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The propagator $K$ from ($x_a,t_a$) to ($x_b,t_b$), as defined by Gottfried, can be written as $$ K(b,a) = F(t_b-t_a)\exp\left(\frac{i}{\hbar}S_{c}(b,a)\right) $$ where $S_c$ is the classical action and $F(t_b-t_a)$ is the integral over all paths from the origin and back, during the interval $t_b-t_a$, and is known as the prefactor in some literatures.

I've noticed that the prefactor for both the "regular" harmonic oscillator and the driven oscillator (with any arbitrary forcing $f(t)$) is the same,

$$ F(t_b-t_a) = \sqrt{\frac{m\omega}{2\pi i\hbar \sin\omega(t_b-t_a)}} .$$

Is there any physical or mathematical reason for this? How can I justify this must be the case for the driven oscillator, without going through a 10 page calculation using Feynman's trick of exploiting the composition law and the fact that $F$ is the propagator from the origin to the origin?

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There is certainly a mathematical reason:

  1. The external force $f$ appears in the term linear in the position variable $x$ of the action $S[x]$.

  2. When we split the path integral variable $x~=~x_{\rm cl}+ y$ in a classical path $x_{\rm cl}$ plus quantum fluctuation $y$, we know that the action $S[x]~=~S[x_{\rm cl}]+ S_q[y]$ has no term linear in the fluctuations (because the Taylor coefficient $\frac{\delta S}{\delta x}|_{x=x_{\rm cl}}~=~0$ of the linear term is the classical EOM).

  3. The quantum action $S_q[y]$ does not depend on the classical path $x_{\rm cl}$ since the action $S[x]$ for the harmonic oscillator has at most quadratic terms in $x$. (This is important since the classical path $x_{\rm cl}$ depends implicitly on the external force $f$.)

  4. Hence the quantum action $S_q[y]$ and the corresponding path integral $$ F(t_b-t_a) = \int_0^0 {\cal D}y~e^{\frac{i}{\hbar}S_q[y]} $$ cannot depend on the external force $f$.

References:

  1. R.P. Feynman and A.R. Hibbs, Quantum Mechanics and Path Integrals, 1965; Problem 3.11.
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