3
$\begingroup$

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?

$\endgroup$
0
2
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ 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 $t_b-t_a$? $\endgroup$ – Tommi Höynälänmaa Mar 26 at 9:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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