Can we study the angular momentum of a plane wave scalar particle?

Question

In a plane wave we are in a $p_z$ eigenstate. I saw that $[L_z,p_z]=0$, but $[L^2,p_z] \ne 0$. Is it enough to say that the particle has a defined angular momentum along $z$?

Answer 1

The well known expansion $$\begin{align}e^{i \mathbf{k} \cdot\mathbf{r}} &= 4 \pi \sum\limits_{\ell=0}^\infty \sum_{m=-\ell}^\ell \!i^\ell j_\ell(kr) Y_\ell^m (\hat{\mathbf{k}})^\ast \, Y_\ell^{m}(\hat{\mathbf{r}})= \sqrt{4 \pi} \sum\limits_{\ell =0}^\infty \! i^\ell \sqrt{2 \ell+1} \, j_\ell(k r) Y_\ell^0(\chi,0), \\ \hat{\mathbf{k}}&= \mathbf{k}/|\mathbf{k}|, \, \hat{\mathbf{r}} = \mathbf{r}/ |\mathbf{r}|, \, \mathbf{k} \cdot \mathbf{r} = k r \cos \chi \end{align}$$ of a plane wave $e^{i \mathbf{k} \cdot \mathbf{r}}$ (with momentum $\mathbf{p} = \hbar \mathbf{k}$ ) into spherical harmonics $Y_\ell^m(\hat{\mathbf{r}})$ (being eigenfunctions of $\mathbf{L}^2$ and $L_z$) answers your question: A plane wave is in general not an eigenfunction of $L_z$, except for $\mathbf{k}= k \, \mathbf{e}_z$.

Alternatively, you could compute $$L_z \, e^{i \mathbf{k} \cdot \mathbf{r}}= - i \hbar \partial_\varphi \, e^{i r(k_x \sin \theta \cos \varphi +k_y \sin \theta \sin \varphi +k_z \cos \theta)}=\hbar r \sin \theta (-k_x \sin \varphi+k_y \cos \varphi)e^{i \mathbf{k}{\cdot \mathbf{r}}},$$ observing that the plane wave $e^{i \mathbf{k} \cdot \mathbf{r}}$ becomes an eigenfunction of $L_z$ (with eigenvalue $0$) only if $k_x=k_y=0$, in agreement with the previous finding.