Today’s Outline - November 02, 2015phys.iit.edu/~segre/phys405/15F/lecture_18.pdf · Today’s...

Today’s Outline - November 02, 2015

• Generating the Clebsch-Gordan coefficients

• Chapter 4 problems

Midterm Exam 2: Monday, November 16, 2015 covers through Chapter4.4.2 (at least!)

Homework Assignment #08:Chapter 4: 10,13,14,15,16,38due Wednesday, November 04, 2015

C. Segre (IIT) PHYS 405 - Fall 2015 November 02, 2015 1 / 17

Deriving the Clebsch-Gordan coefficients

Deriving the Clebsch-Gordan coefficients is non-trivial and there are anumber of ways to do it in the literature.

We will start with the generalderivation, then apply 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. We will start with the generalderivation, then apply 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 different values of s are generated using the unused combinations ofthe 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 is ob-tained

|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 4.8

(a) Check that u = Arj1(kr) satisfies the radial equation with V (r) = 0and l = 1

(b) Determine graphically the allowed energies for the infinite sphericalwell when l = 1. Show that for large n

En1 ≈~2π2

2ma2(n + 1

With V = 0 and l = 1 the radialequation which needs to be verifiedis

with u being

dr2− 2

r2u = −k2u

u = Ar

[sin(kr)

(kr)2− cos(kr)

[sin(kr)

kr− cos(kr)

Problem 4.8

En1 ≈~2π2

2ma2(n + 1

with u being

dr2− 2

r2u = −k2u

u = Ar

[sin(kr)

(kr)2− cos(kr)

[sin(kr)

kr− cos(kr)

Problem 4.8

En1 ≈~2π2

2ma2(n + 1

with u being

dr2− 2

r2u = −k2u

u = Ar

[sin(kr)

(kr)2− cos(kr)

[sin(kr)

kr− cos(kr)

Problem 4.8

En1 ≈~2π2

2ma2(n + 1

with u being

dr2− 2

r2u = −k2u

u = Ar

[sin(kr)

(kr)2− cos(kr)

[sin(kr)

kr− cos(kr)

Problem 4.8

En1 ≈~2π2

2ma2(n + 1

with u being

dr2− 2

r2u = −k2u

u = Ar

[sin(kr)

(kr)2− cos(kr)

[sin(kr)

kr− cos(kr)

Problem 4.8

En1 ≈~2π2

2ma2(n + 1

with u being

dr2− 2

r2u = −k2u

u = Ar

[sin(kr)

(kr)2− cos(kr)

[sin(kr)

kr− cos(kr)

]C. Segre (IIT) PHYS 405 - Fall 2015 November 02, 2015 9 / 17

Problem 4.8 (cont.)

[sin(kr)

(kr)− cos(kr)

[k2r cos(kr)− k sin(kr)

+ k sin(kr)

[cos(kr)

kr− sin(kr)

(kr)2+ sin(kr)

dr2= A

−k2r sin(kr)− k cos(kr)

(kr)2− k3r2 cos(kr)− 2k2r sin(kr)

place2

holder+ k cos(kr)

− sin(kr)

(kr)− cos(kr)

(kr)2− cos(kr)

(kr)2+ 2

sin(kr)

(kr)3+ cos(kr)

(1− 2

)cos(kr) +

(kr)3− 1

)sin(kr)

r2− k2

[sin(kr)

(kr)− cos(kr)

Problem 4.8 (cont.)

[sin(kr)

(kr)− cos(kr)

+ k sin(kr)

[cos(kr)

kr− sin(kr)

(kr)2+ sin(kr)

dr2= A

place2

holder+ k cos(kr)

− sin(kr)

(kr)− cos(kr)

(kr)2− cos(kr)

(kr)2+ 2

sin(kr)

(kr)3+ cos(kr)

(1− 2

)cos(kr) +

(kr)3− 1

)sin(kr)

r2− k2

[sin(kr)

(kr)− cos(kr)

Problem 4.8 (cont.)

[sin(kr)

(kr)− cos(kr)

(kr)2+ k sin(kr)

[cos(kr)

kr− sin(kr)

(kr)2+ sin(kr)

dr2= A

place2

holder+ k cos(kr)

− sin(kr)

(kr)− cos(kr)

(kr)2− cos(kr)

(kr)2+ 2

sin(kr)

(kr)3+ cos(kr)

(1− 2

)cos(kr) +

(kr)3− 1

)sin(kr)

r2− k2

[sin(kr)

(kr)− cos(kr)

Problem 4.8 (cont.)

[sin(kr)

(kr)− cos(kr)

(kr)2+ k sin(kr)

[cos(kr)

kr− sin(kr)

(kr)2+ sin(kr)

dr2= A

place2

holder+ k cos(kr)

− sin(kr)

(kr)− cos(kr)

(kr)2− cos(kr)

(kr)2+ 2

sin(kr)

(kr)3+ cos(kr)

(1− 2

)cos(kr) +

(kr)3− 1

)sin(kr)

r2− k2

[sin(kr)

(kr)− cos(kr)

Problem 4.8 (cont.)

[sin(kr)

(kr)− cos(kr)

(kr)2+ k sin(kr)

[cos(kr)

kr− sin(kr)

(kr)2+ sin(kr)

dr2= A

place2

holder+ k cos(kr)

− sin(kr)

(kr)− cos(kr)

(kr)2− cos(kr)

(kr)2+ 2

sin(kr)

(kr)3+ cos(kr)

(1− 2

)cos(kr) +

(kr)3− 1

)sin(kr)

r2− k2

[sin(kr)

(kr)− cos(kr)

Problem 4.8 (cont.)

[sin(kr)

(kr)− cos(kr)

(kr)2+ k sin(kr)

[cos(kr)

kr− sin(kr)

(kr)2+ sin(kr)

dr2= A

[−k2r sin(kr)− k cos(kr)

− k3r2 cos(kr)− 2k2r sin(kr)

place2

holder+ k cos(kr)

− sin(kr)

(kr)− cos(kr)

(kr)2− cos(kr)

(kr)2+ 2

sin(kr)

(kr)3+ cos(kr)

(1− 2

)cos(kr) +

(kr)3− 1

)sin(kr)

r2− k2

[sin(kr)

(kr)− cos(kr)

Problem 4.8 (cont.)

[sin(kr)

(kr)− cos(kr)

(kr)2+ k sin(kr)

[cos(kr)

kr− sin(kr)

(kr)2+ sin(kr)

dr2= A

place2

holder+ k cos(kr)

− sin(kr)

(kr)− cos(kr)

(kr)2− cos(kr)

(kr)2+ 2

sin(kr)

(kr)3+ cos(kr)

(1− 2

)cos(kr) +

(kr)3− 1

)sin(kr)

r2− k2

[sin(kr)

(kr)− cos(kr)

Problem 4.8 (cont.)

[sin(kr)

(kr)− cos(kr)

(kr)2+ k sin(kr)

[cos(kr)

kr− sin(kr)

(kr)2+ sin(kr)

dr2= A

place2

holder

+ k cos(kr)

− sin(kr)

(kr)− cos(kr)

(kr)2− cos(kr)

(kr)2+ 2

sin(kr)

(kr)3+ cos(kr)

(1− 2

)cos(kr) +

(kr)3− 1

)sin(kr)

r2− k2

[sin(kr)

(kr)− cos(kr)

Problem 4.8 (cont.)

[sin(kr)

(kr)− cos(kr)

(kr)2+ k sin(kr)

[cos(kr)

kr− sin(kr)

(kr)2+ sin(kr)

dr2= A

place2

holder

+ k cos(kr)

− sin(kr)

(kr)− cos(kr)

(kr)2− cos(kr)

(kr)2+ 2

sin(kr)

(kr)3+ cos(kr)

(1− 2

)cos(kr) +

(kr)3− 1

)sin(kr)

r2− k2

[sin(kr)

(kr)− cos(kr)

Problem 4.8 (cont.)

[sin(kr)

(kr)− cos(kr)

(kr)2+ k sin(kr)

[cos(kr)

kr− sin(kr)

(kr)2+ sin(kr)

dr2= A

place2

holder

+ k cos(kr)

[− sin(kr)

(kr)− cos(kr)

− cos(kr)

(kr)2+ 2

sin(kr)

(kr)3+ cos(kr)

(1− 2

)cos(kr) +

(kr)3− 1

)sin(kr)

r2− k2

[sin(kr)

(kr)− cos(kr)

Problem 4.8 (cont.)

[sin(kr)

(kr)− cos(kr)

(kr)2+ k sin(kr)

[cos(kr)

kr− sin(kr)

(kr)2+ sin(kr)

dr2= A

place2

holder

+ k cos(kr)

[− sin(kr)

(kr)− cos(kr)

(kr)2− cos(kr)

(kr)2+ 2

sin(kr)

+ cos(kr)

(1− 2

)cos(kr) +

(kr)3− 1

)sin(kr)

r2− k2

[sin(kr)

(kr)− cos(kr)

Problem 4.8 (cont.)

[sin(kr)

(kr)− cos(kr)

(kr)2+ k sin(kr)

[cos(kr)

kr− sin(kr)

(kr)2+ sin(kr)

dr2= A

place2

holder

+ k cos(kr)

[− sin(kr)

(kr)− cos(kr)

(kr)2− cos(kr)

(kr)2+ 2

sin(kr)

(kr)3+ cos(kr)

(1− 2

)cos(kr) +

(kr)3− 1

)sin(kr)

r2− k2

[sin(kr)

(kr)− cos(kr)

Problem 4.8 (cont.)

[sin(kr)

(kr)− cos(kr)

(kr)2+ k sin(kr)

[cos(kr)

kr− sin(kr)

(kr)2+ sin(kr)

dr2= A

place2

holder

+ k cos(kr)

[− sin(kr)

(kr)− cos(kr)

(kr)2− cos(kr)

(kr)2+ 2

sin(kr)

(kr)3+ cos(kr)

(1− 2

)cos(kr) +

(kr)3− 1

)sin(kr)

r2− k2

[sin(kr)

(kr)− cos(kr)

Problem 4.8 (cont.)

[sin(kr)

(kr)− cos(kr)

(kr)2+ k sin(kr)

[cos(kr)

kr− sin(kr)

(kr)2+ sin(kr)

dr2= A

place2

holder

+ k cos(kr)

[− sin(kr)

(kr)− cos(kr)

(kr)2− cos(kr)

(kr)2+ 2

sin(kr)

(kr)3+ cos(kr)

[(1− 2

)cos(kr)

(kr)3− 1

)sin(kr)

r2− k2

[sin(kr)

(kr)− cos(kr)

Problem 4.8 (cont.)

[sin(kr)

(kr)− cos(kr)

(kr)2+ k sin(kr)

[cos(kr)

kr− sin(kr)

(kr)2+ sin(kr)

dr2= A

place2

holder

+ k cos(kr)

[− sin(kr)

(kr)− cos(kr)

(kr)2− cos(kr)

(kr)2+ 2

sin(kr)

(kr)3+ cos(kr)

[(1− 2

)cos(kr) +

(kr)3− 1

)sin(kr)

r2− k2

[sin(kr)

(kr)− cos(kr)

Problem 4.8 (cont.)

[sin(kr)

(kr)− cos(kr)

(kr)2+ k sin(kr)

[cos(kr)

kr− sin(kr)

(kr)2+ sin(kr)

dr2= A

place2

holder

+ k cos(kr)

[− sin(kr)

(kr)− cos(kr)

(kr)2− cos(kr)

(kr)2+ 2

sin(kr)

(kr)3+ cos(kr)

[(1− 2

)cos(kr) +

(kr)3− 1

)sin(kr)

r2− k2

[sin(kr)

(kr)− cos(kr)

Problem 4.8 (cont.)

[sin(kr)

(kr)− cos(kr)

(kr)2+ k sin(kr)

[cos(kr)

kr− sin(kr)

(kr)2+ sin(kr)

dr2= A

place2

holder

+ k cos(kr)

[− sin(kr)

(kr)− cos(kr)

(kr)2− cos(kr)

(kr)2+ 2

sin(kr)

(kr)3+ cos(kr)

[(1− 2

)cos(kr) +

(kr)3− 1

)sin(kr)

r2− k2

[sin(kr)

(kr)− cos(kr)

]C. Segre (IIT) PHYS 405 - Fall 2015 November 02, 2015 10 / 17

Problem 4.8 (cont.)

According to the boundary conditions im-posed, j1(z) = 0 when z = ka

rearranging, we have

j1 =sin z

z2− cos z

z = tan z

plotting these two func-tions, we can identify thecrossing points as the so-lutions

∼ 1.43π, ∼ 2.46π0 π/2 π 3π/2 2π 5π/2

as n becomes large, the solutions approach z = (n + 12)π and the energies

become

En =~2k2

2ma2≈ ~2π2

2ma2(n + 1

Problem 4.8 (cont.)

j1 =sin z

z2− cos z

z = tan z

∼ 1.43π, ∼ 2.46π0 π/2 π 3π/2 2π 5π/2

become

En =~2k2

2ma2≈ ~2π2

2ma2(n + 1

Problem 4.8 (cont.)

j1 =sin z

z2− cos z

z = tan z

∼ 1.43π, ∼ 2.46π0 π/2 π 3π/2 2π 5π/2

become

En =~2k2

2ma2≈ ~2π2

2ma2(n + 1

Problem 4.8 (cont.)

j1 =sin z

z2− cos z

z = tan z

∼ 1.43π, ∼ 2.46π0 π/2 π 3π/2 2π 5π/2

become

En =~2k2

2ma2≈ ~2π2

2ma2(n + 1

Problem 4.8 (cont.)

j1 =sin z

z2− cos z

z = tan z

∼ 1.43π, ∼ 2.46π0 π/2 π 3π/2 2π 5π/2

become

En =~2k2

2ma2≈ ~2π2

2ma2(n + 1

Problem 4.8 (cont.)

j1 =sin z

z2− cos z

z = tan z

∼ 1.43π, ∼ 2.46π

0 π/2 π 3π/2 2π 5π/2

become

En =~2k2

2ma2≈ ~2π2

2ma2(n + 1

Problem 4.8 (cont.)

j1 =sin z

z2− cos z

z = tan z

∼ 1.43π, ∼ 2.46π0 π/2 π 3π/2 2π 5π/2

become

En =~2k2

2ma2≈ ~2π2

2ma2(n + 1

Problem 4.8 (cont.)

j1 =sin z

z2− cos z

z = tan z

∼ 1.43π, ∼ 2.46π0 π/2 π 3π/2 2π 5π/2

become

En =~2k2

2ma2≈ ~2π2

2ma2(n + 1

Problem 4.8 (cont.)

j1 =sin z

z2− cos z

z = tan z

∼ 1.43π, ∼ 2.46π0 π/2 π 3π/2 2π 5π/2

become

En =~2k2

2ma2≈ ~2π2

2ma2(n + 1

Problem 4.8 (cont.)

j1 =sin z

z2− cos z

z = tan z

∼ 1.43π, ∼ 2.46π0 π/2 π 3π/2 2π 5π/2

become

En =~2k2

≈ ~2π2

2ma2(n + 1

Problem 4.8 (cont.)

j1 =sin z

z2− cos z

z = tan z

∼ 1.43π, ∼ 2.46π0 π/2 π 3π/2 2π 5π/2

become

En =~2k2

2ma2≈ ~2π2

2ma2(n + 1

)2C. Segre (IIT) PHYS 405 - Fall 2015 November 02, 2015 11 / 17

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 −

Problem 4.52 (cont.)

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 2015 November 02, 2015 13 / 17

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 2015 November 02, 2015 13 / 17

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 4.35

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?

(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

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