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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?

wooohooo
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1 Answers1

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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, $$t_n~=~t_0+n\epsilon, \quad \epsilon ~=~ \frac{t_N-t_0}{N}, \quad N~\in~\mathbb{N},\quad n~\in~\frac{1}{2}\mathbb{N}. $$

Qmechanic
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  • 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 $dq$? – wooohooo Oct 21 '21 at 16:57
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    The difference of 1 is independent of $N$, if that's what you're asking. – Qmechanic Oct 21 '21 at 17:14