Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

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 $-i\hat{\mathbf q}^i\cdot\hbar \nabla$?

share|improve this question
    
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

1 Answer 1

up vote 2 down vote accepted

If you use curvilinear coordinates you basically need to transform your derivatives likewise, ie, you need to use the Jacobian of the coordinate transformation you used to go from Cartesian coords to the curvilinear system you have at hands.

More generally, you can think in terms of Quantum Mechanics over a curved manifold, in which case you would simply use the covariant derivative.

share|improve this answer
    
There isn't a unique covariant derivative on a manifold! So which one? –  MBN Mar 21 '11 at 4:32
1  
@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
    
@Deepak: I know that there is unique such connection (plus the condition to be trosion free). The question is not about manifolds with a choice of metric, so the answer should say something about it, otherwise it is incomplete and misleading. –  MBN Mar 21 '11 at 13:55
    
@MBN: the last equation in the question has an inner product, which led me to believe that i could make some "general Physical assumptions", such as the existence of a metric and of a metric-compatible connection. –  Daniel Mar 21 '11 at 15:44
    
Fair enough. –  MBN Mar 21 '11 at 18:06

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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