Operator Theory Problem on Momentum Operator (QM)

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Homework Help Overview

The problem involves proving a commutation relation between the position operator \(\hat{x}\) and a function of the momentum operator \(g(\hat{p})\) in the context of quantum mechanics. The operators are defined as \(\hat{x}=x\cdot\) and \(\hat{p}=-i\hbar \frac{d}{dx}\).

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss starting points for the proof, with one suggesting the expansion of \(g(p)\) in a power series and questioning the commutation relation \([x,p^n]\). Others consider the implications of working in momentum-space instead of position-space.

Discussion Status

The discussion is ongoing, with participants exploring different approaches and considerations. Some guidance has been offered regarding the use of power series and the context of momentum-space, but no consensus has been reached on a specific method.

Contextual Notes

Participants express uncertainty about how to begin the problem and seek hints rather than complete solutions. There is an acknowledgment of the simplicity of the problem in certain contexts.

LolWolf
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Homework Statement



Given the operators [itex]\hat{x}=x\cdot[/itex] and [itex]\hat{p}=-i\hbar \frac{d}{dx}[/itex], prove that:

[itex][\hat{x}, g(\hat{p})]=i\hbar \frac{dg}{d\hat p}[/itex]

Homework Equations



None.

The Attempt at a Solution



I have very little idea on how to begin this problem, but I don't want a solution, I simply want a hint in the right direction.

Thanks, mates.
 
Last edited:
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LolWolf said:

Homework Statement



Given the operators [itex]\hat{x}=x\cdot[/itex] and [itex]\hat{p}=-i\hbar \frac{d}{dx}[/itex], prove that:

[itex][\hat{x}, g(\hat{p})]=i\hbar \frac{dg}{d\hat p}[/itex]

Homework Equations



None.

The Attempt at a Solution



I have very little idea on how to begin this problem, but I don't want a solution, I simply want a hint in the right direction.

Thanks, mates.

Expand ##g(p)## in a power series. What's ##[x,p^n]##?
 
Actually, I realized it was even easier than that, but thank you!

Consider the case in momentum-space rather than position-space, and this reduces nicely using elementary operations.
 
LolWolf said:
Actually, I realized it was even easier than that, but thank you!

Consider the case in momentum-space rather than position-space, and this reduces nicely using elementary operations.

Sure, that works also. x is a differentiation operator in p space.
 

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