# Possible outcomes of measuring $L^2$ and $L_z$?

1. Oct 17, 2016

### Kara386

1. The problem statement, all variables and given/known data
A particle has the wavefunction

$\psi(\theta, \phi) = \sqrt{\frac{5}{2\pi}}\sin(\theta)cos(\frac{\theta}{2})^2\cos(\phi)$

What are the possible outcomes of measuring $L^2$ and $L_z$? And the relative probabilities of each outcome?
2. Relevant equations

3. The attempt at a solution
I think I should try and write the wavefunction as a combination of spherical harmonics so I can find the eigenvalues more easily. Substituting in $cos(\frac{\theta}{2})^2 = \frac{1}{2} +\frac{1}{2}\cos(\theta)$,

$\psi(\theta, \phi) = \sqrt{\frac{5}{8\pi}}\sin(\theta)\cos(\phi)+\sqrt{\frac{5}{8\pi}}\cos(\theta)cos(\phi)$

I found that $\psi(\theta, \phi) = \sqrt{\frac{5}{12\pi}}(Y_{1,-1}-Y_{1,1}) + \sqrt{\frac{1}{12}}(Y_{2,-1}-Y_{2,1})$. Now I don't really know what to do with that! I know that the coefficient of each spherical harmonic squared will be the probability of that particular outcome. And if you square every coefficient and add them then they do add to 1.

The possible outcomes are the eigenvalues of the operator in question, which for $L^2$ are $\hbar^2 l(l+1)$ and for $L_z$ are $\hbar m$. So what I don't understand is, every spherical harmonic will have an eigenvalue of $L^2$ and one for $L_z$, but when I square the coefficients am I finding the probability of the $L^2$ outcome, the $L_z$ outcome or both?

2. Oct 17, 2016

### Staff: Mentor

What you found is that, for instance, the probability of being in state $Y_{1,-1}$ is 5/12. That state will lead to defined values of $L^2$ and $L_z$. You need to go through all states the same way, and figure out the probability of individual outcomes, and you must then combine them to find, say, the total probability of measuring $L = 2 \hbar^2$.