Angular momentum for a given Wavefunction Problem: Give a particle in the state $\Psi = e^{\frac{-(x^{2}+y^{2}+z^{2})}{a^{2}}}(\frac{x}{a} + \frac{yz}{a^{2}})$, what are the allowed values for $l_{x}$ (and later for $l_{y}, l_{z}$).
Attempt: Using the eigenvalue equation $L_{x} \Psi = l_{x} \Psi$, where $L_{x}$ is the angular momentum operator, $y p_{z} - z p_{y}$, and $l_{x}$ is the eigenvalue.
After substituting the for the momentum operators and calculating out, I get $L_{x} \Psi =  \frac{i \hbar}{a^{2}}(z^{2} - y^{2})e^{\frac{-(x^{2}+y^{2}+z^{2})}{a^{2}}}$ which doesn't seem to apply here.
I was then thinking of looking at spherical harmonics. The $(\frac{x}{a} + \frac{yz}{a^{2}})$ term can be expressed as the linear combination $\alpha \frac{r}{a} (Y_{1}^{-1} - Y_{1}^{1}) + i \beta \frac{r^{2}}{a^{2}}(Y_{2}^{1} + Y_{2}^{-1})$, where $\alpha = \sqrt{\frac{2 \pi}{3}}$ and $\beta = 2 \sqrt{\frac{\pi}{30}}$. And I'm using the spherical harmonics about the z-axis for simplicity, as I will have to do all of them eventually. However, I don't like the $r$ dependence of this, as I can only factor out one factor of $r$. Additionally, if I wanted to find the probability for a state it would depend on $r$. Is there a way to get separate the $r$ dependence, or are the probabilities just dependent on $r$?
 A: Let's look at a similar problem. Let $\left|nlm\right\rangle$ be an eigenvector of the hydrogen energy operator $\hat H,$ the angular momentum operator $\hat{L^2},$ and the angular momentum operator $\hat L_z$ (with eigenvalue $\hbar m$).
Now, what if you had a state like $\frac{\sqrt 1}{3}\left|753\right\rangle+\frac{\sqrt 3}{3}\left|752\right\rangle+\frac{\sqrt 5}{3}\left|751\right\rangle$ then the allowed eigenvalues of $\hat L_z$ are clearly $\hbar,$ $2\hbar,$ and $3\hbar.$  But it was only so clear and obvious because I wrote the state as $\frac{\sqrt 1}{3}\left|753\right\rangle+\frac{\sqrt 3}{3}\left|752\right\rangle+\frac{\sqrt 5}{3}\left|751\right\rangle$
What if it was just written as a function like $\Psi(x,y,z)$? Then we could still compute $\frac{\sqrt 1}{3}=\left\langle 753\right|\left.\Psi\right\rangle,$ $\frac{\sqrt 3 }{3}=\left\langle 752\right|\left.\Psi\right\rangle,$ and $\frac{\sqrt 5}{3}=\left\langle 751\right|\left.\Psi\right\rangle$. And once we have those, the fact that the squares add up to one tell us that $\Psi=\frac{\sqrt 1}{3}\left|753\right\rangle+\frac{\sqrt 3}{3}\left|752\right\rangle+\frac{\sqrt 5}{3}\left|751\right\rangle$.
So does this apply to your situation? I can look at it and see it is a gaussian times a second order polynomial in r so while it is not a linear combination of the first three energy levels we don't actually care about the radial part. Just taking the angular parts are enough. So while there are not a finite number of states to check against. You just need the ones that give you the simple combinations of sines and cosines in the angular variables.
Effectively, any nice enough angular function can be written as a linear combination of the different angular parts of the hydrogen eigenfunctions.
Since this is a homework like problem, you should be able to do the rest yourself. But if you get stuck you can do the same angular integrals by just choosing the polar axis to point in whichever directions you want. So just knowing the angular form for $\hat L_z$ is fine if you permute the $x,$ $y,$ and $z$ in your expression. But I'd do $\hat L_z$ first and make sure that's OK.
A: You given wavefunction here is not an eigenfunction of $L_x$. So you can not use eigenvalue equation here. What you can do is to calculate the expectation value of $L_x$ which is $\int d^3r\,\Phi^* L_x \Phi$. And I think it will give you zero(it can be seen that you integrand it odd with respect to x,y,z and integral of an odd function is 0)
