In canonical quantization, we replace the canonical conjugate of position($x$), $p$ by $-i \hbar\frac{d}{dx}$. In general we replace $\mathbf{p}$ by $-i\hbar \nabla$. My question is, what if I wanted to do the opposite thing?

Take the laplacian on a 2D sphere, $\nabla^2_{S^2}=\frac{1}{R^2 \sin ^2 \theta}\frac{\partial ^2}{\partial\phi^2}+\frac{1}{R^2 \sin \theta}\frac{\partial}{\partial \theta}\left(\sin \theta \frac{\partial }{\partial \theta}\right)$ where $R$ is of course the radius of the sphere.

What would be the corresponding classical KE term corresponding to this laplacian?

My naive guess of identification is :

$$p_{\theta} \leftrightarrow -\frac{i \hbar}{R}\frac{\partial}{\partial \theta},\; p_{\phi } \leftrightarrow -\frac{i \hbar}{R \sin \theta }\frac{\partial }{\partial \phi} $$

Then it would mean: $$KE=\frac{1}{2m}\left(p^2_\phi+p^2_\theta-i \hbar\frac{\cot \theta}{R}p_{\theta} \right)$$

which is crazy due to appearances of both the $i$ as well as as $\hbar$ which should be alien to a classical description.

Any thoughts on this?

  • 1
    $\begingroup$ Related: physics.stackexchange.com/q/111216/2451 $\endgroup$
    – Qmechanic
    May 21, 2018 at 18:10
  • $\begingroup$ @Qmechanic The answers in the linked post cleared most of my doubt. $\endgroup$
    – user122637
    May 22, 2018 at 1:20

1 Answer 1


The Lagrangian is just classical kinetic energy $$ L = \frac {mR^2} 2({\dot \theta}^2 +{\rm sin}^2\theta \, {\dot \phi}^2), $$ so $$ p_\theta = \frac{\partial L}{\partial \dot \theta}= mR^2 \dot \theta $$ and $$ p_\phi = \frac{\partial L}{\partial \dot \phi}=mR^2 {\rm sin}^2 \theta \,\dot\phi $$ Then $$ H= p_\theta \dot \theta + p_\phi \dot \phi -L = \frac{1}{2mR^2}\left(p_\theta^2 + \frac {1}{{\rm sin}^2\theta}p^2_\phi\right). $$


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