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In the Book "Quantum Mechanics and Path Integrals" by Feynman & Hibbs the path integral is approximated (page 32 and following) by $$ K(b,a)\approx\int...\int\int\phi[x(t)]dx_1dx_2...dx_{N-1}\tag{2.20} $$ with $b=(x_b,t_b)$ and $a=(x_a,t_a)$ being the start and endpoints of the path and $$\phi[x(t)]=const\cdot e^{(i/\hbar)S[x(t)]}=const\cdot e^{(i/\hbar)\int_{t_a}^{t_b} L[x(t),v(t),t]dt}.\tag{2.15}$$ Now I dont quite get this approximation.

  1. First of all I assume that the $dx_1dx_2...dx_{N-1}$ integrals have to be executed first and only after that the $dt$ integral in $\phi[x(t)]$ (or rather in $S[x(t)]$) should be executed. Is that right?

  2. And the second thing is that I dont get the meaning behind the $dx_1dx_2...dx_{N-1}$ integrals itself (each is integrated from $-\infty$ to $\infty$ according to wikipedia). So in the book the path was divided into straight lines between $x_k$ and $x_{k+1}$ with equal length and $x_0=x_a$ and $x_N=x_b$. That's why I would have thought the integration would not go from $-\infty$ to $\infty$ but rather from $x_k$ to $x_{k+1}$. So the integral would then look sth. like this $$K(b,a)\approx\int_{x_{N-1}}^{x_N}...\int_{x_1}^{x_2}\int_{x_0}^{x_1}\phi[x(t)]dx_0dx_1...dx_{N-1} $$ Could someone explain to me in an easy way why that is not the case?

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  • $\begingroup$ I'm not a theorist (full disclaimer). But I get the impression that Feynman thinks differently than most of us, and in the Lectures, and possibly here, he presents his peculiar take on things. Line integrals can be approached as they are in calculus 101, or they can be approached as pullback. I don't have the book you cite, but it sounds like none of the above. I may be wrong in this case, but I can say that in the case of the Lectures the value comes after you've already been exposed to the material. Feynman provides a different way of looking at things. Maybe start somewhere else? $\endgroup$
    – garyp
    Jun 16, 2022 at 3:26

3 Answers 3

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The whole idea of a path integral is based on the concept of a functional, i.e., a function of functions. Therefore the path integral is really in integration over functions in a functional space rather than the values of a variable. In Feynman's approximation, he breaks the functions over which is being integrated up into all the different function values at all the different points in these functions and then integrates over all those values, to ensure that the path integral actually integrates over all possible functions. That is why the individual $x$-integration runs from $-\infty$ to $\infty$. Therefore one would not be able to evaluate the $t$-integral first. That would require knowledge of the function $x(t)$, but the function can be anything. It is the "variable" over which the integration is performed. Hope this helps.

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  • $\begingroup$ I think I get it now. I was just confused because in the equation I wrote for $K$ there were only $N-2$ points that are integrated over. But a path consists of an infinite amount of points. I forgot that in the end, to transfer this approximation into an exact equation, $N\rightarrow\infty$. After that there indeed are an infinite amount of points. $\endgroup$ Jun 16, 2022 at 7:49
  • $\begingroup$ But with that I now see another problem. Maybe. With this also paths are considered that Einstein would not be a fan of. For example there will also be a path where $x_1=-\infty$ and $x_2=\infty$. If the path integral would be evaluated for a particle moving from $(x_a,t_a)$ to $(x_b,t_b)$ this would assume the particle taking on infinite velocity for some time. $\endgroup$ Jun 16, 2022 at 7:50
  • $\begingroup$ Yes, but the dynamics contained in $\phi[...]$ should suppress those paths. $\endgroup$ Jun 17, 2022 at 4:14
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  1. The action functional $S[t\mapsto x(t)]$ with its $dt$-integration is the continuum limit of a discretion $S[x_0,\ldots, x_N]$ that is implicitly implied on the RHS of eq. (2.20), so in eq. (2.20) there is strictly speaking not a $dt$-integration to perform, cf. OP's question. It only emerges in the continuum limit.

  2. OP's proposal does not make sense as written as integration limits can only refer to later performed integrations, not the other way around. Anyway, Feynman's point is that we should sum over all histories, i.e. the integration variables should cover the whole target space, i.e. position space $\mathbb{R}$ from $-\infty$ to $\infty$.

References:

  1. R.P. Feynman & A.R. Hibbs, Quantum Mechanics and Path Integrals, 1965; chapters 2 + 3 + 4.

  2. W. Dittrich & M. Reuter, Classical and Quantum Dynamics, 6th ed, 2020; chapters 19 + 20.

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  • $\begingroup$ 1. In this paper I found $K$ (there it is named $U$) gets explicitly calculated for a free particle with the same approximation in 2.2. In eq. (5) he splits up the action the same way the coordiantes $x$ get split up. This leads to a summation over $N$ integrals. Is this what you meant? $\endgroup$ Jun 16, 2022 at 8:01
  • $\begingroup$ $\uparrow$ Yes. $\endgroup$
    – Qmechanic
    Jun 16, 2022 at 8:02
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For question 2, the bounds should be from $-\infty$ to $\infty$. That is because at each time, we integrate over every possible position through the use of the identity

$1 = \int _{-\infty}^{\infty} |x \rangle \langle x | dx$

In terms of interpretation, at each point in time $t$, the position on the path can be any real number. So consecutive positions are not bounded by their neighbors.

For question 1, the way I think of it is to think of the $dx_k$ integrals as riemann sums. For each term in the result, you have a piecewise path $\bar{x}(t)$ with which the integral $dt$ can be evaluated. The limit as those riemann sums tend to integrals is the result. In that sense I think of doing the $dt$ integral first.

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  • $\begingroup$ Im not quite sure what you mean with $ |x \rangle \langle x |$ could you formulate this in non bra-ket notation ? I think then I would understand it better. And for question 1, I assumed that the internal integral needs to be evaluated in the end because I was not sure what to substitute for $x(t)$ in $S[x(t),v(t),t]$ since the motivation behind this path integral is to take each possible path into consideration. So I wouldnt know which path to substitute for $x(t)$ $\endgroup$ Jun 15, 2022 at 21:54
  • $\begingroup$ Hi peter, I just checked the source you linked, and on that page it is not so much a derivation of the path integral, but a rough conceptual explanation for the result. To derive the path integral, you do need the identity above, and I'm not sure how to write it without bra-ket notation. Shankar's principles of quantum mechanics does a nice job deriving the path integral in QM. He also introduces that identity in his chapter 1. That identity comes up all the time in quantum mechanics so it will be rewarding to learn if you want to understand QM deeply. $\endgroup$ Jun 16, 2022 at 6:25
  • $\begingroup$ If you are looking for a rough conceptual explanation for the result, then I can't give the justification, but I can loosely say that in response to your question (2), the path integral integrates over all possible paths, e.g. every point on that path could be anywhere in the universe, and the paths are not assumed to be continuous at all. That's the nature of the path integral. $\endgroup$ Jun 16, 2022 at 6:33

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