Physics 472 Problem Set 5 Solutions Spring 2014
19. From the Clebsch-Gordon table, the state |2 1⟩|2−1⟩can be written as
|2 2⟩|2−2⟩=√1
70 |4 0⟩+√1
10 |3 0⟩+√2
7|2 0⟩+√2
5|1 0⟩ − √1
5|0 0⟩.
(a) Since the state is an eigenstate of Jz, measurement of Jzwill yield 0 with probability
1.
(b) The possible results of measuring
J2are: 20¯h2,12¯h2,6¯h2,2¯h2,0, with probabilities:
1
70 ,1
10 ,2
7,2
5,1
5.
20. Quarks and anti-quarks are not identical particles, so we may combine their spins without
worrying about symmetrization.
(a) The two spins combine to form either spin 1 or spin 0.
(b) The triplet and singlet states are
3χ(1,2) =
χ(1)
1
2
χ(2)
1
2
√1
2[χ(1)
1
2
χ(2)
−1
2
+χ(1)
−1
2
χ(2)
1
2
]
χ(1)
−1
2
χ(2)
−1
2
1χ(1,2) = √1
2[χ(1)
1
2
χ(2)
−1
2
−χ(1)
−1
2
χ(2)
1
2
].
To calculate the expectation value of
S1·
S2use
S=
S1+
S2to obtain
S1·
S2=1
2(
S2−
S2
1−
S2
2)=1
2(
S2−3
2¯h2).
Then, the triplet and singlet expectation values are
3χ†(1,2)V0
S1·
S23χ(1,2) = 1
2V03χ†(1,2) (2¯h2−3
2¯h2)3χ(1,2) = 1
4V0¯h2,
1χ†(1,2)V0
S1·
S21χ(1,2) = 1
2V01χ†(1,2) (0¯h2−3
2¯h2)1χ(1,2) = −3
4V0¯h2
21. Griffiths 5.4. Using Ψ±(r1, r2) of equation 5.10, the normalization integral is
∫d3r1∫d3r2|Ψ±(r1, r2)|2=|A|2∫d3r1∫d3r2|ψa(r1)ψb(r2)±ψa(r2)ψb(r1)|2
= 2|A|2[∫d3r1|ψa(r1)|2∫d3r2|ψb(r2)|2
±∫d3r1ψ∗
a(r1)ψb(r1)∫d3r2ψ∗
b(r2)ψa(r2)]
= 2|A|2[1 ±δab] = 1 .
For a=b,|A|2=1
2and for a=b, only the + term survives giving |A|2=1
4.
