- #1

- 1

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[x,p] = i * h-bar,

prove that

<p|X|p'> = [i * h-bar / (p' - p)] * δ(p - p').

I don't understand why commutator matters with this proof?

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- Thread starter fish830617
- Start date

- #1

- 1

- 0

[x,p] = i * h-bar,

prove that

<p|X|p'> = [i * h-bar / (p' - p)] * δ(p - p').

I don't understand why commutator matters with this proof?

- #2

- 239

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Cheers,

Jazz

- #3

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- #4

Avodyne

Science Advisor

- 1,396

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And what's written is not even true for the distributional extensions.

- #5

- 44

- 1

[x,p] = i * h-bar,

prove that

<p|X|p'> = [i * h-bar / (p' - p)] * δ(p - p').

I don't understand why commutator matters with this proof?

Hint: try evaluating <p|[X,p]|p'>, and use the fact (not given, but based on the result this is how |p> is normalized) that <p|p'>= δ(p - p').

- #6

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I don't understand how is this a nonsense? I mean we can arrive at second equation (with minor correction) starting from commutation relation and certain assumptions. Can't we?

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