# Canonical momentum operators in curvilinear coordinates

What is the quantum canonical momentum operator corresponding to arbitrary canonical position. For example, in Cartesian coordinates ($x^i$), the canonical momentum operator with respect to each $x^i$ is $-i\hbar \frac{\partial}{\partial x^i}$. For arbitrary canonical coordinates $q^i$, would the corresponding canonical momentum operators just be $-i\hat{\mathbf q}^i\cdot\hbar \nabla$?

-
In general, there is nothing such as "the" unique canonical momentum for a coordinate $q^i$. A canonical momentum is defined in such a way that its commutator with $q^i$ is nonzero, but it is zero for other $q$'s. One may redefine the momenta and coordinates in many ways, the so-called canonical transformations en.wikipedia.org/wiki/Canonical_transformations , without changing the commutator. The idea that one always has a unique one-to-one map is OK to count the degrees of freedom but it's fundamentally flawed if one wants to study the full physical system. –  Luboš Motl Mar 20 '11 at 19:03
Please define what you mean by $\hat{\mathbf q}^i$. Does your manifold have a metric? Moreover, $-i\hat{\mathbf q}^i\cdot\hbar \nabla$ has unusual "index up", where momentum $p_i$ traditionally has "index down". I recommend you to simply write $-i\hbar\nabla_{i}$ instead. –  Qmechanic Mar 21 '11 at 22:39

@MBN generally one works with a metric compatible connection, i.e. $\nabla g_{\mu\nu} = 0$. Given a metric, this equation determines the corresponding connection. –  user346 Mar 21 '11 at 6:25