How can test functions help understand commutation relations?

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This discussion centers on the use of test functions to understand commutation relations in quantum mechanics, specifically the expression [p^2, x] = p [p, x] + [p, x] p = 2ħ/i p. The participant references Gasiorowicz's quantum mechanics textbook and highlights the importance of applying these expressions to test functions, such as f(x), to derive results accurately. The conversation emphasizes that while using test functions may not be the quickest method, it is essential for grasping the underlying principles of commutation relations.

PREREQUISITES
  • Understanding of quantum mechanics principles, particularly commutation relations.
  • Familiarity with the momentum operator, p = -iħ ∂/∂x.
  • Knowledge of the product rule in calculus.
  • Basic proficiency in applying mathematical operators to functions.
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  • Study the derivation of commutation relations using test functions in quantum mechanics.
  • Learn about the implications of the Heisenberg uncertainty principle in relation to commutation relations.
  • Explore advanced topics in quantum mechanics, such as the role of operators in Hilbert space.
  • Investigate the application of test functions in other areas of physics, such as field theory.
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Students of quantum mechanics, physicists exploring operator theory, and anyone seeking to deepen their understanding of commutation relations and their applications in quantum systems.

silimay
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So I was reading from my quantum book (Gasiorowicz) and I ame across this sentence:

[p^2, x] = p [p, x] + [p, x] p = \frac{2\hbar}{i} p

I don't understand this. I know that p = -i \hbar \frac{\partial}{\partial x}, but I can't see how to get that expression...I just come up with something like x {\hbar}^2 \frac{{\partial}^2}{{\partial x}^2} when I try multiplying it out.
 
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try to derive [AB,C] = ?
then use [x,p] = i\hbar
to find [p^2,x]
 
silimay said:
So I was reading from my quantum book (Gasiorowicz) and I ame across this sentence:

[p^2, x] = p [p, x] + [p, x] p = \frac{2\hbar}{i} p

I don't understand this. I know that p = -i \hbar \frac{\partial}{\partial x}, but I can't see how to get that expression...I just come up with something like x {\hbar}^2 \frac{{\partial}^2}{{\partial x}^2} when I try multiplying it out.

All those expressions only make sense if you imagine applying them to some "test function" f(x). For example,

[x,p_x] f(x) = -i \hbar x \partial_x f(x) + i \hbar \partial_x (x f(x)) apply the product rule on the second term and then something will cancel out. At the very end of the calculation (and only then) you may drop the test function f(x).

Using a test function is not the fastest way to prove complicated commutation relations, however. But it's the only way to make sense of these commutation relations. after you have done a few with a test function you will be able to do the more complex cases without the crutch of a test function.
 

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