The Hamiltonian operator for a free non-relativistic particle looks like

$$ \hat{H} = \frac{\hat{p}^2}{2m} = -\frac{\hbar^2}{2m} \nabla^2. $$

In polar coordinates, the Laplacian expands to

$$ \hat{H} = -\frac{\hbar^2}{2m} \left (\frac{1}{r} \frac{\partial}{\partial r} \left( r \frac{\partial}{\partial r} \right) + \frac{1}{r^2} \frac{\partial^2}{\partial \theta^2} \right). $$

The radial and angular momentum operators are $$ \begin{align*} \hat{p}_r = \frac{\hbar}{i} \left(\frac{\partial}{\partial r} + \frac{1}{2r}\right) && \hat{p}_\theta = \frac{\hbar}{i}\frac{1}{r}\frac{\partial}{\partial \theta}. \end{align*}$$

After squaring, summing and comparing to the Hamiltonian, we find that

$$ \hat{H} = -\frac{\hbar^2}{2m} ( \hat{p}_r^2 + \hat{p}_\theta^2 ) - \frac{\hbar^2}{8mr^2}. $$

In classical mechanics, we expect that $ p^2 = p_r^2 + p_\theta^2 $, does this not hold in quantum mechanics? Why does this curious $- \frac{\hbar^2}{8mr^2}$ potential appear, does it have any significance?


To clarify the choice of radial momentum operator, consider the naive $ \hat{p}'_r = \frac{\hbar}{i} \frac{\partial}{\partial r} $. Taking the adjoint we find that

$$ \langle \phi, \hat{p}'_r \psi \rangle = \int d\theta \int_0^{+\infty} r dr \phi^* \frac{\hbar}{i} \frac{\partial}{\partial r} \psi = - \int d\theta \int_0^{+\infty} r dr \frac{\hbar}{i} \left( \frac{\partial \phi^*}{\partial r} + \frac{\phi^*}{r} \right)\psi = \int d\theta \int_0^{+\infty} rdr \frac{\hbar}{-i} \left( \frac{\partial}{\partial r} + \frac{1}{r} \right) \phi^* \psi \neq \langle \hat{p}'_r \phi, \psi \rangle$$

With an extra term $\frac{\hbar}{i} \frac{1}{2r}$ it is self-adjoint. This term is different from the one in spherical coordinates by a factor of 2.

  • 1
    $\begingroup$ See physics.stackexchange.com/q/9349 $\endgroup$ – G. Smith Sep 10 at 21:10
  • $\begingroup$ @G.Smith It looks, conceptually, like a virtual duplicate. $\endgroup$ – Cosmas Zachos Sep 10 at 21:58
  • $\begingroup$ I'm aware of that question, I did do my research, but I don't see any resemblance. I am not asking anything about how to construct the radial momentum operator, and I use the correct one. I'm asking why a "naive" decomposition of the Hamiltonian does not work, and what the significance is of the inverse square potential-like term. This might have something to do with the radial momentum operator, but this is not discussed in the linked question (which btw uses spherical coordinates, where this extra term does not show up) $\endgroup$ – Kasper Meerts Sep 10 at 22:07
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    $\begingroup$ Since you use a different radial momentum operator than the answers in the other question construct, you need to justify why it is “the correct one”. Says who, besides you? $\endgroup$ – G. Smith Sep 10 at 23:26
  • $\begingroup$ as I said, they use spherical coordinates, making the operator a little different. Again, my question is not about the construction of the radial momentum operator, I do not need to repeat basic quantum mechanics here. Repeat the linked derivation for polar or cylindrical coordinates, or just Google it, if you are curious about it $\endgroup$ – Kasper Meerts Sep 11 at 6:18

The main reason you can't square the components here directly is that the basis vectors $\hat{r}$ and $\hat{\theta}$ are changing with the coordinates. The second reason is that the metric of this coordinate is not $\mathbf{1}$. (This coordinate is a curved orthogonal coordinate but cartesian coordinate is a flat coordinate.) Thus, when taking the derivative of a vector field $\mathbf{F}$, you have $$\partial_j\mathbf{F}=\partial_j F^i\mathbf{g}_i+F^i\partial_j\mathbf{g}_i=\left(\partial_jF^i+F^k\Gamma^i_{jk}\right)\mathbf{g}_i$$ Where the additional second term originates from the difference of basis vectors at different points.

Thus $\nabla\cdot\mathbf{F}=\partial_iF^i=\left(\partial_iF^i+F^k\Gamma^i_{ik}\right)\mathbf{g}_i=\frac{1}{\sqrt{g}}\partial_k(\sqrt{g}F^k)$. The $\nabla^2$ operator is then $\nabla^2=\nabla\cdot\nabla=\partial_i\partial^i=\frac{1}{\sqrt{g}}\partial_k(\sqrt{g}\partial^k)=\frac{1}{\Pi_kA_k}\partial_i(\frac{\Pi_jA_j}{A_i^2}\partial_i)$. Where $A_i$ are the Lame constants.

Here the $A_i$ are $1$ and $r$ respectively, simply plug into the formula for $\nabla\cdot$ and $\nabla^2$ You get $\nabla_r=\frac{1}{r}\partial_r r=\frac{1}{r}+\partial_r$ (you had a coefficient wrong) and $\nabla_\theta=\frac{1}{r}\partial_\theta$. And $$\nabla^2=\frac{1}{r}\left(\partial_r \left(r\partial_r\right)+\partial^2_\theta\right)$$

  • $\begingroup$ I don't think going to 3D is necessary, the Hamiltonian is well-defined for two dimensional spaces. In any case, as $\hat{p}_z$ commutes with everything, that would just add a $\frac{\hat{p}_z^2}{2m}$ term, which wouldn't change anything about the radial potential-like term. $\endgroup$ – Kasper Meerts Sep 11 at 9:26
  • $\begingroup$ @KasperMeerts It's ok to do in 2D, but the coordinates here are curved so the terms can't be multiplied directly. $\endgroup$ – Ladmon Draxngfüskiii Sep 11 at 10:26

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