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Feynman path integral is equivalent to Fokker–Planck equation. This is mentioned here, but it's not clear why.

(This page says Schrodinger equation is also equivalent to Fokker–Planck equation which makes me even more confused).

Basically what I am asking is:

How to show that Fokker-Plank: $\frac{\partial}{\partial t} p(x, t) = -\frac{\partial}{\partial x}\left[\mu(x, t) p(x, t)\right] + \frac{\partial^2}{\partial x^2}\left[D(x, t) p(x, t)\right]$ is equivalent to path integral: $\psi(x,t)=\frac{1}{Z}\int_{\mathbf{x}(0)=x} e^{iS(\mathbf{x},\dot{\mathbf{x}})}\psi_0(\mathbf{x}(t))\, \mathcal{D}\mathbf{x}$?

Where $S(\mathbf{x},\dot{\mathbf{x}})= \int L( \mathbf{x}(t),\dot{\mathbf{x}}(t))dt$ is the action (integral over the Lagrangian) and $\mathcal{D}\mathbf{x}$ stands for the integration over all possible paths?

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    $\begingroup$ Details of this are worked out in "Quantum Field Theory and Critical Phenomena" by J. Zinn-Justin. $\endgroup$ – BobaFret Oct 6 '17 at 15:06
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    $\begingroup$ Related Phys.SE question: physics.stackexchange.com/q/310623/2451 $\endgroup$ – Qmechanic Nov 12 '17 at 14:08
  • $\begingroup$ theory.physics.manchester.ac.uk/~ajm/stoch09.pdf $\endgroup$ – 0x90 Jan 19 '18 at 10:54
  • $\begingroup$ @BobaFret Could you say where in the book it's addressed? $\endgroup$ – J.G. Mar 9 at 8:39
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    $\begingroup$ @J.G. I’ll check when I’m in my office again $\endgroup$ – BobaFret Mar 9 at 18:20
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I will start by apologising for a physicist's level of rigor in the following derivations but I think they give good insight into the connection between what I would call stochastic physics' Trinity of Langevin, Fokker-Planck and Path integral.

Fokker-Planck from Langevin equation

Given a Langevin equation $dZ_t=b(Z_t)dt+\sigma(Z_t)dW_t$ (the most common form found in physics) we can derive the Fokker-Planck equation from the Chapman-Kolmogorov equation in a quick and dirty way:

$$ p(y,t|x)=\int dz\, p(y,t|z)p(z,t'|x) $$

We suppose that $t = \delta t$ and thus we can write $$p(y,t|z) = \left\langle \delta (y-z-h)\right\rangle_h$$ where $h=\delta Z$ is defined by the associated Langevin equation. Now we have

$$ p(y,t|x)=\int dz\, \left\langle \delta (y-z-h)\right\rangle_hp(z,t'|x)\,.\tag{$\ast$} $$

We can then Taylor expand the delta function as

$$ \left(1+\left\langle h\right\rangle \frac{\partial}{\partial y} +\frac{1}{2}\left\langle h^{2}\right\rangle \frac{\partial^2}{\partial y^2} +\ldots\right)\delta(y-z) = (1+\mathscr{L})\delta(y-z)\,, $$

then integrating by parts and the delta function we are left with

$$ p(y,t|x)=(1+\mathscr{L^\dagger})p(y,t'|x)\,. $$

Expanding $p(y,t|x)= p(y,t'|x) + \frac{\partial}{\partial t}p(y,t'|x) + \ldots$ in the limit $t \rightarrow 0$ we arrive at the Fokker-Planck equation

$$ \frac{\partial p(y,t|x)}{\partial t}=\mathscr{L^\dagger}p(y,t|x) $$

relabelling $t'$ as $t$.

Path integral from Langevin

We apply the Chapman-Kolmogorov equation many times:

$$ p(y,t|x)=\int\prod_{i=1}^{N}dz_{i}\, p(y,t_{N}|z_{N})\ldots p(z_{2},t_{2}|z_1)p(z_{1},t_{1}|x) $$

And then use $(\ast)$ to replace each along with the identity $\delta(x) = \int dk \exp(ikx)$ we get a sequence of averages over complex exponentials:

$$ p(y,t|x)=\int Dz Dk\, \left\langle e^{ik_N(z_N-z_{N-1}-h_N)}\right\rangle\ldots \left\langle e^{ik_2(z_2-z_{1}-h_2)}\right\rangle \left\langle e^{ik_1(z_1-x-h_1)}\right\rangle $$

however we can still do better. Knowing the probability distribution of $h$ (usually Gaussian in physics) we can write $$ \left\langle O \right\rangle = \int dh P(h) O $$ thus after replacing every averaged over exponential we get

$$ p(y,t|x)=\int Dz Dk Dh\, e^{i\int dt \,k\dot{z}} P[h_t] $$

and when you perform the integrations over the paths of $k$ and $h$ you will get your path integral.

Path integral from Fokker-Planck

I am less sure about how to go about directly showing the equivalence without the presence of a Langevin equation. However if we create a Lagrangian by multiplying the F-P equation by an auxiliary field and then integrate over all paths of our probability density and our auxiliary field, I have a feeling this will work but I am not certain so I will leave my answer here.

The main point of this is to show that the path integral and the FP equation are essentially two different representations of the Chapman-Kolmogorov equation. As far as I can tell they are basically interchangeable and I hope someone more knowledgeable than myself can come in and explain why you would use one over the other in certain situations.

I hope this has been useful!

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  • $\begingroup$ Can you please define symbols and letters? $\endgroup$ – 0x90 Jan 18 '18 at 3:01
  • $\begingroup$ Almost all symbols and letters used are standard notation. I'm assuming you are familiar with the propagator (little $p$). Big $P$ is the probability distribution of the generic random variable $O$ and in the context of the question would be the probability distribution of each step $h_i$ which in the continuum limit becomes a functional of the path $h_t$ $\endgroup$ – Takoda Jan 18 '18 at 9:47
  • $\begingroup$ Shouldn't the integral in the chapman-kolomogrov function be over dt' as well? What about dW shouldn't it be $dW_t$? $\endgroup$ – 0x90 Jan 20 '18 at 19:20
  • $\begingroup$ No to the first question, because what the C-K equation says is that at some time the particle must be at some point in our space. That is to say, pick an intermediate $t'$ and the particle must be somewhere, thus we integrate over all space, not time. Yes tot he second, I will correct this. $\endgroup$ – Takoda Feb 5 '18 at 8:58
  • $\begingroup$ on the RHS you have an expression with t'. On the LHS there is only t. Without integrating over t' the LHS has to have t'. $\endgroup$ – 0x90 Feb 5 '18 at 11:54

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