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I am unable to understand how the Feynman path integral is a sum of paths and not a product of paths. I understand a sum of paths as follows:

$$ Z=\exp ( i S_1 )+ \exp (i S_2)+... $$

I do not see how such a sum relates to the Feynman path integral:

$$ Z=\int_{-\infty}^\infty \exp(iS(x_1))dx_1\int_{-\infty}^\infty \exp(iS(x_2))dx_2\int_{-\infty}^\infty \exp(iS(x_3))dx_3... $$

which appears as a product. Is the appellation "sum over paths/sum over histories" incorrect?

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  • $\begingroup$ The answers you are getting are good, but note that your expression for the path integral as a product is wrong. The action is a functional over the whole path, not a function of one of the points. $\endgroup$
    – fqq
    Mar 18, 2020 at 23:58
  • $\begingroup$ @fqq can you post the correct notation for my Z as a comment, for reference? $\endgroup$
    – Anon21
    Mar 19, 2020 at 14:05
  • $\begingroup$ @CosmasZachos Your answer is super nice, but it will take me days to understand it line by line. Stay tuned, I'll follow up soon. $\endgroup$
    – Anon21
    Mar 19, 2020 at 16:25
  • $\begingroup$ @AlexandreH.Tremblay It's at the end of Cosmas Zachos' answer, something like $\int exp[i S(x_1, x_2 \dots x_n)]dx_1\dots dx_n$ if you keep the discretised notations. $\endgroup$
    – fqq
    Mar 19, 2020 at 16:41

2 Answers 2

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"Sum over paths" is 150% correct. It, in fact, means product over integrals over each point of a path.

A picture should be worth a thousand words. Discretize the integral by two screens, with 3 and 4 slits, respectively. Ultimately, you'll consider the limit of an infinity of screens, each with an infinity of slits.

enter image description here

Think of your assignment to go from A to B in straight segments through one of the 3 slits at t1, and then the 4 at t2. You have then 3×4=12 options, 12 paths, 12 histories to achieve that. Your first (topmost) sum will then have 12 terms of exponentials of $iS_1, ..., iS_{12}$, the sum of 12 complex amplitudes, each corresponding to a different path. But each path/amplitude is a product of three segments, characterized by their endpoint: that 's why we are skipping the last factor of the segment abutting at B.

Now, each path/history is characterized by the location of its slit on each screen; e.g., calling each slit location 1,2,3, and 1,2,3,4, respectively, $$ S_i=\int dt ~ L_i(x(t)) \leftarrow L(x_1(i)) + L(x_2(i)) \leadsto \\ S_1 = L(x_1=1) +L(x_2=1) , \\ S_2 = L(x_1=2) +L(x_2=1) , \\ S_3 = L(x_1=3) +L(x_2=1) , \\ S_4 = L(x_1=1) +L(x_2=2) ,... \\ S_{12} = L(x_1=3) +L(x_2=4) . $$

The sum of the 12 exponentials then amounts to $$ \sum_{x_1} \sum_{x_2} e^{i L(x_1) +iL(x_2)}= \sum_{x_1} \sum_{x_2} e^{i L(x_1)} e^{iL(x_2)}, $$ a sum over 12 paths.

It should then be evident how to generalize to the continuum path integral.

First, still keeping just two screens, but perforating them with an infinity of slits to the point of evanescence, you get $$ \sum_{x_1} \to \int dx_1 , \sum_{x_1} \to \int dx_1 . $$

You then take the number of screens to N and then infinity, so $$ L(x_1) +L(x_2) \to L(x_1) +L(x_2)+...+ L(x_N) \to \int dt ~L(x(t)), $$ which, times i, is exponentiated to provide the integrand of the functional integral, now over an infinity of screens, $\int dx_1 \int dx_2 ...\int dx_N \to \iint \cal Dx $.

In sum, $$ \iint \cal{D} x ~~ e^{i\int dt ~L(x(t))}. $$ You might review the free particle to fix your notation.

  • Τhis is, of course, all in black and white on pp 68-69 of Dirac's epochal paper where he introduces the path integral Feynman developed.
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    $\begingroup$ The ritual silent downvoter sows his wild oats again. $\endgroup$ Mar 17, 2020 at 0:55
  • $\begingroup$ Does the path integral produces a complex number or a real number. If a complex number, how does it account for the definition of the expectation value as : $ \langle F\rangle = \frac{\int D[\phi] F(\phi) e^{iS(\phi)}}{\int D[\phi]e^{iS(\phi)}} $ ? $\endgroup$
    – Anon21
    Jul 17, 2021 at 13:21
  • $\begingroup$ For a real function, it produces a real expectation value, even though the propagator kernel is complex, of course. Read up and follow the oscillator example. $\endgroup$ Jul 17, 2021 at 14:07
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I am unable to understand how the Feynman path integral is a sum of paths and not a product of paths.

It is actually both a sum and product of paths; the product is taken when we select one specific path; it is then discretised and we take the product over all small paths and then take the limit; the sum is taken over all paths.

It is a famously open problem to frame the Feynman path integral rigorously in exactly these terms; what can be done is to show the Euclideanised version can be stated in such a rigorous fashion; this is done via stochastic integrals where the measure is actually stochastic.

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