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I'm trying to calculate $$<0|e^{a\hat{x}^2}e^{b\hat{x}}e^{c\hat{p}}|0>$$ where $a$, $b$ and $c$ are complex numbers, $\hat{x}$ is the position operator, $\hat{p}$ is momentum operator and $|0>$ refers to vacuum state (i.e. Fock state with 0 photons).

So far, I have tried direct expansion of exponential in Taylor series, playing with Baker-Campbell-Hausdorff sort of formulae, and trying to recognize known states, such as squuezed states in the formula I'm asking about. This efforts were unsuccessful, though. Any ideas?

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closed as off-topic by AccidentalFourierTransform, Kyle Kanos, ZeroTheHero, sammy gerbil, Jon Custer May 31 '17 at 1:24

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Assuming that all quantities are written in units of the harmonic oscillator natural length $\sqrt{\hbar/m\omega}$, and using the ground-state wavefunction $\psi_0(x)=\frac{1}{\pi^{1/4}}e^{-x^2/2}$, one can perform the average by inserting enough resolution of the identity with respect to $x$ and $p$, to obtain (whether all constants are correct is left as an exercise to the reader) : $$\langle 0|e^{a\hat x^2}e^{b\hat x}e^{c\hat p}|0\rangle=\frac{e^{\frac{b^2+c^2-2 i b c-2 a c^2}{4(1-a)}}}{\sqrt{1-a}}.$$

Note that this average is valid only if $a<1$.

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  • $\begingroup$ why the downvotes ? $\endgroup$ – Adam May 31 '17 at 7:42

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