here is my question:

Im trying to find the matrix elements of the momentum operator $\hat{P}$ in the position basis $|x\rangle$:

$$\langle x|\hat{P}|x'\rangle := \langle x|\hat{P}\int dp|p\rangle \langle p|x'\rangle=\langle x|\hat{P}\int dp|p\rangle \frac{1}{\sqrt {2\pi}}e^{-ipx'}$$

where $\hbar=1$. Now acting $\hat{P}$ on the ket $|p\rangle$ and multiplying by the bra $\langle x|$:

$$\tag{1}\langle x|\hat{P}|x'\rangle :=\frac{1}{2\pi}\int dp\space p e^{ip(x-x')}$$

so according to some texts (and remembering QM lessons) the matrix elements should be: $$\tag{2}\langle x|\hat{P}|x'\rangle:=-i\frac{d}{dx}\delta(x-x')$$

so if im not wrong this should mean that (1) its equal to (2). Are these calculations right? How is the integral equal to the expression (2)? Does someone knows some book where they develop the momentum operator in the x basis formally?


2 Answers 2


I'm proceeding the same way as you did: $$\hat{p}|x'\rangle=\int\hat{p}|p\rangle\langle p|x'\rangle dp=\int p|p\rangle\frac{1}{\sqrt{2\pi}}e^{-ipx'}dp$$ $$\langle x|\hat{p}|x'\rangle=\int p\langle x|p\rangle\frac{1}{\sqrt{2\pi}}e^{-ipx'}dp=\frac{1}{2\pi}\int e^{ip(x-x')}pdp$$ $$\langle x|\hat{p}|x'\rangle=-i\frac{d}{dx}\left(\frac{1}{2\pi}\int e^{ip(x-x')}dp\right)$$ Here We will use the identity $$\delta(x'-x)=\delta(x-x')=\frac{1}{2\pi}\int dk \ e^{ik(x'-x)}$$ Which lead to $$\langle x|\hat{p}|x'\rangle=-i\frac{d}{dx}\delta(x-x')$$

I think R. Shankar's book has good proof.

Here $\hbar$ is taken to be 1.


This is a well-known result, though I don't believe it's usually introduced in most introductory level quantum mechanics texts like Griffiths or Shankar. To answer your questions:

  1. Yes, $(1)$ is equal to $(2)$, though you should know that "check-my-work" type problems are generally considered off-topic on this site.

  2. It is equal in the sense of two distributions being equal. In other words, it is equal in the same sense that $$\delta(x) = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^\infty \, \text{d}p \,\,e^{-i px},$$

  3. This topic is usually covered under the more general case of representations of observables with continuous eigenvalues in many books, including Sakurai's Modern Quantum Mechanics (Section 1.6: Position, Momentum, and Translation, a section that has hundreds of questions related to it on this site) and Dirac's Principles of Quantum Mechanics (Chapter III: Representations). But keep in mind, as Sakurai warns, that "the rigorous mathematics of a vector space spanned by eigenkets that exhibit a continuous spectrum is rather treacherous."


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.