In time-dependent perturbation theories, one encounters multi-dimensional time-ordered integrals $$ (-i)^n\int_0^tdt_1\int_0^{t_1} dt_2 \cdots \int_0^{t_{n-1}}dt_n f(t_1,t_2,\cdots,t_n) $$

What is the best way of numerically evaluating such multi-dimensional time-ordered integrals? Monte Carlo integration is the first way that comes into my mind. How does Monte Carlo deal with the time-ordering?


1 Answer 1


The time-ordered integral can be transformed to a normal multi-dimensional integral over a rectangular volume by changing variables. Setting $t=\beta$, and using the following change of variables: $$ \begin{cases} t_1 = y_1\\ t_2 = y_2\frac{y_1}{\beta}\\ t_3 = y_3\frac{y_2y_1}{\beta^2}\\ \vdots\\ t_n = y_n\frac{y_{n-1}\cdots y_1}{\beta^{n-1}} \end{cases} $$ and the corresponding Jacobian $$ \left(\frac{y_1}{\beta}\right)^{n-1}\left(\frac{y_2}{\beta}\right)^{n-2}\cdots \left(\frac{y_{n-1}}{\beta}\right)^{1}, $$ the integral $(-i)^n\int_0^tdt_1\int_0^{t_1} dt_2 \cdots \int_0^{t_{n-1}}dt_n f(t_1,t_2,\cdots,t_n)$ becomes $$ (-i)^n\int_0^\beta dy_1 \int_0^{\beta} dy_2 \cdots \int_0^{\beta} dy_n \, f(y_1, \frac{y_2y_1}{\beta},\cdots,\frac{y_ny_{n-1}\cdots y_1}{\beta^{n-1}}) \left(\frac{y_1}{\beta}\right)^{n-1}\left(\frac{y_2}{\beta}\right)^{n-2}\cdots \left(\frac{y_{n-1}}{\beta}\right)^{1}. $$

It is then easy to use a regular Monte Carlo integrator to complete the integral. For example, one may use the vegas algorithm https://vegas.readthedocs.io/en/latest/tutorial.html .

  • $\begingroup$ is there a typo in your $t_3$? Shouldn't this be $y_3 y_2 y_1/\beta^2$ rather than $y_3 y_2 y_2/\beta^2$ as you have it now? And would you have a source to this? $\endgroup$ Commented Nov 19, 2022 at 19:53
  • $\begingroup$ @ZeroTheHero Thanks! I edited. A similar but not identical transformation was found in exercise 8 in Chapter 2 of this book: Approximating Integrals Via Monte Carlo and Deterministic Methods (Michael Evans & Tim Swartz). Based on the transformation in the book, I came up with the transformation in my answer. $\endgroup$
    – Frank
    Commented Nov 19, 2022 at 20:27

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