[nicwilson]

FT is a 90 degree rotation not a 180, the FT of the FT of a function is the mirror image about the origin, not the function itself.

[nextaccountic]

Ok you're right! I originally wrote down 90 degrees but then I had a conflicting view about being the inverse and then reasoned it must be 180 degrees

So the fourier transform of the fourier transform isn't the same as the inverse fourier transform? (ignoring the scaling bits that can be normalized I think), so I've been lied to?

Anyway here is a funny pair of questions

https://math.stackexchange.com/questions/1472528/why-is-the-...

https://math.stackexchange.com/questions/3922412/why-isnt-th...

[meindnoch]

90 degree rotation? That would imply that the fourier transform is orthogonal to the original function.

[nicwilson]

It is, because wavenumber and position are distinct variables and are orthogonal to each other. FT turn position into wavenumber (positional frequency) and wavenumber into negative position:

[ 0 1] [x] [ ω]

          = 

[-1 0] [ω] [-x]

see also https://en.wikipedia.org/wiki/Linear_canonical_transformatio...

the rotation matrix [[ 0 1], [-1 0]] is a 90 degree rotation.

[meindnoch]

Ok, now I see it!