# How to simplify this expression in Dirac notation

An expression cropped up in a homework problem that I'm not sure how to simplify. Consider the following, where $$|x\rangle$$ is a position eigenstate and $$|p_1\rangle, |p_2\rangle$$ are momentum eigenstates: $$\int dx |x\rangle\langle x|p_1\rangle\langle p_2|x\rangle\langle x|$$ If there were only a single $$|x\rangle\langle x|$$ projector, I'd use the completeness relation $$\int dx |x\rangle\langle x|=\hat{1}$$ and be done with it. But there are two projectors, so I'm not sure how to deal with this - they aren't separable, so far as I can see. I have also tried writing out the position-momentum overlaps explicitly, turning the expression to: $$\int dx e^{i(p_1-p_2)x/\hbar}|x\rangle\langle x|$$ which strongly resembles the usual integral form of the Dirac delta, but with the extra projector factor I don't know how to deal with.

How might I proceed with this?

• It looks meaningless to me. Perhaps a typo? Can you give some context or describe the problem in more detail? Commented Mar 20, 2019 at 12:21
• It's a dopey operator $\cal O$ whose matrix elements between $\langle x_a|$ and $|x_b\rangle$ are $e^{ix_a (p_1-p_2)/\hbar}\delta(x_b-x_a)$, but so what? Commented Mar 20, 2019 at 22:47

Recall that if we have a complete continuous eigenbasis $$\{|a\rangle\}$$ of $$\hat{A}$$, we can write a function of the operator $$\hat{A}$$ as
$$f(\hat{A}) = \int da\ |a\rangle f(a)\langle a|$$
where $$\hat{A}|a\rangle= a|a\rangle$$. So it looks like what you've got is
$$e^{i\frac{\Delta p}{\hbar} \hat{x}} = \int dx\ |x\rangle e^{i\frac{\Delta p}{\hbar} x} \langle x| .$$