Momentum Operators and the Schwartz Integrability Condition

DreadyPhysics
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Hi All,

When computing the commutator \left[x,p_{y}\right], I eventually arrived (as expected) at \hbar^{2}\left(\frac{\partial}{\partial y}\left(\frac{\partial f}{\partial x}\right) - \frac{\partial}{\partial x}\left(\frac{\partial f}{\partial y}\right)\right) and I realized that, as correct as it seems, I couldn't figure exactly why these two values should equal each other and thus set the commutator equal to 0. I know that they do- I knew the answer ahead of time - but at this step, I was at a loss to prove that the differential operators themselves commute. It would seem conceivable to me that they might not in some circumstances.

I looked into it a bit and it turns out that this hinges on the Schwartz Integrability Condition (see http://en.wikipedia.org/wiki/Symmetry_of_second_derivatives). However, the article is a little unclear to me. Is this a general property of two-dimensional functions, or by assuming it are we limiting ourselves to a specific domain of functions?

The article mentions even that in physics specifically, sometimes we suspend the condition and allow functions to violate it. Does anyone have any examples of this specifically in a physical context? And, if we suspend the condition here, won't it make the momentum operators non-commuting? Wouldn't that would have specific physical consequences?
 
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DreadyPhysics said:
it turns out that this hinges on the Schwartz Integrability Condition (see http://en.wikipedia.org/wiki/Symmetry_of_second_derivatives). However, the article is a little unclear to me. Is this a general property of two-dimensional functions, or by assuming it are we limiting ourselves to a specific domain of functions?
The strongest version of this statement "equality of mixed partials" that I know of, i.e. the one with the weakest hypotheses, is stated here.
 
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Landau said:
The strongest version of this statement "equality of mixed partials" that I know of, i.e. the one with the weakest hypotheses, is stated here.

Excellent, that answers my question. Thank you very much.
 
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