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In a text on path integrals, I find the following: \begin{equation} \langle q_{j+1}|e^{-i(\hat{p}/2m)\delta t}|q_j\rangle = \int\frac{dp}{2\pi}\langle q_{j+1}|e^{-i(\hat{p}/2m)\delta t}|p\rangle\langle p|q_j\rangle\ .\tag{1} \end{equation}

Shouldn't it be rather:

\begin{equation} \langle q_{j+1}|e^{-i(\hat{p}/2m)\delta t}|q_j\rangle = \langle q_{j+1}|e^{-i(\hat{p}/2m)\delta t}\int\frac{dp}{2\pi}|p\rangle\langle p|q_j\rangle\ ?\tag{2} \end{equation}

I would have inserted the resolution of identity as in (2), but then I wouldn't have dared to move the integral past the $e^{-i(\hat{p}/2m)\delta t}$ operator, because in (1), $e^{-i(\hat{p}/2m)\delta t}|p\rangle$ is a function of $p$, namely: $e^{-i(\hat{p}/2m)\delta t}|p\rangle=e^{-ip/2m\delta t}|p\rangle$ (no hat anymore). I am not sure about the interplay of that function of $p$ and the integral. Can someone shed some light on this?

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    $\begingroup$ Nothing to do with "operators". You can move an integral through anything that's not dependent on the variable that's integrated. $\hat{p}$ is not a function of $p$ (yeah, it's crappy notation), so you can just move the integral past. $\endgroup$
    – ACuriousMind
    Commented Nov 25, 2015 at 14:52
  • $\begingroup$ OK - then it's just that the notation is somewhat confusing to me... But wait - surely $e^{...}|p\rangle$ is a function of p? $\endgroup$
    – Frank
    Commented Nov 25, 2015 at 14:54

1 Answer 1

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Comments to the question (v1):

  1. The operator $f(\hat{p})$ and the identity operator ${\bf 1}= \int\!\frac{dp}{2\pi}|p\rangle\langle p|$ commute.

  2. The operator $f(\hat{p})$ and the integration $\int\!\frac{dp}{2\pi}$ are independent of each other.

  3. The $q$-ket and the $q$-bra are independent of the integration $\int\!\frac{dp}{2\pi}$.

  4. Of course, if one replaces $f(\hat{p})|p\rangle$ with $f(p)|p\rangle$, it does not make sense to move the function $f(p)$ [and, for that matter, the $p$-ket and the $p$-bra] outside of the $p$-integral.

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  • $\begingroup$ Qmechanic - but the commutativity would move the operator right in front of $|q_j\rangle$, starting from the second expression, not leading to the first expression, right? $\endgroup$
    – Frank
    Commented Nov 25, 2015 at 15:04
  • $\begingroup$ Also, isn't there a contradiction between 2. and 4.? $f(\hat{p})|p\rangle=f(p)|p\rangle$, so there is a function of $p$ at work here - you can't just "formally" ignore it, can you? It's not a matter of choosing to replace or not, it's an equality that must hold throughout (which I guess is the thrust of my initial question) $\endgroup$
    – Frank
    Commented Nov 25, 2015 at 15:08
  • $\begingroup$ Concerning your last comment, the resolution is that one is not allowed to substitute $f(\hat{p})\rightarrow f(p)$ outside of the $p$-integral, so one does not face a contradiction. $\endgroup$
    – Qmechanic
    Commented Nov 25, 2015 at 15:30
  • $\begingroup$ Maybe I can rephrase this: nothing is moving except for the integral sign. If $g$ is linear, as is everything in linear algebra, then $g(x+y) = g(x) + g(y)$, i.e. $g$ commutes with sums, which means $g$ commutes with integrals. Now let $g$ be everything in your expression before the integral sign. $\endgroup$
    – knzhou
    Commented Nov 25, 2015 at 16:02
  • $\begingroup$ Hence you can always move integrals and sums all the way to the left in anything you do in QM/QFT. It's all linear. (Note that this doesn't mean you can always move integrals to the right.) $\endgroup$
    – knzhou
    Commented Nov 25, 2015 at 16:03

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