Decomposing $|\Psi \rangle$ into the $|L=1 \, m_z \rangle$ basis

Question

I have been asked to decompose the state $|\Psi \rangle$, with wavefunction in spherical coordinates $\langle r \, \theta \, \phi|\Psi \rangle = \sqrt{\frac{3}{\pi}}e^{-r}\sin{\theta}\cos{\phi}$ into the $|L = 1 \, m_z\rangle$ basis.

I know $\langle \theta \, \phi|\ell \, m\rangle = Y^m_\ell(\theta, \phi)$, and $\Psi(r,\theta,\phi) = \sqrt{2}e^{-r}(Y_1^{-1} - Y_1^1)$, but I'm unsure whether this is already the proper decomposition into that basis. It honestly sounds like a bogus question to me.

My point is, I can't express $|\Psi\rangle = \frac{1}{\sqrt{2}}(|L = 1 \, m_z=-1\rangle - |L = 1 \, m_z=1\rangle)$ without leaving out the radial part of the wavefunction. Is there a way to do this, or is there no proper answer to this question apart from the decomposition of the wavefunction itself into spherical harmonics?

Answer 1

Your decomposition is fine. Clearly you cannot write down $|\Psi\rangle$ solely by angular momentum eigenstates, or spherical harmonics for that matter, since they give only the angular part of the wavefunction. So either you write your state as a product of radial wavefunction and spherical harmonics, as you did $$\Psi(r,\theta,\phi) = \sqrt{2}e^{-r}\left(Y^{-1}_1(\theta,\phi)-Y^{1}_1(\theta,\phi)\right)$$ or you can use the braket notation $$|\Psi\rangle = N|n\rangle(|1,-1\rangle-|1,1\rangle)$$ where $N$ is some normalization constant.

In any case you'll need to specify three quantum numbers $n,l,m$ which are encoded either into the radial part of the wavefunction or the anguar part. There's no way to get rid of the radial part since by doing so you'll be giving away a quantum number.