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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?

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    $\begingroup$ It looks meaningless to me. Perhaps a typo? Can you give some context or describe the problem in more detail? $\endgroup$ – mike stone Mar 20 '19 at 12:21
  • $\begingroup$ 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? $\endgroup$ – Cosmas Zachos Mar 20 '19 at 22:47
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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

$$ e^{i\frac{\Delta p}{\hbar} \hat{x}} = \int dx\ |x\rangle e^{i\frac{\Delta p}{\hbar} x} \langle x| . $$

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