# A problem of momentum representation

Given
[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?

The commutator tells you what the relationship between p and x is. You are not supposed to use explicit realizations of x or p, but only the commutator.

Cheers,

Jazz

dextercioby
Homework Helper
This is nonsense. The commutator is defined for the H-space operators, the thing you got to prove is for their distributional extensions.

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

Given
[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').

This is nonsense. The commutator is defined for the H-space operators, the thing you got to prove is for their distributional extensions.

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?