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ehrenfest
Sep16-07, 12:31 AM
1. The problem statement, all variables and given/known data
Richard Robinett defined the Fourier transform with an exp(-ikx) and the inverse Fourier transform with an exp(ikx). I have always seen the opposite convention and I thought it was not even a convention but a necessity to do it the other in order to apply it to some Gaussian equations. Has anyone ever seen this sign convention before?




2. Relevant equations



3. The attempt at a solution

dextercioby
Sep16-07, 01:49 AM
It really isn't relevant whether it's with a plus, or with a minus. I've seen in most cases

\phi (x)=\frac{1}{(2\pi)^{3/2}}\int dk \ \tilde{\phi}(k) e^{-ikx} .

HallsofIvy
Sep16-07, 06:02 AM
Note that the integration is from -\infty to \infty. That's why the sign does not matter.

genneth
Sep16-07, 06:10 AM
Furthermore, the constants in front also do not really matter, as long as they combine to give 1/(2 pi). There are a couple of theorems which depend on them (I think Parseval's theorem and the associated ones do), but it's all up to a constant. My supervisor (in physics) recommends just ignoring the constants, and adding them back in if you have to at the end :wink:

Jimmy Snyder
Sep16-07, 09:11 AM
If you define the Fourier transform as dextercioby did:
\phi (x)=\frac{1}{(2\pi)^{3/2}}\int dk \phi(k) e^{-ikx}
then the inverse transform is:
\phi (k)=\frac{1}{(2\pi)^{3/2}}\int dx \phi(x) e^{ikx}
It is merely a matter of convention which is called which. There's no 'wrong' convention as long as you remain consistent.

On page 11 of 'Photons and Atoms' by Claude Cohen-Tannoudji et. al. the convention above is used. On page 97 of 'The Principles of QM' by P.A.M. Dirac, the convention is left deliciously ambiguous.
These formulas have elementary significance. They show that either of the representations is given, apart from numerical coefficients, by the amplitudes of the Fourier components of the other.