Momentum Operators and the Schwartz Integrability Condition

In summary, the article explains that the commutator between differential operators must be equal to 0 if the operators are to be Schwartz Integrable. This condition is based on a symmetry of second derivatives. If the operators do not have this symmetry, then the commutator may not be equal to 0.
  • #1
DreadyPhysics
21
0
Hi All,

When computing the commutator [tex]\left[x,p_{y}\right][/tex], I eventually arrived (as expected) at [tex]\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)[/tex] 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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  • #2
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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  • #3
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.
 

1. What is a momentum operator?

A momentum operator is a mathematical operator that is used to describe the momentum of a particle in quantum mechanics. It is represented by the symbol p and is defined as the rate of change of the wave function with respect to position.

2. What is the significance of momentum operators in quantum mechanics?

Momentum operators are crucial in quantum mechanics because they help describe the motion and behavior of particles at the atomic and subatomic level. They are used to calculate the momentum of a particle, which is essential in determining its position and energy.

3. What is the Schwartz integrability condition?

The Schwartz integrability condition is a mathematical condition that must be satisfied for a function to be considered "well-behaved" and integrable. It states that the function must decrease faster than any polynomial as the variable approaches infinity.

4. How are momentum operators and the Schwartz integrability condition related?

In quantum mechanics, the momentum operator is defined as the derivative of the wave function with respect to position. In order for this derivative to exist and be well-defined, the wave function must satisfy the Schwartz integrability condition.

5. How is the Schwartz integrability condition used in quantum mechanics?

The Schwartz integrability condition is used to ensure that the wave function, which describes the state of a particle, is well-behaved and can be used to calculate its momentum. It is also used in the mathematical formulation of quantum mechanics to ensure that certain equations and operators are well-defined.

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