friend
- 1,448
- 9
Thank you, vanhees71. But I don't know what question you were trying to answer.vanhees71 said:Well, let's do the calculation
Thank you, vanhees71. But I don't know what question you were trying to answer.vanhees71 said:Well, let's do the calculation
friend said:I've tried to do a Fourier Xform the old fashion way, and I get pretty much the same think you got except for factors of 2π and a minus sign in the exponent. The difference seems to be that you do not use the 2π in your definition of the Fourier Xform. And you do not use the minus sign in the exponent in the second Fourier Xform. Could you please tell me why you seem to be defining your Fourier transforms differently than what I see in wikipedia.org? Thanks.
stevendaryl said:If you have \langle x | U(t) | x' \rangle, then you can compute \langle p| U(t)|p'\rangle as follows:
\langle p |U(t) |p' \rangle = \int dx \int dx' \langle p|x \rangle \langle x|U(t)|x'\rangle \langle x'|p'\rangle
where \langle x' | p' \rangle = \sqrt{\frac{1}{2\pi}} e^{i p' x'} and \langle p|x \rangle = (\langle x | p \rangle)^* = \sqrt{\frac{1}{2\pi}} e^{-i p x}
(again, letting \hbar = 1 for simplicity).
I wrote the complete derivation, but it's very long, so I deleted it. The result is:
\langle p |U(t) |p' \rangle = e^{-i \frac{p^2}{2mt}} \delta(p'-p)