Plane waves, angular momentum, and the 2D Schrödinger equation

Question

I've been thinking about the 2D Schrödinger equation for a free particle, particularly in polar coordinates, and the particular solution $$\Psi(r, \phi) = J_1(r)e^{i\phi}.$$

This solution has an angular momentum of $\hbar$. Now, my understanding is that the collection of plane wave solutions to the S.E. constitute a complete set, so this particular solution should be expressible as an integral of plane waves. However, in 2D, the angular momentum operator on any plane wave gives zero. So how could it be possible to add up a bunch of plane waves and get this solution?

If I could find the 2D Fourier transform of this $\Psi$ that might answer my question. But, well, it's complicated. Also, there is a great deal on the internet about expressing plane waves as sums of Bessel functions but not much going in the other direction, which is what I need.

So where am I going wrong? Do plane waves not form a complete set? Do some plane waves have angular momentum, somehow? Is there maybe a singularity at the 2D origin I am forgetting? Any other possibilities?

Answer 1

I believe the answer to your question is simply that plane waves are not angular momentum eigenfunctions, as a direct calculation shows: \begin{align} (x\partial_y - y\partial_x)e^{ik_x x+ ik_y y} = (xk_y - yk_x)e^{ik_x x+ ik_y y} \end{align} This is only an eigenfunction if $k_x = k_y = 0$, as pointed out by another answer in the 3d case.

A superposition of plane waves can however be an angular momentum eigenfunction, in the same way that $\cos(m\phi)$ and $\sin(m\phi)$ are not eigenfunctions but \begin{align} e^{im\phi} = \cos(m\phi) + i \sin(m\phi) \end{align} is.

While I was thinking about this question, I tried to calculate the Fourier transform of your wavefunction \begin{align} \psi(\vec{r}) = J_1(qr)e^{i\phi} \end{align} I don't know how illuminating the result is, but I thought I would include what I got for the sake of completeness. Actually there is one step here I find a little dubious (tagged below as $(*)$). I would appreciate if anyone could point out whether this step is legal or not.

I took the liberty of inserting a factor of $q=\sqrt{2mE}/\hbar$ so that $r$ has units of length. Since your wavefunction is already not normalized, I didn't bother keep track of factors of $2\pi$. (Hence the $\sim$.) We have \begin{align} \tilde{\psi}\left(\vec{k}\right) &\sim \int e^{-i\vec{k}\cdot\vec{r}}\psi(\vec{r}) d^2\vec{r}\\ &\sim \int_0^{\infty}rdr\int_0^{2\pi} d\phi e^{-ikr\cos(\phi - \theta)}J_1(qr) e^{i\phi} \end{align} The $\phi$ integral is \begin{align} \int_0^{2\pi} d\phi e^{-ikr\cos(\phi - \theta) + i \phi} &= e^{i\theta}\int_\theta^{\theta + 2\pi}d\chi e^{-ikr\cos\chi + i \chi}\\ &= -ie^{i\theta}J_1(kr) \tag{*} \end{align} so \begin{align} \tilde{\psi}\left(\vec{k}\right) &\sim -i e^{i\theta}\int_0^{\infty} dr\, r J_1(qr) J_1(kr) \\ &\sim - i e^{i\theta}\frac{\delta(k -q)}{k} \end{align}