2
$\begingroup$

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.


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..

$\endgroup$
3
  • $\begingroup$ r commutes with angular momentum, so it integrates out as a variable. Visualize the angular equivalence of all concentric spheres. (1) and the one that follows are equivalent. $\endgroup$ Nov 15 '20 at 15:58
  • $\begingroup$ So is it wrong to have it as a coordinate in the base? Or is it useless? Also we define angular momentum as $L_i=\varepsilon _{ijk}x^jp^k$ and then we perform a change in coordinate system from cartesian to spherical, so the fact that $r$ commutes and then this implies that we don't need it in the base seems non trivial to me.. Can you expand on this with an answer to my question instead of a comment? It would really help me to understand properly this topic. $\endgroup$
    – Noumeno
    Nov 15 '20 at 16:06
  • $\begingroup$ Commutativity of r with L is a basic fact of angular momentum theory and is covered in all QM texts, normally near hydrogen atoms. If you are on this chapter discussing this issue, you've clearly skipped/misunderstood factorization and how the radial wave functions have dropped off (factored off) the problem. This is what QM texts may be good at: you don't expect a bespoke tutorial, here! $\endgroup$ Nov 15 '20 at 16:49
2
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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 :)

$\endgroup$
1
  • $\begingroup$ Unfortunately this does not help me a lot.. Mainly because we can think of $Y$ as a wave function in 3D, and so I would expect it to be a function with 3 arguments and with square that we can integrate in 3d to get a probability. Also it would be nice to show that (1) and the following are equivalent starting from the definition of the angular momentum operator. $\endgroup$
    – Noumeno
    Nov 15 '20 at 16:25

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.