The propagator of a quantum system is defined by $$\mathcal{K}(t,x;\,t_{0},x_{0})\,\equiv\,\left\langle x\right|\hat{U}(t,\,t_{0})\left|x_{0}\right\rangle.$$ In this notation, the unitarity demands that

\begin{equation}\begin{split} &\hat{U}^{\dagger}(t,\,t_{0})\,\hat{U}(t,\,t_{0})\,=\,\left(\left|x_{2}\right\rangle \left\langle x_{2}\right|\hat{U}(t,\,t_{0})\left|x\right\rangle \left\langle x\right|\right)^{\dagger}\,\left|x_{1}\right\rangle \left\langle x_{1}\right|\hat{U}(t,\,t_{0})\left|x_{0}\right\rangle \left\langle x_{0}\right|\\ &=\,\left|x\right\rangle \mathcal{K}^{*}(t,x_{1};\,t_{0},x)\,\mathcal{K}(t,x_{1};\,t_{0},x_{0})\left\langle x_{0}\right|=\boldsymbol{\hat{1}}. \end{split}\end{equation}

with implicit summation over any repeated index. So, unitarity is translated directly to the condition

\begin{equation} \int_{-\infty}^{\infty}dx_{1}\mathcal{K}^{*}(t,x_{1};\,t_{0},x)\,\mathcal{K}(t,x_{1};\,t_{0},x_{0})\,=\,\delta(x-x_{0}), \end{equation}

for any given $t$ and $t_{0}$ (the last condition can also be found in W. Dittrich and M. Reuter, Classical and Quantum Dynamics, 6th ed, 2020; eq. (19.68)). However, for the Harmonic oscillator,

\begin{equation}\begin{split} &\mathcal{K}(t,x_{1};\,t_{0},x_{0})=\,\sqrt{\frac{m\omega}{2\pi i\sin[\omega(t-t_{0})]}}\exp\left(\frac{im\omega^{2}}{2\sin[\omega(t-t_{0})]}\left[(x_{0}^{2}+x_{1}^{2})\cos[\omega(t-t_{0})]-2x_{1}x_{0}\right]\right). \end{split}\end{equation}

By using the above result,

\begin{equation}\begin{split} &\int_{-\infty}^{\infty}dx_{1}\mathcal{K}^{*}(t,x_{1};\,t_{0},x)\,\mathcal{K}(t,x_{1};\,t_{0},x_{0})=\,\frac{m\omega}{2\pi\sin[\omega(t-t_{0})]}\int_{-\infty}^{\infty}dx_{1}\exp\left(\frac{im\omega^{2}}{2\sin[\omega(t-t_{0})]}\left[(x_{0}^{2}-x^{2})\cos[\omega(t-t_{0})]+2x_{1}(x-x_{0})\right]\right). \end{split}\end{equation}

How can I show that the above relation is satisfied? I noticed that there are other posts discussing similar issues, but I'm particularly interested in $t>t_{0}$, which does not seems to be discussed.

  • $\begingroup$ The condition you are trying to prove is wrong. An integration over all $x_1$s is warranted, as you missed the forrest for the trees. Simplify m=$\hbar$=ω=1, rationalize, and skip superfluous circular steps. The answer is a triviality, if only you take care to set up your problem properly. It won't cost you a thing to take $t_0=0$. WP. $\endgroup$ Sep 26, 2023 at 19:32
  • $\begingroup$ @CosmasZachos: can you please explain why this condition that I'm trying to prove is wrong? it seems to be a trivial requirement. $\endgroup$
    – Yair
    Sep 26, 2023 at 19:48
  • $\begingroup$ @Buzz: No, this is what you get by multiplying two propagators. Half of the terms cancel due to the conjugation so it looks similar.. There was indeed a typo which I corrected. Thanks. $\endgroup$
    – Yair
    Sep 26, 2023 at 20:19
  • 1
    $\begingroup$ Now that you adduced the intermediate point integration, it's right, and toboggans to the simple answer breathlessly. See answer. $\endgroup$ Sep 26, 2023 at 21:26

1 Answer 1


You are essentially there: $$ \int_{-\infty}^{\infty}dx_{1}\mathcal{K}^{*}(t,x_{1};\,t_{0},x)\,\mathcal{K}(t,x_{1};\,t_{0},x_{0})\\ = \exp\left(\frac{im\omega}{2\sin[\omega(t-t_{0})]} (x_{0}^{2}-x^{2})\cos[\omega(t-t_{0})] \right) \\ \times \int_{-\infty}^{\infty}\!\!dx_{1}~~\frac{m\omega}{2\pi\sin[\omega(t-t_{0})]}~\exp\left(\frac{im\omega}{\sin[\omega(t-t_{0})]} x_{1}(x-x_{0}) \right). $$ The last line, the integral, is a plain Dirac δ, in turn trivializing the middle line, which is thus 1, netting you $$ =\delta(x-x_0). $$ You must not let the trees make you miss the forest. You had a superfluous $\omega$ in the exponent, which I corrected.

Clarification w.r.t. comments and draft: It is evident from the sophisticated draft you sent me that the question operates on a higher, unacknowledged, level of qualms. The simple-minded, conventional answer here assumes the customary $\lim_{\epsilon\to 0} \int\!\!\frac{dx}{\epsilon}~ e^{-ixy/\epsilon} =2\pi \delta (y)$ which clearly troubles you. This is a far more sophisticated sub-question which possibly merits its own, self-standing question.

  • $\begingroup$ I don't think that what you say is correct: there is clearly a time dependent coefficient. $\endgroup$
    – Yair
    Sep 26, 2023 at 21:27
  • $\begingroup$ ? where? have you absorbed it to the dummy integration variable? $\endgroup$ Sep 26, 2023 at 21:28
  • $\begingroup$ There is nothing that cancels the exponent in the second line that you wrote in your answer. $\endgroup$
    – Yair
    Sep 26, 2023 at 21:30
  • $\begingroup$ ? The prefactor of the cosine is collapsed to 0 by the δ, no? $\endgroup$ Sep 26, 2023 at 21:31
  • $\begingroup$ Not identically: there is a measure 0 set of times in which it won't. $\endgroup$
    – Yair
    Sep 26, 2023 at 21:36

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