(no subject)

[impossiblewizardry]

The Fourier transform of a standard Gaussian is itself. That is, the standard Gaussian is a fixed point of the Fourier transform.

That's how I thought about it, until I considered units, and I realized they're not really the same thing. Like, a Gaussian wavefunction could map positions to amplitidues in position space. Then, its Fourier transform maps wavenumbers to amplitudes in wavenumber space. This actually made more sense when, instead of considering a standard Guassian, I considered a Gaussian with standard deviation σ, since σ has a unit.

Let φ be a standard Gaussian wavefunction, with unitless input and output. Then the Gaussian wavefunction is

ψ(x) = (1/√σ) φ(x/σ)

The constant takes care of the units. Like, if x is in meters, σ is also in meters, and the input x/σ is unitless. To see that the output is in the right units, consider calculating a probability:

∫ ψ(x)² dx = ∫ (1/σ) φ(x/σ)² dx

The units in 1/σ and dx cancel, and the output is unitless.

Then, the Fourier transform of ψ(x) is

(F ψ)(k) = √σ φ(σ k)

(I use k to mean angular wavenumber.) So again the units in the input cancel. For example if k is in inverse meters, then σ k = meters * inverse meters is unitless.

So ψ and F ψ are not really the same function for units reasons, but they're both expressible in terms of this one unitless function φ.