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I am reading Feynman and Hibbs on Path Integrals. In section 3.5, they show that the kernel for a lagrangian of the form $L=a(t)\dot{x}^2+b(t)\dot{x}x+c(t)x^2+d(t)\dot{x}+e(t)x+f(t)$ is $K(b,a)=e^{\frac{i}{\hbar}S_{cl}[b,a]}F(t_a,t_b)$. In general, how do I calculate the factor $F(t_a,t_b)$. In the problems after the section, I have calculated the classical action, for the particle in a magnetic field, and the forced harmonic oscillator. But I don't know how to calculate the prefactors. For e.g. this is the problem 3-11 from Feynman and Hibbs asks you to calculate the kernel of the harmonic oscillator driven bby an external force $f(t)$. The Lagrangian is $L=\frac{m}{2}\dot{x}^2-\frac{m\omega^2}{2}x^2+f(t)x$. The answer is $$K=\sqrt{\frac{m \omega}{2 \pi i \hbar \sin{\omega T}}}e^{\frac{i}{\hbar}S_{cl}}$$

where $T=t_f-t_i$ and $S_{cl}$ is the classical action. How can I see that the above is the factor multiplying the exponent directly or via a calculation.

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2 Answers 2

The prefactor $F(t_f,t_i)$ is given in eq.(3-50) of Ref. 1 as

$$ F(t_f,t_i)~=~\qquad $$ $$\tag{3-50'} \int_{y(t_i)=0}^{{y(t_f)=0}}\!\!\! {\cal D}y~\exp\left\{\frac{i}{\hbar} \int_{t_i}^{t_f} \!\! dt[a(t) \dot{y}(t)^2+ b(t) y(t)\dot{y}(t)+c(t) y(t)^2 ] \right\}. $$

I doubt that there exists a closed formula for the path integral (3-50') for arbitrary coefficients $a(t)$, $b(t)$, and $c(t)$ with explicit time dependence.

For time-independent coefficients $a$, $b$, and $c$, the evaluation of the Gaussian path integral (3-50') is shown in many textbooks, e.g. in Section 3-11 of Ref. 1 or Appendix A of Ref. 2.

References:

  1. R.P. Feynman & A.R. Hibbs, Quantum Mechanics and Path Integrals, 1965.

  2. J. Polchinski, String Theory Vol. 1, 1998.

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I have not the Feynman and Hibbs book, but I think, that, the action being quadratic, you will have to use the square root of the determinant of the second (coordinates) derivative matrix of the classical action.

You will confirm this by testing that, when $t_b \rightarrow t_a$, then $K(b, a, t_b, t_a) \rightarrow \delta(b - a)$

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