I am new to QFT. In books like Fradkin's QFT an integrated approach, and Stefan's Gauge field theories 2nd Ed., they derive the path integral from first writing down the integral over the phase space, $$ \lim_{N\to +\infty} \int \left\{\prod_{n=1}^{N-1} \mathrm{d}q_n\right\} \left\{\prod_{n=1}^{N} \frac{\mathrm{d}p_n}{2\pi\hslash}\right\} \exp\left[{\frac{i}{\hslash} \varepsilon \sum_{n=1}^N p_n \frac{q_n - q_{n-1}}{\varepsilon} - H(\overline{q}_n, p_n)}\right] $$ And then proceed with $$ \int \frac{\mathcal D q \mathcal D p}{2\pi \hbar} \exp\left[{\frac{i}{\hslash} \varepsilon \sum_{n=1}^N p_n \frac{q_n - q_{n-1}}{\varepsilon} - H(\overline{q}_n, p_n)}\right] $$ with the notation of $ \mathcal D q \mathcal D p $ meaning the equal number of product of $dq$ and $dp$ going to infinity, $$\mathcal D q \mathcal D p = \prod_{i=1}^{\infty} dp_i dq_i/2\pi \hbar \tag{1}$$

I am wondering if that missing $dq$ matters. Dimension-wise, at least the Stefan book made amendments to the dimension, while the Fradkin book didn't. I am not sure if that's how we write the functional integral.

Edit: I mean I do agree that there should be one less $dq$. And I see how that comes about. But why are we ignoring this missing $dq$ (or discarding the extra $dp$) when we go to the equation 1?

  • 3
    $\begingroup$ If your amplitude takes you from 0 to $q_N$, you might not want to sum over all end points? $\endgroup$ Oct 21, 2021 at 16:08
  • $\begingroup$ @CosmasZachos yes I know we SHOULD have one less. I guess my question is: is it correct to assume an equal number of $dp$ and $dq$ when we go to $N\rightarrow \infty$ $\endgroup$
    – wooohooo
    Oct 21, 2021 at 16:59
  • $\begingroup$ Why not? isn't this what infinity means? $\endgroup$ Oct 21, 2021 at 17:24
  • $\begingroup$ related: physics.stackexchange.com/q/134714/226902 $\endgroup$
    – Quillo
    Sep 19, 2022 at 7:57

1 Answer 1


This is because there are 1 less position integration due to

  1. the Dirichlet boundary conditions $$q(t_0)~=~q_0\quad\text{and}\quad q(t_N)~=~q_N,$$

  2. and the fact that the insertion of complete sets of position resp. momentum eigenstates in phase space path integral alternates temporally $$ p(t_{1/2}),\quad q(t_1),\quad p(t_{3/2}),\quad q(t_2),\quad \ldots,\quad p(t_{N-3/2}), \quad q(t_{N-1}),\quad p(t_{N-1/2}),$$ along the time discretization, $$t_n~=~t_0+n\epsilon, \quad \epsilon ~=~ \frac{t_N-t_0}{N}, \quad N~\in~\mathbb{N},\quad n~\in~\frac{1}{2}\mathbb{N}. $$

  • $\begingroup$ So there is INDEED one less $dq$, and the books saying that when we take $N$ to infinity we still want to keep that difference in number of $dp$ and $dq$? $\endgroup$
    – wooohooo
    Oct 21, 2021 at 16:57
  • 1
    $\begingroup$ The difference of 1 is independent of $N$, if that's what you're asking. $\endgroup$
    – Qmechanic
    Oct 21, 2021 at 17:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.