 − 3)]

= −λ[−λ3 + 7λ]−[3λ2 − 9] = λ4 − 10λ2 + 9

0 = λ4 − 10λ2 + 9

= (λ2 − 9)(λ2 − 1) = (λ+ 3)(λ− 3)(λ+ 1)(λ− 1)

12~,−

Sx = 12(S+ + S−) =

3 0 0√3 0 2 0

0 2 0√

0 0√

∣∣∣∣∣∣∣∣−λ

√3 0 0√

3 −λ 2 0

0 2 −λ√

0 0√

3 −λ

∣∣∣∣∣∣∣∣ = −λ

∣∣∣∣∣∣−λ 2 0

2 −λ√

3 −λ

∣∣∣∣∣∣−√3

∣∣∣∣∣∣√

0 −λ√

3 −λ

∣∣∣∣∣∣= −λ[−λ(λ2 − 3)− 2(−2λ)]−

√3[√

3(λ2 − 3)]

= −λ[−λ3 + 7λ]−[3λ2 − 9] = λ4 − 10λ2 + 9

0 = λ4 − 10λ2 + 9 = (λ2 − 9)(λ2 − 1)

= (λ+ 3)(λ− 3)(λ+ 1)(λ− 1)

12~,−

Sx = 12(S+ + S−) =

3 0 0√3 0 2 0

0 2 0√

0 0√

∣∣∣∣∣∣∣∣−λ

√3 0 0√

3 −λ 2 0

0 2 −λ√

0 0√

3 −λ

∣∣∣∣∣∣∣∣ = −λ

∣∣∣∣∣∣−λ 2 0

2 −λ√

3 −λ

∣∣∣∣∣∣−√3

∣∣∣∣∣∣√

0 −λ√

3 −λ

∣∣∣∣∣∣= −λ[−λ(λ2 − 3)− 2(−2λ)]−

√3[√

3(λ2 − 3)]

= −λ[−λ3 + 7λ]−[3λ2 − 9] = λ4 − 10λ2 + 9

0 = λ4 − 10λ2 + 9 = (λ2 − 9)(λ2 − 1) = (λ+ 3)(λ− 3)(λ+ 1)(λ− 1)

12~,−

Sx = 12(S+ + S−) =

3 0 0√3 0 2 0

0 2 0√

0 0√

∣∣∣∣∣∣∣∣−λ

√3 0 0√

3 −λ 2 0

0 2 −λ√

0 0√

3 −λ

∣∣∣∣∣∣∣∣ = −λ

∣∣∣∣∣∣−λ 2 0

2 −λ√

3 −λ

∣∣∣∣∣∣−√3

∣∣∣∣∣∣√

0 −λ√

3 −λ

∣∣∣∣∣∣= −λ[−λ(λ2 − 3)− 2(−2λ)]−

√3[√

3(λ2 − 3)]

= −λ[−λ3 + 7λ]−[3λ2 − 9] = λ4 − 10λ2 + 9

0 = λ4 − 10λ2 + 9 = (λ2 − 9)(λ2 − 1) = (λ+ 3)(λ− 3)(λ+ 1)(λ− 1)

12~,−

Problem 4.52 (addendum)

We can now compute the eigenvectors of Sx in the Sz basis set

startingwith the eigenvalue λ = 3

Sx |α〉

3 0 0√3 0 2 0

0 2 0√

0 0√

= λ|α〉

this gives four equations

√3b = 3a → b =

√3a + 2c = 3b →

√3a + 2

√3d = 3

2b +√

3d = 3c → 2√

3a +√

3d = 3√

3c = 3d → c =√

|α〉 =1√8

1√3√3

and so with all foureigenvalues

We can now compute the eigenvectors of Sx in the Sz basis set

startingwith the eigenvalue λ = 3

Sx |α〉 =~2

3 0 0√3 0 2 0

0 2 0√

0 0√

= λ|α〉

√3b = 3a → b =

√3a + 2c = 3b →

√3a + 2

√3d = 3

2b +√

3d = 3c → 2√

3a +√

3d = 3√

3c = 3d → c =√

|α〉 =1√8

1√3√3

We can now compute the eigenvectors of Sx in the Sz basis set startingwith the eigenvalue λ = 3

Sx |α〉 =~2

3 0 0√3 0 2 0

0 2 0√

0 0√

= λ|α〉

√3b = 3a → b =

√3a + 2c = 3b →

√3a + 2

√3d = 3

2b +√

3d = 3c → 2√

3a +√

3d = 3√

3c = 3d → c =√

|α〉 =1√8

1√3√3

Sx |α〉 =~2

3 0 0√3 0 2 0

0 2 0√

0 0√

= λ|α〉

√3b = 3a → b =

√3a + 2c = 3b →

√3a + 2

√3d = 3

2b +√

3d = 3c → 2√

3a +√

3d = 3√

3c = 3d → c =√

|α〉 =1√8

1√3√3

Sx |α〉 =~2

3 0 0√3 0 2 0

0 2 0√

0 0√

= λ|α〉

√3b = 3a → b =

√3a + 2c = 3b →

√3a + 2

√3d = 3

2b +√

3d = 3c → 2√

3a +√

3d = 3√

3c = 3d → c =√

|α〉 =1√8

1√3√3

Sx |α〉 =~2

3 0 0√3 0 2 0

0 2 0√

0 0√

= λ|α〉

√3b = 3a

→ b =√

3a + 2c = 3b →√

3a + 2√

3d = 3√

2b +√

3d = 3c → 2√

3a +√

3d = 3√

3c = 3d → c =√

|α〉 =1√8

1√3√3

Sx |α〉 =~2

3 0 0√3 0 2 0

0 2 0√

0 0√

= λ|α〉

√3b = 3a

→ b =√

√3a + 2c = 3b

→√

3a + 2√

3d = 3√

2b +√

3d = 3c → 2√

3a +√

3d = 3√

3c = 3d → c =√

|α〉 =1√8

1√3√3

Sx |α〉 =~2

3 0 0√3 0 2 0

0 2 0√

0 0√

= λ|α〉

√3b = 3a

→ b =√

√3a + 2c = 3b

→√

3a + 2√

3d = 3√

2b +√

3d = 3c

→ 2√

3a +√

3d = 3√

3c = 3d → c =√

|α〉 =1√8

1√3√3

Sx |α〉 =~2

3 0 0√3 0 2 0

0 2 0√

0 0√

= λ|α〉

√3b = 3a

→ b =√

√3a + 2c = 3b

→√

3a + 2√

3d = 3√

2b +√

3d = 3c

→ 2√

3a +√

3d = 3√

√3c = 3d

→ c =√

|α〉 =1√8

1√3√3

Sx |α〉 =~2

3 0 0√3 0 2 0

0 2 0√

0 0√

= λ|α〉

√3b = 3a → b =

√3a + 2c = 3b

→√

3a + 2√

3d = 3√

2b +√

3d = 3c

→ 2√

3a +√

3d = 3√

√3c = 3d

→ c =√

|α〉 =1√8

1√3√3

Sx |α〉 =~2

3 0 0√3 0 2 0

0 2 0√

0 0√

= λ|α〉

√3b = 3a → b =

√3a + 2c = 3b →

√3a + 2

√3d = 3

2b +√

3d = 3c

→ 2√

3a +√

3d = 3√

√3c = 3d → c =

|α〉 =1√8

1√3√3

Sx |α〉 =~2

3 0 0√3 0 2 0

0 2 0√

0 0√

= λ|α〉

√3b = 3a → b =

√3a + 2c = 3b →

√3a + 2

√3d = 3

2b +√

3d = 3c → 2√

3a +√

3d = 3√

3c = 3d → c =√

|α〉 =1√8

1√3√3

Sx |α〉 =~2

3 0 0√3 0 2 0

0 2 0√

0 0√

= λ|α〉

√3b = 3a → b =

√3a + 2c = 3b →

√3a + 2

√3d = 3

2b +√

3d = 3c → 2√

3a +√

3d = 3√

3c = 3d → c =√

|α〉 =1√8

1√3√3

Sx |α〉 =~2

3 0 0√3 0 2 0

0 2 0√

0 0√

= λ|α〉

√3b = 3a → b =

√3a + 2c = 3b →

√3a + 2

√3d = 3

2b +√

3d = 3c → 2√

3a +√

3d = 3√

3c = 3d → c =√

|α〉 =1√8

1√3√3

Sx |α〉 =~2

3 0 0√3 0 2 0

0 2 0√

0 0√

= λ|α〉

√3b = 3a → b =

√3a + 2c = 3b →

√3a + 2

√3d = 3

2b +√

3d = 3c → 2√

3a +√

3d = 3√

3c = 3d → c =√

|α〉 =1√8

1√3√3

Two-particle systems

For a two-particle system, we havea wave function

and Schrodinger equation

Ψ(~r1,~r2, t)

i~∂Ψ

∂t= HΨ

H = − ~2

2m1∇2

1 −~2

2m2∇2

2 + V (~r1,~r2, t)

for the time independent wave function, ψ(~r1,~r2), the wave equationbecomes

− ~2

2m1∇2

1ψ −~2

2m2∇2

2ψ + Vψ = Eψ

Ψ(~r1,~r2, t)

i~∂Ψ

∂t= HΨ

H = − ~2

2m1∇2

1 −~2

2m2∇2

2 + V (~r1,~r2, t)

− ~2

2m1∇2

1ψ −~2

2m2∇2

2ψ + Vψ = Eψ

Ψ(~r1,~r2, t)

i~∂Ψ

∂t= HΨ

H = − ~2

2m1∇2

1 −~2

2m2∇2

2 + V (~r1,~r2, t)

− ~2

2m1∇2

1ψ −~2

2m2∇2

2ψ + Vψ = Eψ

Ψ(~r1,~r2, t)

i~∂Ψ

∂t= HΨ

H = − ~2

2m1∇2

1 −~2

2m2∇2

2 + V (~r1,~r2, t)

− ~2

2m1∇2

1ψ −~2

2m2∇2

2ψ + Vψ = Eψ

Ψ(~r1,~r2, t)

i~∂Ψ

∂t= HΨ

H = − ~2

2m1∇2

1 −~2

2m2∇2

2 + V (~r1,~r2, t)

− ~2

2m1∇2

1ψ −~2

2m2∇2

2ψ + Vψ = Eψ

Ψ(~r1,~r2, t)

i~∂Ψ

∂t= HΨ

H = − ~2

2m1∇2

1 −~2

2m2∇2

2 + V (~r1,~r2, t)

− ~2

2m1∇2

1ψ −~2

2m2∇2

2ψ + Vψ = Eψ

Ψ(~r1,~r2, t)

i~∂Ψ

∂t= HΨ

H = − ~2

2m1∇2

1 −~2

2m2∇2

2 + V (~r1,~r2, t)

− ~2

2m1∇2

1ψ −~2

2m2∇2

2ψ + Vψ = Eψ

Bosons and Fermions

Suppose that particle 1 is in theone-particle state ψa while particle2 is in ψb. The full two-particlewavefunction is

ψ(~r1,~r2) = ψa(~r1)ψb(~r2)

but only if the particles can be dis-tinguished!

if the particles are identical, we need to have a wavefunction which doesnot choose between states for the two particles

ψ±(~r1,~r2) = A [ψa(~r1)ψb(~r2)± ψb(~r1)ψa(~r2)]

The two signs imply very different behavior and are characteristic of twodifferent types of particles, bosons (“+”) and fermions (“–”).

bosons have integer spinfermions have half-integer spin

Bosons and Fermions

ψ(~r1,~r2) = ψa(~r1)ψb(~r2)

Bosons and Fermions

ψ(~r1,~r2) = ψa(~r1)ψb(~r2)

Bosons and Fermions

ψ(~r1,~r2) = ψa(~r1)ψb(~r2)

Bosons and Fermions

ψ(~r1,~r2) = ψa(~r1)ψb(~r2)

Bosons and Fermions

ψ(~r1,~r2) = ψa(~r1)ψb(~r2)

Bosons and Fermions

ψ(~r1,~r2) = ψa(~r1)ψb(~r2)

Bosons and Fermions

ψ(~r1,~r2) = ψa(~r1)ψb(~r2)

bosons have integer spin

fermions have half-integer spin

Bosons and Fermions

ψ(~r1,~r2) = ψa(~r1)ψb(~r2)

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