The wave function of a particle in a spherically symmetric potential $V(r)$ is given by: $$ \psi(\vec{r})=(x+y+3z)f(r).$$ Determine the probabilities to find $\psi$ in the different eigenstates of $\hat{L}_z$.

My attempt:

It's best to rewrite $\psi(\vec{r})$ in terms of spherical harmonics: $$\psi(r,\theta,\phi) = \left(\frac {4\pi}{3}\right)^{1/2}\left( -\frac{i-1}{\sqrt 2} Y_{1,1}(\theta,\phi)+\frac{1+i}{\sqrt 2}Y_{1,-1}(\theta,\phi)+3Y_{1,0}(\theta,\phi)\right)rf(r).$$ This is clearly an eigenstate of $\hat{\vec{L}}^2$ (with eigenvalue $2\hbar^2$) and therefore of $\hat{L}_z$. The possible eigenstates of $\hat{L}_z$ in which we can observe $\psi$ are $Y_{1,1},Y_{1,-1}$ and $Y_{1,0}$, so we need to find $\operatorname{Pr}(m=\pm 1) $ and $\operatorname{Pr}(m=0)$.

Now, $\operatorname{Pr}(m=1)=\frac{|\langle 1,1|\psi\rangle|^2}{\langle\psi\mid\psi\rangle}$. To find these probability amplitudes, I would use the resolution of the identity w.r.t. $|\theta,\phi\rangle=|\vec{\Omega}\rangle$, i.e. $$ \langle 1,1\mid\psi\rangle=\int d\vec{\Omega} \langle 1,1\mid\vec{\Omega}\rangle\langle\vec{\Omega}\mid\psi\rangle=\int_0^{2\pi}d\phi\int_0^{\pi}\sin\theta d\theta Y^{*}_{1,1}(\theta,\phi)\psi(\theta,\phi).$$ As you can see, the $r$-dependence of $\psi$ does not appear in this integral. Can I just take the first factor of $\psi$ and use it in this integral? Is this still the same wave function that I'm working with?

  • $\begingroup$ Yes, you may completely ignore (factor out) anything about the radial dependence... $\endgroup$ Commented Nov 13, 2020 at 16:11
  • $\begingroup$ It seems to me a duplicate of physics.stackexchange.com/questions/691732/… $\endgroup$ Commented Feb 5, 2022 at 20:51
  • $\begingroup$ Well, my question was asked 2 years ago, so I was not the one duplicating ;) $\endgroup$
    – Zachary
    Commented Feb 7, 2022 at 17:04

1 Answer 1


This problem is best addressed in bra-ket notation, where you can write the angular part of the state as:

$$ \psi_{\Omega} = c_{1,1}|1,1\rangle + c_{1,-1}|1,-1\rangle + c_{1,0}|1,0\rangle $$

where you have already computed the $c_{1,m}$.

It is an eigenstate $\hat L^2$ because each basis state in the expansions is an eigenstate ($l=1$) with the same eigenvalue.

It is not an eigenstate of $L_z$, though, as:

$$ L_z|1,m\rangle = m\hbar|1,m\rangle $$

so by inspection:

$$L_z|\psi_{\Omega}\rangle \propto c_{1,1}|1,1\rangle - c_{1,-1}|1,-1\rangle $$

which is not proportional to $\psi_{\Omega}$.


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