Arguments of the eigenfunctions of angular momentum

Question

Once we have defines the angular momentum operators $L_z,L_y,L_x,L^2$ ($L^2=L_z^2+L_y^2+L_x^2$) suppose we focus on the eigenstates $|l \ m \rangle$ common to both $L_z$ and $L^2$: $$L_z|l \ m\rangle =\hbar m |l \ m\rangle$$ $$L^2|l \ m\rangle =\hbar ^2 l(l+1) |l \ m\rangle$$ suppose now that we want to talk about eigenfunctions of the angular momentum operator instead of the eigenstates of it, every source that I could find performs this switch in the following way: $$\langle \theta \ \phi |L_z|l \ m\rangle =\hbar m \langle \theta \ \phi |l \ m\rangle \ \Rightarrow \ -i\hbar \frac{\partial}{\partial \phi} Y_{l,m}(\theta,\phi) =\hbar m Y_{l,m}(\theta,\phi) \ \ \ \ \ \ (1)$$ My question is: why are we putting ourselves in the base of $\theta \ \phi$ instead of $\theta \ \phi \ r$? In other words why do we have equation (1) and not the following: $$\langle \theta \ \phi \ r |L_z|l \ m\rangle =\hbar m \langle \theta \ \phi \ r |l \ m\rangle \ \Rightarrow \ -i\hbar \frac{\partial}{\partial \phi} Y_{l,m}(\theta,\phi , r) =\hbar m Y_{l , m}(\theta,\phi,r)$$ Since we are working in 3D space I would expect the wave function $Y_{l,m}$ to represent the probability amplitude in 3D space; I would like $Y_{l,m}$ to be something that I can square and then integrate over 3D space to get a probability, right? Seems strange to switch from cartesian to spherical coordinates and then ignore one of the spherical coordinates..
In every lecture I could find everybody uses only $\theta$ and $\phi$ without $r$ but then fails to explain why.

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P.S.
I have noticed that $Y_{l,m}$ are usually called "spherical harmonics", I'm not sure that this is relevant here but for now I have failed to understand the reason behind this name, so maybe this has something to do with this other thing I don't understand..

Answer 1

Look at the angular momentum operators in spherical coordinates ($r$, $\theta$, $\phi$). They are given by $$\begin{align} L_z&=-i\hbar\frac{\partial}{\partial\phi} \\ L^2&=-\hbar^2 \left(\frac{1}{\sin\theta}\frac{\partial}{\partial\theta}\sin\theta\frac{\partial}{\partial\theta} +\frac{1}{\sin^2\theta}\frac{\partial^2}{\partial\phi^2} \right) \end{align}$$

You see, they are made of $\theta$, $\phi$, $\frac{\partial}{\partial\theta}$ and $\frac{\partial}{\partial\phi}$, but not $r$ and $\frac{\partial}{\partial r}$.

The well-known simultaneous eigenfunctions of $L_z$ and $L^2$ are the spherical harmonics $Y_{l,m}(\theta,\phi)$.
But this is not the complete truth. Actually the functions $f(r)Y_{l,m}(\theta,\phi)$ (with any arbitrary radial function $f(r)$ ) are eigenfunctions of $L_z$ and $L^2$ as well.

You can check this by verifying the equations $$\begin{align} L_z\ f(r)Y_{l,m}(\theta,\phi) &= m\hbar\ f(r)Y_{l,m}(\theta,\phi) \\ L^2\ f(r)Y_{l,m}(\theta,\phi) &= l(l+1)\hbar^2\ f(r)Y_{l,m}(\theta,\phi) \end{align}$$ with the operators given above. That is why you don't need to care about the radial dependency of these functions, and usually you just omit it in writing.

Answer 2

As you have noticed, $Y_{l,m}$ are called "spherical harmonics". The reason why we are in the base of ${\phi},{\theta}$ is due to the fact that spherical harmonics are basis functions for irreducible representations of the group SO(3), that is the 3D rotation group. In rotations you need these two parameters and not $r$. Hope this helps :)