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Suppose that $$ \psi = \frac12 Y_{00}+\frac1{\sqrt 3}Y_{11}+\frac 12 Y_{1,-1}+\frac1{\sqrt6}Y_{22}.$$ This wave function is not an eigenstate of $\hat{L}_z$. If a measurement of the $z$-component of the orbital angular momentum is carried out, what is the probability to find $0$?

I would try and look at $|\langle l,0| \psi\rangle|^2$, but as there are three different $l$-values now, maybe $|\sum_l \langle l,0|\psi\rangle|^2$ is more appropriate, correct? But eventually this reduces to only one term with $l=0$, so the probability would be $1/2$. Is this problem just this straightforward, or am I missing possible interferences (f.ex. that $|1,1\rangle+|1,-1\rangle$ has total $z$-projection $0$ as well?).

Suppose that now $\psi=\frac12 Y_{00}+\frac1{\sqrt 3}Y_{\mathbf{10}}+\frac 12 Y_{1,-1}+\frac1{\sqrt6}Y_{22}$. Then $Pr(L_z=0)=1/4+1/3$, right?

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    $\begingroup$ You probably mean $\sum_l |\langle l ,0 | \psi \rangle |^2$ then, as opposed to $| \sum_l \langle l ,0 | \psi \rangle |^2$. $\endgroup$
    – secavara
    Jan 8 '21 at 14:58
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In General

For an observable $\hat{M}$ with possible measurement outcomes $\{m\}$, we have the spectral decomposition $\hat{M}=\sum_m m \hat{P}_m$ where $\hat{P}_m = \sum_\mu|\mu\rangle\langle\mu|$ is the projector onto the eigenspace with eigenvalue $m$ (i.e. the $|\mu\rangle$ are all the eigenstates of $\hat{M}$ with eigenvalue $m$). We can calculate probabilities with the following: For a system in state $|\psi\rangle$, the probability of measuring outcome $m$ is $p(m)=\langle\psi|\hat{P}_m|\psi\rangle$.

Angular Momentum

The $\hat{L}_z$ operator has spectral decomposition $$ \hat{L}_z =\sum_{\ell,m_\ell}\:\hbar m_\ell\:|\ell,m_\ell\rangle\langle \ell,m_\ell|$$ So for a given eigenvalue $\hbar m_\ell$, we can rearrange this to identify the projector: $$ \hat{L_z} = \sum_{m_\ell}\;\hbar m_\ell\;\underbrace{\sum_\ell |\ell,m_\ell\rangle\langle \ell,m_\ell|}_{\hat{P}_{\hbar m_\ell}}$$ In particular, the projector onto all angular momentum eigenstates with $L_z=0$ is $$ \hat{P}_0=\sum_\ell |\ell,0 \rangle\langle \ell,0| $$ So for any state $|\psi\rangle$, the probability of measuring $L_z=0$ is $$ p(L_z=0) = \langle \psi|\hat{P}_0|\psi\rangle = \sum_\ell\langle \psi|\ell,0\rangle\langle\ell,0|\psi\rangle = \sum_\ell |\langle\ell,0|\psi\rangle|^2 $$ Notice this is in general $p(L_z=\hbar m_\ell) = \sum_\ell \left|\langle \ell,m_\ell|\psi\rangle\right|^2$ which is not the same as the expression you have.

Consider a general angular momentum state $|\psi\rangle = \sum_{\ell',m_{\ell'}} c_{\ell',m_{\ell'}}|\ell',m_{\ell'}\rangle$. By the orthogonality relation $\langle\ell,m_{\ell}|\ell',m_{\ell'}\rangle = \delta_{\ell\ell'}\delta_{m_{\ell}m_\ell'}$, we calculate $$ p(L_z=\hbar m_\ell) = \sum_\ell \left|\sum_{\ell',m_{\ell'}} c_{\ell',m_{\ell'}} \langle \ell,m_\ell | \ell',m_{\ell'}\rangle\right|^2$$ $$ = \sum_\ell \left| c_{\ell,m_{\ell}}\right|^2$$

Your Particular State

Your state of interest is $|\psi\rangle = \frac{1}{2}|0,0\rangle + \frac{1}{\sqrt{3}}|1,1\rangle + \frac{1}{2}|1,-1\rangle + \frac{1}{\sqrt{6}}|2,2\rangle$. We then have for the probability of measuring $\hat{L}_z=0$: $$ p(L_z=0) = \sum_\ell \left| \frac{1}{2}\langle\ell,0|0,0\rangle + \frac{1}{\sqrt{3}}\langle\ell,0|1,1\rangle + \frac{1}{2}\langle\ell,0|1,-1\rangle + \frac{1}{\sqrt{6}}\langle\ell,0|2,2\rangle \right|^2 $$

Since $\langle\ell,0|\ell',m_{\ell'}\rangle = \delta_{\ell\ell'}\delta_{m_{\ell'}0}$ the only non-zero term is the first term leaving $$ p(L_z=0) = \left| \frac{1}{2} \right|^2 = \frac{1}{4} $$

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