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The unnormalized ground state of the harmonic oscillator (choosing units such that $m = \hbar = \omega = 1)$ is $$\tag{1}\psi(q,t) = \exp(-q^2/2-it/2).$$

The transition function is $$\tag{2}W(q_2,t_2,q_1,t_1) = \dfrac{ 1}{ \sqrt{2\pi i S} }\exp \left[ \dfrac{i}{2S}\left((q_1^2 + q_2^2)C - 2q_1 q_2 \right) \right],$$ where $S = \sin(t_2 - t_1)$ and $C = \cos(t_2 - t_1)$.

From general considerations, we should have

$$\tag{3}\psi(t_2,q_2) = \int_{-\infty}^{\infty}\! dq_1\, W(q_2,t_2,q_1,t_1)\psi(t_1,q_1).$$

Can we also show this by calculating the integral explicitly for the given state? My attempts at this have failed; in particular, I never get the correct time dependence $\propto \exp(-it_2/2)$ in the end result.

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Hints:

  1. OP's exercise is essentially a matter of checking an oscillatory Gaussian integral (3) over the initial position $q_1$.

  2. Let $\Delta t_M:=t_2-t_1>0$. To render the Gaussian integral convergent, insert Feynman's $i\epsilon$ prescription $\Delta t_M\to\Delta t_M-i\epsilon$. Or equivalently, Wick-rotate $\Delta t_E:=i\Delta t_M$, where ${\rm Re}(\Delta t_E)>0$. Here the letters $M$ and $E$ stands for Minkowski and Euclid, respectively.

  3. Note that $iS:=i\sin\Delta t_M=\sinh\Delta t_E$ and $C:=\cos\Delta t_M=\cosh\Delta t_E$.

  4. Perform the convergent Gaussian integral (3) over the real variable $q_1$.

  5. After the Gaussian integration, the new square root factor $\dfrac{ 1}{ \sqrt{(C+i S}) }$ will yield the sought-for $t_2$ dependence.

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    $\begingroup$ Comment to the answer (v4): The answer assumes implicitly that we haven't passed the first caustic/turning point $\Delta t_M < \pi $. $\endgroup$ – Qmechanic Sep 24 '14 at 19:30

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