# Fourier transform of the product of two wave functions

Good day, I have a wave function in the coordinate representation $$\Psi(x)$$. The Fourier transform I am interested in is: $$\int e^{-i a x} \Psi(x) \Psi^{\star}(x-b) dx,$$ where $$a$$, $$b$$ are some real, positive constants. I wonder if I can say something about the sign of this integral whithout concretization of the wave function?

The Bochner's theorem states that the Fourier transform is positive if it is taken from the positive semi-definite function. However I dont see a way to proof that $$\Psi(x) \Psi^{\star}(x-b)$$ is a positive semidefinite.

• Is $|Psi(x)$ just a scalar function and is the integral $\int dx$ just over the real axis? Or is it a more complicated integral over a group, as in the general description of Bochner's theorem: en.wikipedia.org/wiki/Bochner%27s_theorem ? Apr 16 at 9:56
• Yes, its simple integration over the real axes. I thought that the Bochner theorem can be used not only for the probability measure integrals.
– Kim
Apr 16 at 11:09
• This isn't directly related to quantum computing. It might make more sense to ask on physics SE. Apr 16 at 15:08

If we assume that $$\Psi(x)$$ is just a scalar function and the integral $$\int dx$$ is over the real axis, then the result can be any complex number. And if we consider more complicated cases, as in the description of Bochner's theorem that you mention (see Wikipedia,) then this will in general not be different.

For the simplest case, take a very simple (normalized) wave function: $$\Psi(x) = \pi^{-1/4}\,e^{-x^2/2}$$. The integral in question then gives us: \begin{align} & \int_{-\infty}^{\infty} \frac{dx}{\sqrt{\pi}} \, e^{-iax} \, e^{-x^2/2} \, e^{-(x-b)^2/2} = \int_{-\infty}^{\infty} \frac{dx}{\sqrt{\pi}} \, e^{-x^2+(b-ia)x-b^2/2} \\ & = \int_{-\infty}^{\infty} \frac{dx}{\sqrt{\pi}} \, e^{-(x-(b-ia)/2)^2 -(a^2+b^2+2iab)/4} \\ & \quad = \int_{-\infty}^{\infty} \frac{dy}{\sqrt{\pi}} \, e^{-y^2 -(a^2+b^2+2iab)/4} = \frac{\pi^{-1/2}}{e^{(a^2+b^2)/4}}\ e^{-i a b/2}, \end{align} where in the second line we just complete the square for $$x$$ and subsequently use the fact that a change in integration variable $$x\rightarrow x-(b-ia)/2$$ does not change the result for the integral from $$-\infty$$ to $$\infty$$. So this gives us a positive real number times a phase factor $$e^{-i a b/2}$$, which by choice of $$a$$ and/or $$b$$ can be given any desired phase so you can't say anything about the sign, the expression is in general not positive or negative.

Still, if this phase $$e^{-i a b/2}$$ would be present for any wave function $$\Psi$$, we could still say that the "complex sign" of this expression is known in advance, independent of $$\Psi$$. This is, however, surely not the case, since we can always define another wave function $$\chi(x) = \Psi(x-c)$$ and then the result for $$\chi$$ would be: \begin{align} & \int_{-\infty}^{\infty} dx\ e^{-iax}\ \chi(x)\ \chi^*(x-b) = \int_{-\infty}^{\infty} dx\ e^{-iax}\ \Psi(x-c)\ \Psi^*(x-c-b) \\ & = \int_{-\infty}^{\infty} dy\ e^{-ia(y+c)}\ \Psi(y)\ \Psi^*(y-b) \\ & = e^{-iac}\ \int_{-\infty}^{\infty} dy\ e^{-iay}\ \Psi(y)\ \Psi^*(y-b). \end{align} So by choosing $$c$$ we can make the result for $$\chi$$ differ by an arbitrary phase $$e^{-iac}$$ from the result for $$\Psi$$, without having to change $$a$$ or $$b$$. So the result cannot be predicted based on knowledge of $$a$$ and $$b$$.

• Thank you, I had the same impression, that nothing can be concluded.
– Kim
Apr 17 at 7:30
• It looks a bit like the Fourier transform of the autocorrelation (which is the spectral density, so there you would have a positive result). But that of course would have two integrals, to compute the autocorrelation and subsequently to do the Fourier transform, something like: $$\int db\ e^{-i a b} \int dx\ \Psi(x) \Psi^{\star}(x-b)$$ which is clearly something different. Apr 17 at 7:49
• Hm, the full rpoblem of mine is: $\frac{1}{2\pi}\int \phi(1;\mu,\nu)e^{i\mu\nu/2}e^{-i\mu x}\chi^{\star}(x)\chi(x-\nu)d\mu d\nu dx$, where $\phi(1;\mu,\nu)$ is a characteristic function of a distribution in point one. What you say is even sort of simmilar view.
– Kim
Apr 17 at 12:51
• Yes, but still in that case you can't split off a factor that is exactly a spectral density. And in any case the part you factored out in your original question doesn't give a specific sign. So the best path forward is still to be found... Apr 17 at 13:53