I was going through the calculation of the free-particle kernel in Feynman and Hibbs (pp 43). The book describes $$ \left(\frac{m}{2\pi i\hbar\epsilon}\right)\int_{-\infty}^{\infty}\exp\left(\frac{im}{2\hbar\epsilon}[(x_2-x_1)^2+(x_1-x_0)^2]\right)dx_1 \tag{3.4} $$ as a Gaussian integral. And the result of this integration is given as $$ \left(\frac{m}{2\pi i\hbar \cdot 2\epsilon}\right)^{1/2}\exp\left(\frac{im}{2\hbar\cdot 2\epsilon}(x_2-x_0)^2\right). \tag{3.4} $$ The only way I could do this integral was treating it as a Gaussian integral over $ix_1$, in which case I get the required solution but with a negative sign upfront. Is this the right way to do it, or is there something very obvious that I am not seeing?

On a related note, problem 3-8 (pp 63) also involves a similar "Gaussian" integral in order to obtain the kernel for the quantum harmonic oscillator. My technique of performing a Gaussian integral over $ix$ goes horribly wrong there and I am not even near the correct answer which should be of the form $$ \frac{1}{\sqrt{2\pi\sin\epsilon}}\exp\left(\frac{i}{2\sin\epsilon}[(x_0^2+x_2^2)\cos\epsilon-2x_2x_2]\right) $$ after performing the integration, $$ \exp\left(\frac{i\cot\epsilon}{2}(x_0^2+x_2^2)\right)$$ $$\times \int\exp\left(\frac{i}{2\sin\epsilon}[2x_1^2\cos\epsilon-2x_1(x_0+x_2)]\right)dx_1. $$ Any help will be greatly appreciated.


1 Answer 1

  1. The variable $$\epsilon\equiv\Delta t^M~>~0$$ in eq. (3.4) is the change in Minkowski time.

  2. In order not to deal with purely oscillatory integrands, Feynman uses a prescription $$\Delta t^M\quad\longrightarrow\quad \Delta t^M-i0^+ $$ where an infinitesimal negative imaginary part is added to make the integrand exponentially decaying.

  3. In other words, under a Wick rotation $$\Delta t^E ~\equiv~ i \Delta t^M$$ to Euclidean time, $\Delta t^E$ should have a positive real part.

  4. Eq. (3.4) in Ref. 1 then follows from the Gaussian integral $$\begin{align} &\sqrt{\frac{m}{2\pi \hbar\Delta t^E_{21}}} \sqrt{\frac{m}{2\pi \hbar\Delta t^E_{10}}}\cr &\int_{\mathbb{R}}\!\mathrm{d}x_1~ \exp\left\{-\frac{m}{2\hbar}\left[\frac{\Delta x_{21}^2}{\Delta t^E_{21}} +\frac{\Delta x_{10}^2}{\Delta t^E_{10}}\right]\right\}\cr ~=~&\sqrt{\frac{m}{2\pi \hbar\Delta t^E_{20}}}\exp\left\{-\frac{m}{2\hbar}\frac{\Delta x_{20}^2}{\Delta t^E_{20}}\right\},\end{align}$$ where $$\begin{align} \Delta x_{ab}~:=~&x_a-x_b, \cr \Delta t_{ab}~:=~&t_a-t_b, \cr a,b~\in~&\{0,1,2\}.\end{align} $$

  5. The above square root factor is the famous Feynman's fudge factor, which can be understood in the Hamiltonian phase space path integral as arising from a Gaussian momentum integration, cf. e.g. my Phys.SE answer here.


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

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