1
$\begingroup$

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?

$\endgroup$
2
  • 1
    $\begingroup$ 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. $\endgroup$
    – Philip
    Feb 2, 2021 at 6:51
  • $\begingroup$ 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? $\endgroup$ Feb 2, 2021 at 6:55

1 Answer 1

5
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.