I have the propagator for the harmonic oscillator.

$$K(x_f,x_0,t)=\sqrt{\frac{m\omega}{2 \pi \hbar \sin{wt}}}\exp\left(\frac{i}{\hbar}\frac{m\omega}{2 \sin{\omega t}}((x_0^2+x_f^2)\cos\omega t-2x_0x_f)-\frac{i\pi}{4} \right)$$


$$K(x_f,x_0,t)=<x_f |e^{-\frac{i}{\hbar}\hat{H}t} |x_0 > $$

and the Schrodinger equation

$$\hat{H}\psi(\textbf{q},t)=i\hbar \frac{\partial}{\partial t}\psi(\textbf{q},t) $$

This may seem silly but I just cannot see from my notes how you can the propagator satisfies this Schrodinger equation. I think it may come from some lack of understanding as to what the propagator means.

  • $\begingroup$ How did you derive the propagator? $\endgroup$
    – Ryan Unger
    Feb 4, 2015 at 23:43
  • $\begingroup$ Using the action of the Lagrangian, and a couple of clever tricks. This is stuff from lecture notes $\endgroup$
    – Trajan
    Feb 4, 2015 at 23:48
  • $\begingroup$ For a general proof that path integrals reproduce the Schroedinger equation, see physics.stackexchange.com/q/163075 $\endgroup$
    – Ryan Unger
    Feb 4, 2015 at 23:52
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    $\begingroup$ @ACuriousMind: I think he means "satisfy" in the sense that the convolution of the propagator with an initial solution yields another solution. $\endgroup$
    – Ryan Unger
    Feb 5, 2015 at 0:50
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    $\begingroup$ The propagator doesn't follow the Schroedinger equation, it obeys $\psi(x,t)=\int\psi(x',t')K(x,t;x't')dx'$. To check your propagator, you need to solve that integral. $\endgroup$
    – Kyle Kanos
    Feb 8, 2015 at 15:29

1 Answer 1


The propagator does satisfy the Schrodinger equation for most values of x and t...

The easiest way to show this is to let $\hbar=m=\omega=1$. Also, we can work with $\sqrt{2\pi}e^{i\pi/4}K\to K$ instead of $K$ to clean up the mess a little further. Further, let $x_f\to x$ and $x_0 \to 0$.

Then let: $K=fe^{ig}$ with




And, since $m=\omega=1$, we want to show that:

$i\dot K = -\frac{1}{2}K''+\frac{x^2}{2}K$


$\dot K=\frac{K}{2}\left({\frac{-\cos(t)}{\sin(t)}-i\frac{x^2}{\sin^2(t)}}\right)$



And... I'll leave the rest up to you. But, I assure you, you will find that: $HK=i\dot K$

...of course, this is not actually, true since actually: $HK=i\dot K+i\delta(x_f-x_0)\delta(t)$

And... I'll leave it up to you to reconcile this issue...

  • $\begingroup$ Sorry its not clear "for most values". I was expecting this to be some kind of "simple" substitution. Your step $\hbar=m=\omega=1$ is very useful. $\endgroup$
    – Trajan
    Feb 8, 2015 at 14:40
  • $\begingroup$ Ah, I said "for most values" because K only really satisfies the schrodinger equation if the delta functions (see the very last equation) are zero. For example, you have problems at "t=0". $\endgroup$
    – hft
    Feb 9, 2015 at 2:15