# Discretizing the Action for Feynman Path Integral

I am trying to understand how to compute path integrals, given a Lagrangian. I understand how it is done for the free particle, but I am confused for other actions. I am having trouble understanding how to decompose the action into subintervals. I understand how to do this for the free particle: $$S=\frac{m}{2}\frac{(x_b-x_a)^2}{t_b-t_a}\to\frac{m}{2}\sum_{i=1}^N\frac{(x_{i+1}-x_i)^2}{\epsilon}\tag{1}$$ where, \begin{align} \epsilon & = t_{i+1}-t_i \\ N\epsilon & = t_b - t_a \\ t_0 &= t_a \\ t_N & = t_b \\ x_0 &= x_a \\ x_N &= x_b \end{align}\tag{2} However, I am having trouble understanding how to do this for more general cases. For example, the classical action of the harmonic oscillator is $$S=\frac{m\omega}{2\sin\omega T}\left(\left(x_1^2+x_2^2\right)\cos\omega T -2x_1x_2\right).\tag{3}$$ I tried to make this discrete as follows: \begin{align}S=&\frac{m\omega}{2\sin\omega T}\left(\left(x_1^2+x_2^2\right)\cos\omega T -2x_1x_2\right)\cr \to&\frac{m\omega}{2\sin\omega\epsilon}\sum_{i=1}^N\left(\left(x_{i+1}^2+x^2_i\right)\cos\omega\epsilon-2x_{i+1}x_i\right).\end{align}\tag{4} Therefore, the path integral becomes \begin{align}U(x_b,t_b,x_a,t_a)=&\lim_{\epsilon\to 0}\frac{1}{A}\int\cdots\iint\exp\left(\frac{i}{\hbar}\frac{m\omega}{2\sin\omega\epsilon}\sum_{i=1}^N\left(\left(x_{i+1}^2+x^2_i\right)\cos\omega\epsilon-2x_{i+1}x_i\right)\right)\cr &\times\frac{\mathrm{d}x_1}{A}\frac{\mathrm{d}x_2}{A}\cdots\frac{\mathrm{d}x_{N-1}}{A}\tag{5}\end{align} $$A=\left(\frac{2\pi i\hbar\epsilon}{m}\right)^{\frac{1}{2}}.\tag{6}$$

However, when I evaluated the integral for $$x_1$$, I got the following: $$\sqrt{\frac{m\sin\omega\epsilon}{2i\pi\hbar~2\epsilon^2\omega\cos\omega\epsilon}}\exp\left(\frac{i\omega m}{2\hbar \sin(2\omega\epsilon)}\left(\left(x_0^2+x_2^2\right)\cos2\omega\epsilon-2x_0x_2\right)\right).\tag{7}$$ Comparing to the actual propagator, which is below, I see I got some things right, but the normalization factor is all wrong. Where did I go wrong? $$U(x_b,t_b,x_a,t_a)=\sqrt{\frac{m\omega}{2\pi i\hbar\sin\omega T}}\exp\left(\frac{im\omega}{2\hbar\sin\omega T}\left(\left(x_a^2+x_b^2\right)\cos\omega T -2x_ax_b\right)\right).\tag{8}$$ P.S. I am using Quantum Mechanics and Path Integrals, Feynman and Hibbs to learn this.

The action that goes in the path integral is a functional of the path $$x(t)$$. Namely, $$S[x(t)]=\int_0^T dt \frac{m}{2}\left(\dot{x}^2-\omega^2 x^2\right)$$ which can be discretized very simply $$S=\frac{m}{2}\sum_{i=1}^N\frac{(x_{i+1}-x_i)^2}{\epsilon}-\epsilon\omega^2 x_i^2$$

The $$x$$ in this action is not necessarily a classical trajectory, but this is still the action which appears in classical mechanics and leads to the equation of motion of the harmonic oscillator through the Euler-Lagrange equation.

You were confusing this action functional with the numerical value of the action functional evaluated at the classical path $$S[x_{cl}]=\frac{m\omega}{2\sin\omega T}\left(\left(x_1^2+x_2^2\right)\cos\omega T -2x_1x_2\right)$$

This is just a number (albeit a number that depends on boundary conditions). There is no $$x(t)$$ appearing in it so you can't vary it to get Euler-Lagrange equations, and you can't integrate it in a path integral.

1. As far as the single Gaussian $$x_1$$ integration goes, OP is doing the correct thing (up to possible typos) in eq. (7). However since the time-increment $$\epsilon~\ll~ \omega^{-1}\tag{i}$$ is supposed to be small (in order for Feynman's fudge factor $$1/A$$ to be valid), we have under the square root $$\frac{\sin (\omega \epsilon)}{2\epsilon^2\omega\cos (\omega \epsilon)}~\approx~\frac{1}{2\epsilon}~\approx~\frac{\omega }{\sin (\omega 2\epsilon)}, \tag{ii}$$ in agreement with Feynman's formula (8) for the harmonic oscillator. See also this related Phys.SE post.

2. It is possible to extend the result from 2 to $$N$$ time-increments, see e.g. this Phys.SE post. Feynman's formal conceptional idea is to replace the off-shell action $$I[x;t_a;t_b]$$ in the path integral with the discretized sum of on-shell actions $$\sum_{i=1}^N S(x_i,t_i;x_{i-1},t_{i-1})$$; then integrate over all intermediate positions $$x_1, \ldots, x_{N-1}$$; and in the end take the continuum limit $$N\to\infty$$.

• So, I was correct in plugging in the classical action as discrete, and it should not be done as in the other answer? Nov 6, 2019 at 14:37
• I updated the answer. Nov 6, 2019 at 15:45
• @Tesseract, No, despite Qmechanics answer here perhaps being another correct way to treat the path integral, my answer was correct and is how the path integral is normally treated. Nov 22, 2019 at 12:50