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Physics 325 Solution to Problem Set # 2
1. Problem 4.37 in your textbook.
(a) We must write the total spin state (coupled representation), |s, m;s1,s
2i=|3,1; 2,1i, in terms of the
uncoupled states, |s1,m
1i|s2,m
2i. To do this, we use Table 4.7 in the textbook and look for the table 2 ×1.
Across the top, we find the state 3,+1, and read off the corresponding uncoupled state:
|3,1; 1,2i=r1
15 |2,2i|1,−1i+r8
15 |2,1i|1,0i+r6
15 |2,0i|1,1i
From the expansion in the uncoupled representation we can read off that Szof the spin 2 particle can
have values 2¯h, 1¯h, 0 with the corresponding probabilties of 1/15,8/15,6/15 respectively.
(b) We are told that the electron is in the ψ510 state, ie l=1,m
l= 0 of hydrogen, with s=1/2 and
ms=−1/2. The uncoupled total angular momentum is then |1,0i|1/2,−1/2i. We need to express this
state in the coupled representation to determine the possible values for the total angular momentun. Again,
we use Table 4.7 and look for the table 1 ×1/2 and find the uncoupled state, 0,−1/2 along the left edge.
Reading across the table we find:
|1,0i|1/2,−1/2i=r2
3|3/2,−1/2; 1,1/2i+r1
3|1/2,−1/2; 1,1/2i
The total angular momentum will then be 3/2(3/2 + 1) = (15/4)¯h2or 1/2(1/2 + 1) = (3/4)¯h2with
probabilities 2/3 and 1/3 respectively.
2. Problem 4.48 in your textbook.
ψ(r,θ,φ,s,m
s)=R21(r)r1
3Y0
1(θ, φ)χ++r2
3Y1
1(θ, φ)χ−
a.) ψis already written as a linear combination of eigenstates of L2:Ym
lare eigenstates of L2with eigenvalue
l(l+ 1)¯h2. Therefore you will measure 2¯h2with a probability of 1.
b) Ym
lare eigenstates of Lzwith eigenvalue m¯h. Therefore you will measure 0 with a probability of 1/3
(the modulus squared of the component of ψthat contains Y0
1) and you will measure ¯hwith a probability
of 2/3.
c) χ±are both eigenstates of S2with eigenvalue ¯h2(1/2)(1 + 1/2) = 3¯h2/4. Therefore you will measure
3¯h2/4 with a probability of 1.
d) χ±are eigenstates of Szwith eigenvalues ±¯h/2. Therefore you will measure ¯h/2 with a probability of
1/3 and −¯h/2 with a probability of 2/3.
e) We need to write ψin terms of the eigenstates of J2and Jz,|jm
ji. From Table 4.7 in the text, we find
that:
Y0
1χ+=r2
3
j=3
2mj=1
2−r1
3
j=1
2mj=1
2
Y1
1χ−=r1
3
j=3
2mj=1
2+r2
3
j=1
2mj=1
2
Therefore r1
3Y0
1χ++r2
3Y1
1χ−=2√2
3
j=3
2mj=1
2+1
3
j=1
2mj=1
2
