# Probabilities to find $\psi$ in the different eigenstates of $\hat{L}_z$: how to handle the $r$-dependence?

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?

• Yes, you may completely ignore (factor out) anything about the radial dependence... – Cosmas Zachos Nov 13 '20 at 16:11

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 $$\psi_{\Omega}$$.