# On the prefactor in the path integral formulation

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?

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.