# Behaviour of the path integral kernel $K(x_b,t_b;x_a,t_a)$ for the harmonic oscillator when $t_b \to t_a$

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?

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 $$t\to -it$$.
To show that this sequence is indeed a $$\delta$$-function, you could employ a Stationary Phase Approximation to the integral.
• 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 $t_b-t_a$? – Tommi Höynälänmaa Mar 26 at 9:09