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When deriving the path integral formulation at one reaches an expression of the form

$$ \intop \prod_n dx_n \prod_n dp_n e^{\frac{i}{\hbar } \Delta t \sum_{n} (p_{n+1} \frac{x_{n+1}-x_{n}}{\Delta t}-T(p_n) -\frac{V(x_{n+1})+V(x_n)}{2})} $$

Where $V(x)$ ar the matrix elements of the potential operator $V(\hat{X})$, and similary for $T(\hat{P})$ the kinetic energy operator. Many texts simply state that at the continuum limit this becomes

$$ \intop D x D p \, e^{\frac{i}{\hbar} \intop dt(p\dot{x}- H(p,x)}. $$

What is not clear to me is the justification to take the limit $$\frac{x_{n+1}-x_{n}}{\Delta t} \rightarrow \dot{x}$$ since $x_{n+1}$ and $x_{n}$ are two independent integration variables and there is no obligation that they approach each other when the limit is taken.

Some sources (for example Altland and Simons p.99) mention briefly that paths where $x_{n+1} - x_{n} \not\rightarrow 0$ cancel each other due to the rapid phases approximation. Other simply avoid the issue ( or Kleinert ch. 2, Feznman and Hibbs ch. 2). Is there a way to see it clearly? What assumption suffice to validly take the limit?

proton
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    The continuum path integral is a sum over all paths. This includes paths of arbitrarily large slopes at any given point. Arbitrarily large slopes correspond to arbitrarily large differences between $x_{n}$ and $x_{n+1}$. – Ryder Rude Dec 05 '22 at 17:13
  • The continuum version is a shorthand. $\dot x$ is just a notational place-holder that has to be translated to the discrete version for any actual computation. The discrete version is the definition of the path integral. – Ryder Rude Dec 05 '22 at 17:22
  • @RyderRude What you say makes sense to me, however see for example Altland & Simons p. 99 :"We first notice that rapid fluctuations of the integration arguments xn as a function of the index n are strongly inhibited by the structure of the integrand. ..." – proton Dec 05 '22 at 17:32
  • "... When taken together, contributions for which $(q_{n+1} − q_{n})p{n+1} > O(\hbar)$ (i.e. when the phase of the exponential exceeds 2π) tend to lead to a “random phase cancellation.” In the language of wave mechanics, the superposition of partial waves of erratically different phases destructively interferes. The smooth variation of the paths that contribute significantly motivates the application of a continuum limit " – proton Dec 05 '22 at 17:33
  • Interesting...I would say that the $n\rightarrow \infty$ limit is taken just because it gives the right predictions. Any discrete version is just wrong. Also, if you take Schrodinger equation as your starting point, you can just derive that the $n\rightarrow \infty$ limit must be there. However, writing the path integral in terms of $\dot x$ is nothing but notational convenience. That notation is just a short-hand for the discrete version with $n\rightarrow \infty$ . See, e.g. Shankar's book. So the $\dot x$ does not have any physical relevance anyway. – Ryder Rude Dec 05 '22 at 17:52
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    @RyderRude It seems like you say the descrete version is both meaningless and the only meaningful expression. If the descrete version is wrong, why every textbook contains is? – proton Dec 05 '22 at 18:01
  • @proton I'm sorry. I mean that the discrete version, with the continuum limit $n\rightarrow \infty$, is the only experimentally correct and mathematically meaningful version. Think about the continuum version involving the $\dot x$, which just says to "sum over all continuous paths". Do you of any mathematical definition of that other than to translate it into the discrete version with $n\rightarrow \infty$? – Ryder Rude Dec 05 '22 at 18:10
  • @RyderRude A derivative implies the existance of limit for $x_{n+1}-x_n / \Delta t$ which here does not necesrally exist since $\intop \prod_n dx_n$ allows summation over non-smoot paths and evev non-continuous. – proton Dec 05 '22 at 18:17
  • I think you are correct that there is a subtlety about the contribution of non-differentiable paths in the path integral. I thought this was treated carefully in Kleinert, but I'll have to double check. Another source that might talk about this is Zinn-Justin. – Andrew Dec 05 '22 at 19:04
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    Related: https://physics.stackexchange.com/q/56070/2451 and links therein. – Qmechanic Dec 05 '22 at 19:17
  • It looks like Section 2.2 of Zinn-Justin's book discusses this. – Andrew Dec 05 '22 at 19:37

2 Answers2

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There can be no good justification for $\frac{x_{n-1} - x_n}{\Delta t}\to \dot{x}$ because that's not what happens in the actual mathematically rigorous path integral! You can try to do various physical handwaves for this, but in the end the truth is that the derivative notation here used by physicists is just a shorthand, not something that has an actual formal meaning.

The proper mathematical definition of the integration in the quantum mechanical path integral is by defining the Wiener measure as representing the path integral of the kinetic part, i.e. there is a measure $\mathrm{d}W_{q,q'}$ induced by the free Hamiltonian $H \propto p^2$ on all continuous paths with endpoints $q,q'$ and one indeed defines this for finite $n$ by a very similar formula but one never needs to take an actual limit $n\to \infty$ where the difference quotients become derivatives. Instead the argument that this defines measure everywhere involves standard mathematical arguments about the uniqueness of Gaußian measures, which "in spirit" correspond to this limit but cruically do not do so formally.

The Wiener measure works like this: The kernel of the free Hamiltonian is $$K_t(q,q') = (2\pi t)^{3/2}\mathrm{e}^{-(q -q')^2/2t}$$ meaning a solution $\psi(x,t)$ to the free Schrödinger equation fulfills $$ \psi(q,t) = \int K_t(q,q')\psi(q',0)\mathrm{d}q'$$ and we define the measure of a set $S_{I,t_1}$ of continuous paths $q(t)$ with $t\in[0,t_0]$ and with $q(t_1)\in I\subset \mathbb{R}^3$ for $I$ a nice (Borel) set to be $$ \int_{S_{I,t_1}}\mathrm{d}W_{q,q'} := \int_I K_{t_0/2 + t_1}(q,q_1)K_{t_0/2-t_1}(q_1,q')\mathrm{d}q_1.$$ Physicists write $\mathrm{d}W_{q,q'} = \mathcal{D}q\mathrm{e}^{-S_0[q(t)]}$ where $S_0$ is the action of a free particle. For an interacting action with potential $V(q)$, we simply integrate $\mathrm{e}^{-V(q(t))}\mathrm{d}W_{q,q'}$ - as long as the potential does not involve derivatives, this works just fine without ever talking about differentiability.

Note that this definition does not involve anywhere the actual paths $q(t)$ - it's just a specific convolution of the function $K_t(q,q')$. This might strike you as strange, but there is nothing mathematically wrong with this definition. What this "means" is that we're saying that the paths that pass through $I$ at an instant $t_1$ are weighted by the probability first for them to evolve from $q(0) = q$ and then by the probability to then evolve from some point $q_1 = q(t_1)\in I$ to the fixed endpoint $q(t_0) = q'$. The integral is just the convolution of these two probability densities.

Crucially, the Wiener measure is a measure on all continuous paths and not on differentiable paths (in fact, the differentiable paths have measure zero), and so the object $\dot{x}$ doesn't actually make any sense as something that appears in the integrand - a merely continuous path doesn't have a $\dot{x}$ everywhere.

While the proper mathematical construction of the Wiener measure and relating it to quantum mechanical amplitudes (the Feynman-Kac formula) does involve a limit that takes $n\to\infty$, it never needs to assume differentiability of the path - there is only an argument about Riemann integrability of the potential $V[q(t)]$, which is much weaker.

For the actual details on how to construct the path integral rigorously instead of the vague outline above, I recommend chapter 3 of "Quantum Physics" by Glimm and Jaffe.

ACuriousMind
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  • Thank you for the enlightening answer. What about discontinuous pahs? – proton Dec 05 '22 at 22:20
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    @proton Nothing. They are not the kind of path we consider in any version of the path integral I know of. Note that the choice of continuous paths here isn't arbitrary or grounded in any kind of physical argument, it's because we can prove that with this and the Wiener measure we can write down a path integral that is fully equivalent to the operator formalism of quantum mechanics. – ACuriousMind Dec 05 '22 at 22:31
  • Do we consider any kind of path at all? I understand the integration to be $\intop_{-\infty}^{\infty}dx_n$. Nothing in the above forces a continuous path in the limit $n \rightarrow 0$. – proton Dec 05 '22 at 22:35
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    @proton I've edited the answer to give the actual definition of the (conditional) Wiener measure for a set of continuous paths. The continuity has nothing to do with the integral or any limiting procedure, it is simply part of the definition in the rigorous approach. – ACuriousMind Dec 05 '22 at 22:56
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What is not clear to me is the justification to take the limit $$\frac{x_{n+1}-x_{n}}{\Delta t} \rightarrow \dot{x}$$ since $x_{n+1}$ and $x_{n}$ are two independent integration variables and there is no obligation that they approach each other when the limit is taken.

If they don't, then the velocity is infinite. For finite spacing, the terms where $x_{n+1}$ is very far away from $x_n$ have very high "velocity."

Some sources (for example Altland and Simons p.99) mention briefly that paths where $x_{n+1} - x_{n} \not\rightarrow 0$ cancel each other due to the rapid phases approximation. Other simply avoid the issue ( or Kleinert ch. 2, Feznman and Hibbs ch. 2). Is there a way to see it clearly?

When the velocity is infinite (or very high), you can see or expect that the $e^{ip\dot x}$ term has a "rapidly changing" phase.

hft
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  • How do you know that the rapidly changing phases all end up to cancel each other? Also, I don't see a problem with infinte velocities. Sometimes functions or series do diverge to infinity. Maybe now it's obvious for some people that everythings ends up find, but to me it isn't. – proton Dec 05 '22 at 19:41
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    I don't think it makes a lot of sense to say that they "all... cancel each other." Rather, for each individual $p_i$ integration, if you could just integrate $e^{ipv}$, you would get something that looks like $\frac{\sin(\Lambda v)}{v}$, where $v\to\infty$. The numerator can only be as large as 1, but the denominator can be as large as you like. So the contributions from paths with very high velocity is very small. – hft Dec 05 '22 at 19:52
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    As a quick aside, I'd say don't worry too much if this all doesn't seem super mathematically rigorous. Rigor is overrated. – hft Dec 05 '22 at 19:55
  • But don't the exponents usually look like $e^{i(\frac{p^2}{2m}+vp)}$ so the integration gives something like proportional to $\frac{1}{\sqrt{m}e^{iv^2/m}}$ ? – proton Dec 05 '22 at 22:25
  • Yeah, something like that. And if I squint my eyes, that almost looks like $\propto e^{-\alpha v^2/m}$, for some constant $\alpha$, which looks pretty small for large $v$. But remember, I'm squinting very hard. – hft Dec 06 '22 at 03:45
  • The exponent is complex, so it isn't small. – proton Dec 06 '22 at 10:07
  • That's true. If you are looking for a less "hand waving" answer, I would suggest the other answer. – hft Dec 06 '22 at 17:30