# Taking Fourier Transform to momentum space

I have a wave function $$\psi(x) = \frac{1}{\sqrt{\sigma\sqrt{\pi}}} \exp \left (\frac{-x^2}{2 \sigma^2} \right ) \exp \left (\frac{ipx}{\hbar} \right )$$ 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: $$Q(p) = \frac{1}{\sqrt{2\pi\hbar}}\int_{-\infty}^{\infty} \psi(x) e^{-ipx/\hbar} dx$$ 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?

• The variable "$p$" in the definition of the Fourier Transform is not the same variable $p$ in the definition of the wavefunction. It's best if you call the variable $p$ in the wavefunction some other constant, say, $p_0$, since it actually represents the expectation value of the momentum of the gaussian wavepacket. – Philip Feb 2 at 6:51
• So if I were to do that, I can have an integral of the form $\int_{-\infty}^{\infty}\exp(-ax^2+bx)dx$, where $a=1/2\sigma^2$ and $b=(i/\hbar)(p_0-p)$, right? – AntimoniumHeptadiene Feb 2 at 6:55

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 $$Q(p) \propto \int_{-\infty}^\infty \exp\left( -\frac{x^2}{2\sigma^2}\right) \exp\left( -\frac{i (p-p_0) x}{\hbar}\right) \text{d}x.$$
When you integrate this (by completing the square, etc.) you will get a function of $$p$$ which is the momentum space wavefunction.