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I'm reading the book on Quantum Field Theory by Anthony Duncan, and I'm a little lost with something of propagators.

He first define the propagator $K(q_f,T;q_i,0)$ as the amplitude of detecting a particle that initially was at $q_i$ at time 0 at another place $q_f$ at some time t, ie:

$$K(q_f,t;q_i,o) = <q_f | e^{-iHt} |q_i>$$

And then for the simple harmonic oscillator:

$$H = \frac{p^2}{2m} + \frac{1}{2}mw^2 q^2$$

He says that the propagator satisfies the differential equation:

$$i \frac{\partial}{\partial t} K(q_f,t;q_i,0) = -\frac{1}{2m}\frac{\partial^2 K(q_f,t;q_i,0)}{\partial q^2_f} + \frac{1}{2}mw^2 q_f^2 K(q_f,t;q_i,0)$$

And I've no clue were that equation comes from. I'm guessing it's a Schrodinger like equation but $K$ is an amplitude not a state ket so I'm lost.

Does anyone has a clue?

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    $\begingroup$ From one perspective, you just ask how $\frac{∂}{∂t}\langle x|e^{At}|y\rangle=\langle x|Ae^{At}|y\rangle=A\langle x|e^{At}|y\rangle$ should be made sense of. Is the case without the term $\frac{p^2}{2m}$ clear? $\endgroup$ – Nikolaj-K Sep 13 '15 at 19:29
  • $\begingroup$ Related: physics.stackexchange.com/q/65489/2451 $\endgroup$ – Qmechanic Sep 13 '15 at 20:27
  • $\begingroup$ Yes, now it's really clear! $\endgroup$ – Jasimud Sep 13 '15 at 23:41
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The matrix element $\langle x |\psi\rangle$ where $x$ is allowed to vary is just the wavefunction

$$\psi(x) = \langle x |\psi\rangle.$$

So the matrix element $\langle x |p|\psi\rangle$ is just the wavefunction representation of the state $p|\psi\rangle$. We know that the momentum operator in the wavefunction representation acts as a derivative, so

$$\langle q_f |p|\psi\rangle = -i\frac{\partial}{\partial q_f}\psi(q_f).$$

Also $|q_f\rangle$ is an eigenstate of the $q$ operator (which is self-adjoint) so

$$\langle q_f |q|\psi\rangle = q_f \,\psi(q_f).$$

Now to show the differential equation you are looking for just take $|\psi\rangle=e^{-iHt}|q_i\rangle,$ and notice

$$i\frac{\partial}{\partial t}\langle q_f|\psi\rangle = \langle q_f|H|\psi\rangle = \frac{1}{2m}\langle q_f|p^2|\psi\rangle + \frac{1}{2}m\omega^2 \langle q_f|q^2|\psi\rangle.$$

Now apply the above identities twice.

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  • $\begingroup$ I'd say the second equation with the derivative is not so evident, is it? $\endgroup$ – Nikolaj-K Sep 13 '15 at 22:38
  • $\begingroup$ And whats the best way to solve the partial differential equation for the propagator K? $\endgroup$ – Jasimud Sep 14 '15 at 15:46
  • $\begingroup$ Well, it is the unique fundamental solution of the Schr eqn for the oscillator, with the canonical initial condition for the propagator (δ fctn for t=0) so you instantly get the Mehler kernel in a dozen different ways, as per WP article $\endgroup$ – Cosmas Zachos Jan 17 '17 at 1:27

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