# Spherical harmonics and angular momentum

I'm studying the relation between rotation matrices and spherical harmonics $\langle\hat{r}|l,m\rangle = Y_{lm}(\theta,\phi)$. In one demonstration it is stated that if $|\hat{r}\rangle=|\hat{e}_z\rangle$, then the angular momentum in $z$ must be zero and so, for $\theta=0$, $Y_{lm}=0$ if $m\ne 0$. Why is that?

• It's because of the angular momentum barrier: the same reason that comets don't hit the sun. They get flung out if they're even a little off-target. Apr 19, 2017 at 23:10
• Sorry but what's the question here? Specifically what is $\langle r\vert$? One often writes $Y_{\ell m}(\theta,\varphi)=\langle \theta,\phi\vert \ell m\rangle$ but the spherical harmonics are angular functions only, so what is the meaning of $\langle r\vert$ or $\vert \vec{r}\rangle$, where in the latter the argument in the ket is now a vector. Finally, how is this related to rotation matrices? Apr 20, 2017 at 2:23

Set $\hbar=1$ for simplicity. You are then inspecting $\langle \hat{e}_z|l m\rangle=\langle (\theta=)0,\phi| lm\rangle=Y_{lm}(0,\phi)$.

Now $L_z=-i\partial_\phi$ and ${\mathbf L}\cdot {\mathbf L}= -\left (\frac {1}{\sin^2\theta}\partial^2_\phi+ \frac{1}{\sin \theta}\partial_\theta \sin\theta \partial_\theta\right )$, so for this to be finite, $\theta =0$ dictates null action of $\partial_\phi$, as you posited.

Consider the matrix element representing the incremental change in a rotation about the z-axis, $$\langle 0,\phi |L_z|lm\rangle = m \langle 0,\phi |lm\rangle =mY_{lm}(0,\phi),$$ where $L_z$ acted on the right.

But it also always produces zero acting on the left, as posited above. So, for nonvanishing m, $Y_{lm}(0,\phi)=0$.

In effect, a rotation of the globe leaves the pole (actually poles: south as well as north) invariant!

The spherical harmonics are essentially defined by their action under rotations: if you rotate the state $|l,m\rangle$ by an angle $\alpha$ about its axis of quantization ($z$ for simplicity), it will transform as $$R(\alpha\,\hat{e}_z)|l,m\rangle = e^{-im\alpha}|l,m\rangle. \tag 1$$ This is equivalent to the eigenvalue relationship $$\hat{L}_z|l,m\rangle = \hbar m |l,m\rangle,\tag 2$$ since the rotation operator is given by the exponentiated angular momentum as $$R(\alpha\,\hat{e}_z) = e^{-i\alpha\hat{L_z}/\hbar}; \tag3$$ to translate back and forth between $(1)$ and $(2)$, you just put in $(3)$ and differentiate w.r.t. $\alpha$, or you use the fact that functions of an operator preserve eigenvectors with the function applied to the eigenvalue.

With that in mind, if you choose some finite angle $\alpha$ and you rotate the state $|l,m\rangle$ by that angle, the position-representation wavefunction at the pole can be calculated via two different ways (rotating the state, or rotating the coordinates), which by $(1)$ need to be equal: $$Y_{lm}(0,\phi+\alpha) = \langle (\theta=)0,\phi|R(\alpha\,\hat{e}_z)|l,m\rangle = e^{-im\alpha}\langle (\theta=)0,\phi||l,m\rangle = e^{-im\alpha}Y_{lm}(0,\phi).$$ However, because of the coordinate singularity at the origin, both $(0,\phi+\alpha)$ and $(0,\phi)$ are identical points, so you also need to have $Y_{lm}(0,\phi+\alpha)=Y_{lm}(0,\phi)$, and that then requires you to have $$Y_{lm}(0,\phi) = e^{-im\alpha}Y_{lm}(0,\phi),$$ for all real $\alpha$. The only way this can be true is if $m=0$ or if $Y_{lm}(0,\phi)$ vanishes.