Solutions to extra problems associated with Clebsch-Gordon tables. (Don't bother looking at these solutions without really trying the problems first - at least, I claim it won't do you any good!)

i) ``p orbit" means l=1. Adding l=1 and l=1 to get ${\vert{ 1,-1}\rangle }$ means we have to go to the C-G 1x1 table, and read down the ${\vert{ 1,-1}\rangle }$column.

$${\vert{ 1,-1}\rangle }\ =\ {1\over\sqrt{2}} {\vert{ 1,0}\rang... ... {1\over\sqrt2}{\vert{ 1, -1}\rangle }{\vert{ 1, 0}\rangle }$$ (1)

So there's a 50% chance of m₁=0, and a 50% chance of m₁=-1.

ii) We're adding l=1 and l=1 and l=1. The first pair of them will add up to something we might call the ``total angular momentum of the first two", which in turn we might label l_{1 2}. There are three possibilities for l_{1 2}. It is 1+1 = 0, 1, or 2. Now, the grand total is the quantum sum of l_{1 2} and the remaining, third angular momentum, namely 1.

If l_{1 2} is 0, you get l_tot= l₁₂ +1= 0+1 = 1.

If l_{1 2} is 1, you get l_tot = l₁₂+1 = 1 + 1 = 0, 1 or 2.

If l_{1 2} is 2, you get l_tot = l₁₂+1 = 2 + 1 = 1, 2, or 3.

Putting all this together, I see that we can get a final, total angular momentum of 0, 1, 2, or 3. (1 can be built in several ways, 2 can be built two different ways, 0 and 3 can each be built only one way)

iii) Spin 3 has 2*3+1=7 possible states. Spin 1 has 1*1+1=3 states. The combined system can have 7*3=21 possible basis states.

iv) 1+2 = 1, 2, or 3. a) We are given the total state is ${\vert{ 1,-1}\rangle }$. So go to the 2x1 C-G table, and read down the ${\vert{ 1,-1}\rangle }$column. This gives the answers directly. If you're measuring the z-component of the spin two particle, focus on the first entry in the row names:

3/5 chance of +2, 3/10 chance of +1, and 1/10 chance of +0.

iv-b) Now we're given the microscopic state, so we must read across. We're still in the 2x1 table, and we read acros the row corresponding to ${\vert{ 2 -1}\rangle }{\vert{ 1 -1}\rangle }$ (which of course in the table is simply given by the m components, i.e. it's labelled -1 -1. Reading across, I see our state is

$${\vert{ 2 -1}\rangle }{\vert{ 1 -1}\rangle } = \sqrt{2\over3} {\vert{ 3,-2}\rangle } + \sqrt{1\over3}{\vert{ 2,-2}\rangle }.$$ (2)

So, there is a 2/3 chance of S=3 (or, $S^2=3*4=12\hbar^2$), and a 1/3 chance of S=2 (or, $S^2 = 2*3=6\hbar^2$.) Both cases have S_z=-2$\hbar$, so there's a 100% chance of that.

By the way, the order doesn't matter. S² and S_z commute, so measuring one won't mess up the other.

v) In all these questions, we will use S_tot=S₁+S₂, which means

$${{S_1\cdot S_2}\over{\hbar^2}} = \left( S_{tot}^2-S_1^2-S_2^2\right)/(2\hbar^2)$$ (3)

a) We're adding spin 1 (So $S_1^2=1*2\hbar^2$) and spin 1/2 (So S₂²= 3/4$\hbar^2$). S_tot can clearly be 3/2 or 1/2 here. We're told we're getting S_tot=1/2, and m_tot=1/2, so

$${{S_1\cdot S_2}\over{\hbar^2}} \ \left( S_{tot}^2-S_1^2-S_2^2\right) /(2\hbar^2)$$ = (4)

$$\ (1/2*3/2-1/2*3/2-1*2)/2 = -1$$ = (5)

This gives us an energy of H = A + B(-1)+C*m_tot = A-B+C/2.

b) Now S₁=S₂=1/2, so S_tot = 0 or 1. We're told S_tot=1, and m_tot=0. So

$${{S_1\cdot S_2}\over{\hbar^2}} \ \left( S_{tot}^2-S_1^2-S_2^2\right) /(2\hbar^2)$$ = (6)

$$\ (1*2-1/2*3/2 - 1/2*3/2)/2 = 1/4$$ = (7)

This gives us an energy of H = A + B/4 +C*m_tot = A+B/4.

c) If the particles are identical and spin 1/2, Pauli says they must be in an antisymmetric state. Since spin is all we have to work with here, they must be in the antisymmetric spin state, which is the S_tot=0 one. (And of course, if S=0, m_tot=0 is required as well) So we have

$${{S_1\cdot S_2}\over{\hbar^2}} \ \left( S_{tot}^2-S_1^2-S_2^2\right) /(2\hbar^2)$$ = (8)

$$\ (0*1-1/2*3/2 - 1/2*3/2)/2 = -3/4$$ = (9)

This gives us an energy of H = A - 3B/4 +C*m_tot = A-3B/4.