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

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 $$z$$?

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.