1
$\begingroup$

I was solving a problem from the problem sets provided by MIT OCW. Here's the problem set.

I was stuck in the problem 2(g)

enter image description here

So, I looked at the solution to the problem and couldn't understand it.

enter image description here

Link to the entire solution set.

Things which I didn't get from the solution provided:

  1. The Fourier transform should be: $$\tilde{f}(k) = \frac{1}{\sqrt{2\pi}}\int dx e^{-ikx}f(x)$$ There should be $dx$ but in the solution it is written $dk$.

  2. If, say, there is a typo, and it should be $dx$, then we can't take the position operator inside the integral, now how to approach the problem.

  3. And if it isn't a typo ( I need to learn more!!!), how were the following calculations obtained. $$\hat{x}e^{-ikx} = i\frac{\partial}{\partial k}e^{ikx}$$ and then in the next step there is again a minus sign in the exponential.

$\endgroup$
2
  • 2
    $\begingroup$ It seems that there should be a $\mathrm dx$, indeed. But where is the problem? $\endgroup$ Commented Jun 8, 2022 at 7:06
  • $\begingroup$ Possibly related /of interest: this and the links therein. $\endgroup$ Commented Jun 8, 2022 at 7:08

1 Answer 1

3
$\begingroup$

I'll try my best to answer each of your questions in the following:

  1. Yes, it should be $dx$ not $dk$ (it's a typo)
  2. No, you can still take the position operator inside the integral. This is probably a bit clearer if we re-write the steps as follows \begin{align} \hat x \tilde f(k) &= \hat x \int dx' e^{-ik x'} f(x')\\ &= \int dx' \hat x e^{-ik x'} f(x')\\ &= \int dx' x' e^{-ik x'} f(x') \end{align}
  3. The missing minus sign in the exponential is a typo. The solutions are using the following: $$i \frac{\partial}{\partial k}e^{-ikx} = i (-ix) e^{-ikx} = x e^{-ikx} = \hat x e^{-ikx} $$

Looks like the solutions just have a couple of typos. Otherwise you seem fine.


Throughout this answer I have just accepted the definition \begin{equation} \hat{x} f(x) = x f(x), \end{equation} This definition is a little loose, see my answer to another question here for more details.

$\endgroup$

Your Answer

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

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