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The exact propagator for the harmonic oscillator in atomic units in the path integral formulation is given by $$K(x_b,t_b;x_a,t_a) = \left( \frac{m\omega}{2 \pi i \sin(\omega t)} \right)^{1/2} \exp({iS_{\mathrm{cl}}}) \tag{1} $$ where $$ S_{\mathrm{cl}} = \frac{m\omega}{2\sin(\omega t)} \left( (x_b^2 + x_a^2) \cos(\omega t)-2 x_a x_b\right) \tag{2} $$ and $t = t_b - t_a$. The time evolution of the wave function is given by $$ \psi(x_b,t_b) = \int K(x_b,t_b;x_a,t_a) \psi(x_a,t_a) dx_a \tag{3} $$ See Commun. Comput. Phys. vol. 18 no. 1 pp. 91-103 (pdf). When $t_b \to t_a$ does the quantity (1) tend to infinity? Does this mean that the final wavefunction $\psi(x_b,t_b)$ gets farther from the initial wavefunction $\psi(x_a,t_b)$ when the time interval approaches zero?

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No, that would not be sensible. Consistency requires that as $t\to 0$, $K(x_b,t_b,x_a,t_a) \to \delta(x_b-x_a)$. And indeed this is true.

In the limit of small $t$, the propagator becomes $$ K(x_b,x_a;t) \simeq \sqrt{\frac{m}{2\pi it}}\exp\left\{-\frac{m}{2it}(x_b-x_a)^2\right\} $$ which by itself is perfectly divergent as you say. It is however a well-known representation of the delta function. You may be familiar with the real version obtained by sending $t\to -it$.

Remember that the propagator is to be integrated against, and delta functions only make sense under an integral.
To show that this sequence is indeed a $\delta$-function, you could employ a Stationary Phase Approximation to the integral.

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  • $\begingroup$ Does this affect the numerical computation of integral (3) with Monte Carlo integration when $t_b - t_a$ is small? I.e. does this kind of numerical integration give sensible results with small $t_b-t_a$? $\endgroup$
    – tohoyn
    Commented Mar 26, 2021 at 9:09

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