# Martin-Siggia-Rose action corresponding to Langevin equation

What is the Martin-Siggia-Rose-DeDominicis-Janssen action corresponding to the overdamped Langevin stochastic equation $$\frac{d\mathbf{x}}{dt} = -\mathbf{\nabla}V + \mathbf{\eta}$$ where V is the external potential and $$\mathbf{\eta}$$ a Gaussian noise corresponding to a thermal bath. We assume that $$\mathbf{\eta}$$ has zero mean and constant variance : $$\left<\eta_\alpha(t)\eta_\beta(t')\right> = 2T\delta(t-t')\delta_{\alpha\beta}$$

• @Kostas it's not in MSR, but it is in Janssen's paper (and why his name is associated to the formalism). – fqq Nov 8 '19 at 19:43
• Thank you for your answers. Searching a little more, I found answers in that blog post : inordinatum.wordpress.com/2012/09/27/…  I think I understand most of the derivation, except the identity : $$\delta (\partial_tx(t) - F(x(t), t) - \xi(x(t), t)) = \int\mathcal{D}[\tilde{x}]\exp\left({-\int_t i\tilde{x}(t)(\partial_tx(t) - F(x(t), t) - \xi(x(t), t)})\right)$$ How can the minus sign and the $t$ integral come from the formula : $$\delta(q) \propto \int_\tilde{x}e^{i\tilde{x}q}$$ – Emmy Nov 8 '19 at 20:40
• @Emmy I think the Langevin equation should have $dv/dt$ on the left, not $dx/dt$, after all if $\eta=0$ this should be the Newtons Law. – Kostas Nov 8 '19 at 20:55
• I'm working with the overdamped Langevin equation, in other words Newton's equation : $$m\frac{d^2\mathbf x}{dt^2} + \gamma\frac{d\mathbf x}{dt} = -\mathbf\nabla V + \mathbf\eta$$ for which inertia has been neglected. Please excuse me for the confusion – Emmy Nov 8 '19 at 21:51
• @Emmy You should edit your question to reflect the change. My posted answer would change very little, $\nabla V -> \nabla V + \gamma/m p$ – Kostas Nov 10 '19 at 17:42

This so-called Langevin equation is just the Newtons second Law with an extra random force: $$\frac{d\mathbf{p}(t)}{dt} = -\mathbf{\nabla}V(x) + \mathbf{\eta}(t)$$ together with $$\mathbf{p}=\frac{d\mathbf{x}(t)}{dt}$$ obviously.
It is better behaved mathematically if you write this in the integral form: $$\mathbf{p}(t) = \int dt ~ \big[-\mathbf{\nabla}V(x) + \mathbf{\eta}(t)\big]$$, which is the starting point for Ito calculus.
The action from which stuff can be computed by perturbation theory is: $$L = T \frac {\tilde{\mathbf{x}}^2} 2 - \tilde{\mathbf{x}} \big[\frac{d\mathbf{p}(t)}{dt}+\mathbf{\nabla}V(x) \big]$$ Here $$\tilde{\mathbf{x}}$$ is conjugate variable to $$\mathbf{p}$$.
I am not sure about the coefficient of the first term, it may be $$T$$ or $$iT$$ depending on conventions.
• Hi @Kostas, thank you for the nice answer. However, what is $T$?It seems there is an extra harmonic potential for $T \neq 0$. – Quillo Jun 3 at 14:28