Understanding <q|p>=exp(ip.x): The Explanation You've Been Searching For

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Discussion Overview

The discussion revolves around the expression =exp(ip.x) and its implications in quantum mechanics, particularly in relation to eigenstates of the momentum and position operators. Participants seek clarification on the nature of this expression and its derivation.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant requests an explanation for the expression =exp(ip.x), indicating a lack of understanding.
  • Another participant suggests that the explanation depends on the nature of the function and mentions the orthogonality condition as a relevant factor.
  • A different participant questions whether represents an eigenstate of the momentum operator or the position operator, providing details about the normalization condition for the momentum operator.
  • This participant also asserts that the initial expression is incorrect, implying a misunderstanding of the context.
  • A subsequent post corrects the expression to =exp(ip.x) and elaborates on the relationship between the position operator and the momentum operator, including a brief derivation.
  • The same participant expresses confusion about the source of the expression as quoted in a quantum field theory text, indicating a desire for further clarification.

Areas of Agreement / Disagreement

There is no consensus on the correctness of the initial expression, as participants present competing views regarding its interpretation and context. The discussion remains unresolved with differing opinions on the nature of the expression.

Contextual Notes

Participants reference concepts such as eigenstates, normalization conditions, and orthogonality, but the discussion lacks clarity on certain assumptions and definitions, which may affect the understanding of the expressions involved.

Colin
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Can anyone explain why?

<q|p>=exp(ip.x)

thanks
 
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Depends on the nature of your function to begin with.. more detail needed.. probably due to orthogonality condition
 
Colin said:
Can anyone explain why?

<q|p>=exp(ip.x)

thanks


What is [itex]<q|[/itex],an eigenstate of the momentum operator,or of the position operator (actually of their adjoints,but,because they are selfadjoint,you can think that way as well)?
If it's for the momentum operator,then [tex]|p> [/itex] is an eigenstate as well,and the normalization condition reads:<br /> [tex]<q|p> =2\pi\hbar \delta(q-p)[/tex]<br /> ,but if it's for the coordinate operator,then the scalar product is zero,since one of them is an eigenvector from a Hilbert space and the other is a linear functional over another Hilbert space.<br /> <br /> So,anyway,what u've written there is wrong.<br /> <br /> Daniel.[/tex]
 
correction <x|p>=exp(ip.x)

sorry I meant <x|p>=exp(ip.x)

and I think I have it

P|p>=p|p>

<x|P|p> =p<x|p>=-id/dx<x|p>
solution
<x|p>=exp(ipx)

It's a long time since I did QM, and I started reading Prof Zee's Quantum field theory in a nut shell. He quotes this very early on without proof and I couldn't see where it came from. Thanks for comments.
 

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