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.


1 Answer 1



  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.

  • 1
    $\begingroup$ 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, 2014 at 19:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.