Obviously, if $x$ is a Cartesian coordinate, then the corresponding momentum operator is $-i \hbar \partial_x$. But what if $x$ is something more complicated, like some sort of curvilinear coordinate in 3D space?

There is this question elsewhere on this site. Quantum mechanical analogue of conjugate momentum But it's not clear to me that the accepted answer there is really answering the question I am asking. Nothing in the question or answer specifically addresses non-Cartesian coordinates, certainly not in any detail.


1 Answer 1


If an $n$-dimensional configuration manifold $M$ of some physical system is orientable and endowed with a positive volume form $$\Omega~=~\rho(x)\mathrm{d}x^1\wedge\ldots\wedge\mathrm{d}x^n, \qquad \rho(x)~>~0,$$ we can define a sesquilinear form $$\langle \phi | \psi\rangle~:=~\int_M \! \Omega ~\phi^{\ast} \psi. $$ The Schroedinger representation of the self-adjoint phase space operators is $$ \hat{x}^j~=~x^j,\qquad \hat{p}_k~=~\frac{\hbar}{i\sqrt{\rho(x)}} \frac{\partial}{\partial x^k} \sqrt{\rho(x)}, \qquad [\hat{x}^j,\hat{p}_k]~=~i\hbar\delta^j_k \mathbb{1}. $$ See also this related Phys.SE post.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.