Is a finite function with finite Fourier transform possible?

[Cruikshank]

Clarification: I have seen in quantum mechanics many examples of wavefunctions and their Fourier transforms. I understand that a square pulse has a Fourier transform which is nonzero on an infinite interval. I am curious to know whether there exists any function which is nonzero on only a finite interval, whose Fourier transform is also nonzero on only a finite interval. Is it impossible? I have not been able to find any theorem about it. If an example exists I would be very interested in seeing it because I think it would make a great wavefunction to think about.

 

[Mute]

Hm, I haven't seen any theorems describing exactly this situation, but I suspect it isn't possible. The frequency and time (or momentum and position) variables of the Fourier transform are conjugate variables whose widths are roughly inversely to each other - that is, the more peaked a function is in one variable (e.g., time), the wider the corresponding transform will be. (For an example, a delta function is in some sense "infinitely peaked", and its Fourier transform is a constant which is infinitely wide, and vice versa).

In fact, Fourier transform variables are subject to an uncertainty principle. See here.

I suspect that if you had a function with support only on a finite interval and you assumed that the Fourier transform also only had finite support that you would violate the uncertainty inequality, but I don't know for sure.

 

[A. Neumaier]

Clarification: I have seen in quantum mechanics many examples of wavefunctions and their Fourier transforms. I understand that a square pulse has a Fourier transform which is nonzero on an infinite interval. I am curious to know whether there exists any function which is nonzero on only a finite interval, whose Fourier transform is also nonzero on only a finite interval. Is it impossible? I have not been able to find any theorem about it. If an example exists I would be very interested in seeing it because I think it would make a great wavefunction to think about.

It is not possible. The Fourier transform of a function with bounded support has necessarily unbounded support.