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.


2 Answers 2


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 $N\to\infty$.

  • $\begingroup$ So, I was correct in plugging in the classical action as discrete, and it should not be done as in the other answer? $\endgroup$ Commented Nov 6, 2019 at 14:37
  • 1
    $\begingroup$ I updated the answer. $\endgroup$
    – Qmechanic
    Commented Nov 6, 2019 at 15:45
  • $\begingroup$ @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. $\endgroup$
    – octonion
    Commented Nov 22, 2019 at 12:50

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