Re: Quantum mechanics and operators

From: Edward Green (spamspamspam3_at_netzero.com)
Date: 11/19/04 

Date: 19 Nov 2004 15:36:01 -0800

glhansen@steel.ucs.indiana.edu (Gregory L. Hansen) wrote in message news:<cnh1rg$t3s$1@hood.uits.indiana.edu>...

> With the definitions P psi(p) = p psi(p) and X psi(x) = x psi(x), I
> thought I could Fourier transform one of them, integrate by parts, and
> derive the position operator in the momentum representation. I haven't
> figured that out, and I'm not sure why it doesn't seem to work.

Hmm... I was able to make something like that work on my first
blundering attempt, though I'm not sure if I have grown in wisdom by
it.

Consider a pair of functions related by:

  f(x) = Int[+/-oo] dk exp(ikx)g(k)

then...

  f'(x) = Int[+/-oo] dk exp(ikx)(ik)g(k)

Apparently differentiating one side of the transform pair is
equivalent to multiplying the other side by i times the argument.

(And may I say by carefully writing "P" vs. "p" and "X" vs. "x" above
you are making significant progress in your fight with notational
abuse, and may someday look forward to unsupervised visits? ;-)

Quantcast