Here is a mystery I'm trying to understand. Let ##\hat{U}(t) = \exp[-i\hat{H}t]## is an evolution operator (propagator) in atomic units ([itex]\hbar=1[/itex]). I think I'm not crazy assuming that ##\hat{U}(-t)\hat{U}(t)=\hat{I}## (unit operator). Then I would think that the following should hold(adsbygoogle = window.adsbygoogle || []).push({});

[tex]\left\langle x\right|\hat{U}(-t)\hat{U}(t)\left|x^\prime\right\rangle = \left\langle x\right|\hat{I}\left|x^\prime\right\rangle = \left\langle x|x^\prime\right\rangle= \delta(x-x^\prime)[/tex]

However, when using the resolution of identity in the coordinate representation ##\hat{I}=\int\left|y\rangle\langle y\right|dy## and well known expressions for propagators (for free particle or harmonic oscillator) in the coordinate representation, I get a coordinate dependent phase factor in front of the delta function:

[tex]\left\langle x \right | \hat{U}(-t)\hat{U}(t) \left | x^\prime\right\rangle = \int dy\left\langle x\right|\hat{U}(-t)\left|y\right\rangle\left\langle y\right|\hat{U}(t)\left|x^\prime\right\rangle = e^{i\phi(x,x^\prime)}\delta(x-x^\prime)[/tex]

Could someone, please, comment on that? I'm lost although something tells me the explanation must be trivial.

THANKS!

P. S. The integral over y becomes a Fourier integral, so you can easily verify all this.

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# Product of two propagators U(-t)U(t) in coord representation

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