Commutator Relations: [x,p]=ih, Proof of p=-iħ∂/∂x+f(x)

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Homework Help Overview

The discussion revolves around the commutation relation between position and momentum operators in quantum mechanics, specifically focusing on the representation of the momentum operator given the relation [x,p]=ih. The original poster attempts to show that the momentum operator can be expressed as p=-iħ∂/∂x+f(x), where f(x) is an arbitrary function of x.

Discussion Character

  • Conceptual clarification, Problem interpretation

Approaches and Questions Raised

  • Participants question the phrasing "if x=x" and its implications, suggesting it may be a misunderstanding of the coordinate representation. There is a discussion about the Hilbert space context and the realization of the position operator.

Discussion Status

The conversation is exploring the definitions and representations involved in the problem. Some participants provide clarifications regarding the spectral equation related to the position operator, indicating a productive direction in understanding the problem's context.

Contextual Notes

Participants note that the original poster's phrasing may lead to confusion, as "if x=x" is inherently true by definition. The discussion also emphasizes the importance of the Hilbert space framework in which these operators are defined.

alisa
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given that [x,p]=ih, show that if x=x, p has the representation p=-iħ∂/∂x+f(x) where f(x) is an arbitrary function of x
 
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alisa, you're supposed to show an attempted solution at the problem. That goes for your other threads as well.

alisa said:
given that [x,p]=ih, show that if x=x,

What do you mean " if x=x". x=x by definition.
 
Tom Mattson said:
What do you mean " if x=x". x=x by definition.

It's the coordinate representation in which the Hilbert space is L^{2}(\mathbb{R},dx). The "x" operator is realized by a multiplication by "x". She's asked to prove that the most general representation of the momentum operator in this Hilbert space is the one written there.
 
OK, so then it should read something like "If \hat{x}|\psi>=x|\psi>...", right?
 
Last edited:
Exactly. That's the spectral equation, but nonetheless, yes.
 

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