I was doing this question:

Using $\left< x \middle| p\right> = \frac{1}{\sqrt{2 \pi \hbar}}e^{ipx/\hbar}$ show that:

$$ \left<x \middle| \hat{p} \middle| \psi \right> = -i\hbar \frac{d}{dx} \left< x \middle| \psi\right> $$ for a general $\psi$.

Method 1 (how my lecturer did it)

\begin{align*} \left<x \middle| \hat{p} \middle| \psi \right> &= \int dp \, \left<x \middle| p\right> \left<p \middle| \hat{p} \middle| \psi \right> \\ &= \int dp \, p \frac{1}{\sqrt{2 \pi \hbar}}e^{ipx/\hbar} \left<p \middle|\psi\right> \\ &= \int dp \, (-i\hbar) \frac{d}{dx}\left<x \middle| p\right>\left<p\middle|\psi\right> \\ &= -i\hbar \frac{d}{dx}\left<x \middle|\psi\right> \end{align*}

Here I want to ask:

  1. Why do put it in integral form? (See below why I think it's unnessessary)
  2. Why are we allowed to swap the operator order like we did in line 3?

Method 2 (how I did it seeing that we can just swap the order)

\begin{align*} \left<x \middle| \hat{p} \middle| \psi \right> &= \left<x \middle| p\right> \left<p \middle| \hat{p} \middle| \psi \right> \\ &= \hat{p} \left<x \middle|p\right>\left<p\middle|\psi\right> \\ &= \hat{p} \left<x \middle| \psi\right> \\ &= - i \hbar \frac{d}{dx} \left< x\middle| \psi\right> \end{align*}

I don't understand why putting it in integral form is even correct?

  • $\begingroup$ Method 2 is an incorrect version of Method 1... $\endgroup$ – Phoenix87 May 15 '15 at 11:02
  • $\begingroup$ Can you expand? How so? $\endgroup$ – turnip May 15 '15 at 11:02
  • 1
    $\begingroup$ Method 1 is incorrect anyway. On the third line, the $\hat{p}$ should just be $p$, i.e. the eigenvalue of the operator $\hat{p}$ corresponding to the eigenvector $\lvert p\rangle$. However Method 2 is doubly incorrect, since you need to integrate over all $p$ to use the resolution of identity $1 = \int\mathrm{d}p\,\lvert p \rangle\langle p \rvert$. $\endgroup$ – Mark Mitchison May 15 '15 at 11:06
  • $\begingroup$ You're right, that was just me mistyping it. I'll edit it. $\endgroup$ – turnip May 15 '15 at 11:07
  • 1
    $\begingroup$ @PPG So do you understand now why there is no "swapping of operator order"? $p$ is not an operator. $\endgroup$ – Mark Mitchison May 15 '15 at 11:09

Your lecturer got the eigenvalue using the fact that the operator $\hat{p}$ is Hermitian so you can do this:

\begin{align} \langle p| \hat{p} &= \left( \hat{p}^\dagger |p\rangle\right)^\dagger\\ &= \left( \hat{p} |p\rangle\right)^\dagger\\ &= \left( p |p\rangle\right)^\dagger\\ &= \langle p| p \end{align}

I think it becomes a bit neater if you put the projector $|p\rangle\langle p|$ after the operator $\hat{p}$ because then you don't have to do the gymnastics with the Hermitian conjugate. This would be my attempt (being as explicit as possible at each step):

\begin{align} \langle x|\hat{p} |\psi\rangle &= \int dp \langle x|\hat{p} |p\rangle\langle p|\psi\rangle\\ &= \int dp p \langle x |p\rangle\langle p|\psi\rangle\\ &= \int dp p e^{\frac{i}{\hbar}xp}\langle p|\psi\rangle\\ &= \int dp \left(-i\hbar \frac{d}{dx}\right) e^{\frac{i}{\hbar}xp}\langle p|\psi\rangle\\ &= \left(-i\hbar \frac{d}{dx}\right)\int dp e^{\frac{i}{\hbar}xp}\langle p|\psi\rangle\\ &= \left(-i\hbar \frac{d}{dx}\right)\int dp \langle x|p\rangle\langle p|\psi\rangle\\ &= \left(-i\hbar \frac{d}{dx}\right)\langle x|\psi\rangle\\ \end{align}

Note that it is the number $p$ that is brought to the front in line two not the operator $\hat{p}$. We choose the states $|p\rangle$ so that we have the following equality: $\hat{p}|p\rangle = p|p\rangle$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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