# [QM] spin and orbital angular momentum: most efficient method?

1. Jan 24, 2012

### nonequilibrium

1. The problem statement, all variables and given/known data
Consider a spin-1/2 particle whose state is $|\psi \rangle = \psi_+(\textbf r) |+\rangle + \psi_-(\textbf r) |- \rangle$. Let $\hat{ \textbf S }$ be the spin observable and $\hat{ \textbf L }$ the orbital angular momentum. We assume that
$$\psi_+(\textbf r) = R(r) \left( Y_{0,0}(\theta,\phi) + \frac{1}{\sqrt 3} Y_{1,0}(\theta, \phi) \right),$$
$$\psi_-(\textbf r) = \frac{R(r)}{\sqrt 3} \left( Y_{1,1}(\theta,\phi) - Y_{1,0}(\theta, \phi) \right).$$
a. What is the normalization condition on R(r)?
b. What are the possible results of a measurement of $L_z$? Give the corresponding probabilities.

2. Relevant equations
N/A

3. The attempt at a solution
a. is easy enough, the answer is $\langle R | R \rangle = \int_0^{+ \infty} R(r) r^2 \mathrm d r = \frac{1}{2}$.

For b. I think I can solve it, but I want to know whether or not I'm using the most efficient method or even thinking process.
At first I doubted it a bit. I could find the correct answer (the numerical results are in the back of my book) myself (answer: m=0 with probability 5/6; m=1 with probability 1/6), but my reasoning was a bit hand-wavy, and I don't want that.
So when in doubt, I returned to the definition: I need to project it and then calculate its modulus. It seems to me that the projector for a given m is $\hat P_m = \sum_{n=|m|+1}^{+\infty} \sum_{l=|m|}^{+\infty} |n,l,m,+ \rangle \langle n,l,m,+| + |n,l,m,- \rangle \langle n,l,m,-|$
correct?
Let's take the case of m=0.
Looking at this projector and to the specific form of psi+ and psi- it is not hard to see that $\hat P_0 | \psi \rangle = \psi_+ |+ \rangle - \frac{R(r)}{\sqrt 3} Y_{1,0}(\theta,\phi)|-\rangle$ and calculating the norm of this is easy (using normalization conditions for the spherical harmonic functions). This gives the correct answer 5/6.

First of all: do you agree?
Second of all: is there a conceptually and/or technically easier method, that doesn't involve projectors explicitly (but which still isn't hand-wavy)? Or do you think that this way of solving it is good?