Taking Fourier Transform to momentum space

Question

I have a wave function
$$psi(x) = frac{1}{sqrt{sigmasqrt{pi}}} exp left (frac{-x^2}{2 sigma^2} right ) exp left (frac{ipx}{hbar} right )$$
And I have to convert this to $Q(p)$, in momentum space, by taking the Fourier transform. So I use this to perform the transform:
$$Q(p) = frac{1}{sqrt{2pihbar}}int_{-infty}^{infty} psi(x) e^{-ipx/hbar} dx$$
However, when I do, the imaginary parts in the exponential cancel, and I'm left with a constant which would not yield the original wave function upon taking the inverse Fourier transform. How do I proceed in the right way?

Philip · Accepted Answer

The variable "$p$" in the definition of the wavefunction is not the same as the variable $p$ in the definition of the Fourier Transform. It's best if you call the first $p$ some other constant, say, $p_0$. This also makes sense physically, since it actually represents the expectation value of the momentum of the Gaussian wavepacket. (You should be able to show that $langle hat{p} rangle_psi = p_0$.)
If you do this, the integral reduces to $$Q(p) propto int_{-infty}^infty expleft( -frac{x^2}{2sigma^2}right) expleft( -frac{i (p-p_0) x}{hbar}right) text{d}x.$$
When you integrate this (by completing the square, etc.) you will get a function of $p$ which is the momentum space wavefunction.

Taking Fourier Transform to momentum space

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