I am considering the infinite square well $[0,a]$, where I know the stationary states to be given by $\psi_n(x)=\sqrt{\frac{2}{a}}\sin(\frac{n\pi x}{a})$. Then, using Plancherel's theorem to translate to momentum space, $$\phi_n(k)=\frac{1}{\sqrt{\pi a}}\int_0^a\sin\left(\frac{n\pi x}{a}\right)e^{-ikx}dx=\frac{1}{\sqrt{\pi a}}\frac{a}{a^2k^2-n^2\pi^2}e^{-iak}(n\pi(-e^{iak})+iak\sin(n\pi)+n\pi\cos(n\pi))$$ where I omitted the actual computation of the integral, it's just applying integration by parts twice and then solving for the integral expression. This then gives $$\phi^*_n(k)=\frac{1}{\sqrt{\pi a}}\frac{a}{a^2k^2-n^2\pi^2}e^{iak}(n\pi(-e^{-iak})-iak\sin(n\pi)+n\pi\cos(n\pi))$$. So far so good (I think - if I made an error, please let me know). Regardless of the fact that the expression for $\phi^*_n\phi_n$ is now going to become quite convoluted, I am not sure how to compute the expectation of the momentum from this, since momentum is supposed to be an operator in quantum mechanics, and I have not worked with momentum space before. Does it simply become $\langle p\rangle=\int\hbar k\phi^*_n\phi_ndk$, because $p=\hbar k$?
I would appreciate it if someone could let me know whether my approach to this problem is correct, and if it is not, how else to proceed. I do not want a full solution.

  • $\begingroup$ So you are just asking what the momentum operator is in the momentum basis? $\endgroup$ – BioPhysicist Sep 14 '20 at 22:31
  • $\begingroup$ Also whether the way in which I obtained $\phi_n(k)$ is correct, because the book only states the result for going from $\psi$ to $\phi$, not specifically the stationary states $\psi_n$ to $\phi_n$. $\endgroup$ – Quaere Verum Sep 14 '20 at 22:32

Since we know that $P|p\rangle=p|p\rangle$, and if we express the momentum wave function of our state $|\psi\rangle$ as $\langle p|\psi\rangle=\phi(p)$, then we can easily determine what the momentum operator does when expressed in its own eigenbasis:

$$\langle p|P|\psi\rangle=p\langle p|\psi\rangle=p\phi(p)$$

In other words, it serves to just multiply the momentum wave function by $p$.

This can then easily show you how to find the momentum expectation value:

$$\langle p\rangle=\langle\psi|P\psi\rangle=\iint\langle \psi|p'\rangle\langle p'|P|p\rangle\langle p|\psi\rangle\,\text dp\,\text dp'=\iint p\phi^*(p')\langle p'|p\rangle\phi(p)\,\text dp\,\text dp'=\iint p\phi^*(p')\delta(p'-p)\phi(p)\,\text dp\,\text dp'=\int p\phi^*(p)\phi(p)\,\text dp$$

  • $\begingroup$ Bra-ket notation is somewhat foreign to me, but I think I can follow along and see how it is indeed just multiplication by $p$. Thank you, that's instructive. As for the other half of my question, was my usage of Plancheret's theorem justified? $\endgroup$ – Quaere Verum Sep 14 '20 at 22:46
  • $\begingroup$ @QuaereVerum Yes; it is just a change of basis that applies to any state $\endgroup$ – BioPhysicist Sep 14 '20 at 22:52
  • $\begingroup$ Thanks. The mathematical formalism is rather absent in the early chapters of Griffiths, so I was not too sure about what operations are "legal". $\endgroup$ – Quaere Verum Sep 14 '20 at 22:54

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