# Probability of finding certain value of the $z$-component of the angular momentum when the wave function contains multiple $l$-values [closed]

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?

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

## 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 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 $$p(L_z=0) = \left| \frac{1}{2} \right|^2 = \frac{1}{4}$$