Inverse Fourier Transfrom of a wavefunction

Question

I was reading about how a Fourier transform yields the wave-function expressed in terms of the momenta which constitute it, i.e. the wave-function in momentum space.

I'm not so good at calculus yet so I couldn't do this yet: what would happen, qualitatively, if we took the inverse Fourier transform of a wave-function? I figure it would yield something which is constituted by the probability amplitudes of a wave-function, does that mean it will yield 1?

Relevant information: http://en.wikipedia.org/wiki/Fourier_transform

http://en.wikipedia.org/wiki/Wave_function

Answer 1

Let's drop the scaling constants and say that the wavefunction in momentum co-ordinates $\psi_p(\vec{p})$ is the FT of that $\psi_x(\vec{x})$ in position co-ordinates , i.e.

$$\psi_p(\vec{p}) = \mathfrak{F}_{\vec{p}}(\psi_x(\vec{x}))$$

where $\mathfrak{F}$ is the Fourier transform. Now you propose to take the inverse Fourier transform of $\psi_x$. The inverse FT is simply the same as the Fourier transform but with the sign of the transform variable switched. So we have, from the equation above:

$$\mathfrak{F}_{\vec{p}}^{-1}(\psi_x(\vec{x}))=\mathfrak{F}_{-\vec{p}}(\psi_x(\vec{x})) = \psi_p(-\vec{p})$$

So your operation simply calculates the state written in momentum co-ordinates, but with all the co-ordinates multiplied by $-1$. In one dimension, this would simply mean reflexion of the function $\psi_p(p)$ in the ordinate axis.