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I think I understand the derivation of the expression for the path integral $$ \int \mathcal D q = C\lim_{N\rightarrow+\infty} \prod_{k=1}^{N}\int_{-\infty}^{+\infty}dq_k $$ as is done in Srednicki - by cutting time into slices and using $N$ completeness relations.
I do not understand the interpretation that I hear people discuss and is presented for example in Peskin.

I understand that, by integrating a functional over all possible paths starting at $x_a$ and ending at $x_b$, we are integrating (summing) over all dots of the picture below, and then change the positions of those dots and summing again, and so on over all possible configurations of such dots. Alternatively, this can be seen integrating over all the horizontal lines (except the first and last lines) of the picture (from Peskin).

enter image description here

From what I understand, people say that what we are doing with the path integral above is precisely changing to this last way of conceptualizing the sum over all paths.

My problem with this: The second way of conceptualizing this suggests me that we should sum all the integrals over the horizontal lines, not multiply them, so it seems that it should be

$$ \int \mathcal D q = C\lim_{N\rightarrow+\infty} \sum_{k=1}^{N}\int_{-\infty}^{+\infty}dq_k $$

What am I understanding wrongly?

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  • $\begingroup$ An extremely simple comparison: $\int dx + \int dy + \int dz \; $ vs $\; \int dx dy dz$. In the latter the integration is over all possible configurations of positions. The former, which is analogical with what you expect, is completely irrelevant. $\endgroup$ – Oktay Doğangün May 18 '18 at 18:49
  • $\begingroup$ @OktayDoğangün That was precisely the case in my mind, though: I don't know why you say the sum one is irrelevant. We want to integrate over several lines, not under an n-dimensional cube of some sort... $\endgroup$ – Soap May 18 '18 at 19:28
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    $\begingroup$ Try this: in a discrete version, suppose a particle can be at 3 different points at a given time. How many different configurations you can have for two time slices? $\endgroup$ – user110373 May 18 '18 at 23:54
  • $\begingroup$ @user110373 If I understood your wording correctly, then $9$. I think I got it: I was thinking I should integrate over the "area where the paths will be", but now I see that this is not enough: many paths pass through many of the same points as others did, of course. So, if I take the product of sums (integrals in this case) I'll get all the possible configurations. In your case: $3^2$ configurations. Thanks! $\endgroup$ – Soap May 19 '18 at 8:25

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