inordinatum

Physics and Mathematics of Disordered Systems

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A quick introduction to the Martin-Siggia-Rose formalism

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The Martin-Siggia-Rose (MSR) formalism is a method to write stochastic differential equations (SDEs) as a field theory formulated using path integrals. These are familiar from many branches of physics, and are simple to treat perturbatively (or, in some cases, even to solve exactly).

We start from an SDE of the form:

\displaystyle \partial_t x(t) = F(x(t),t) + \xi(x(t),t),

where F is a deterministic function and \xi(x(t),t) is Gaussian noise with an arbitrary correlation function G:

\displaystyle \left\langle \xi(x,t) \xi(x',t') \right\rangle = G(x,t,x',t').

MSR recognized that observables, averaged over solutions of this SDE, can be written as a path integral with a certain action:

\left\langle \mathcal{O}[x(t)] \right\rangle = \int \mathcal{D}[x,\tilde{x}]\,\mathcal{O}[x(t)]\,e^{-S[x,\tilde{x}]},
where
\displaystyle S[x,\tilde{x}] = \int_t i\tilde{x}(t)\left[\partial_t x(t) - F(x(t),t)\right] + \frac{1}{2}\int_{t,t'} G(x(t),t,x(t'),t')\tilde{x}(t)\tilde{x}(t').

A useful special case is an automonous (time-independent) SDE with multiplicative noise:

\displaystyle \partial_t x(t) = F(x) + H(x)\xi(t),

where \overline{\xi(t)\xi(t')} = 2\sigma \delta(t-t') is Gaussian white noise. The resulting action for the path integral is local in time:

\displaystyle S[x,\tilde{x}] = \int_t i\tilde{x}(t)\left[\partial_t x(t) - F(x(t))\right] + \frac{\sigma}{2} H(x(t))^2\tilde{x}(t)^2.

I will now show a few concrete examples, and then explain how this mapping is derived.

Example: Ornstein-Uhlenbeck process

The Ornstein-Uhlenbeck process is one of the few stochastic processes treatable analytically. Its SDE is:
\displaystyle \partial_t x(t) = -\alpha x(t) + \xi(t),
where \alpha > 0 (else the process blows up) and \xi(t) is white noise:
\left\langle \xi(t)\xi(t') \right\rangle = 2\sigma \delta(t-t').

In this section, I will show that the generating functional for this Ornstein-Uhlenbeck process is given by:
\displaystyle \left\langle e^{\int_t \lambda(t) x(t)} \right\rangle = \exp \left[ \frac{\sigma}{2\alpha}\int_{-\infty}^\infty \mathrm{d} s_1 \int_{-\infty}^\infty \mathrm{d} s_2 \,e^{-\alpha|s_1-s_2|}\, \lambda(s_1) \lambda(s_2) \right] .

To show this, we apply the MSR formula to express the generating functional as a path integral:

\displaystyle \left\langle e^{\int_t \lambda(t) x(t)} \right\rangle = \int \mathcal{D}[x,\tilde{x}] \exp\left\{-\int_t i\tilde{x}(t)\left[\partial_t x(t) +\alpha x(t)\right] - \sigma\int_{t} \tilde{x}(t)^2 + \int_t \lambda(t) x(t)\right\}.

This path integral can be solved exactly. The key observation is that the exponent is linear in x(t), and hence one can integrate over this field. Using \int_x e^{i x q} \propto \delta(q), this gives (up to a normalization factor \mathcal{N}) a \delta-functional:

\displaystyle \left\langle e^{\int_t \lambda(t) x(t)} \right\rangle = \mathcal{N}\int \mathcal{D}[\tilde{x}] e^{ \sigma\int_{t} \tilde{x}(t)^2 } \delta\left[i\left(\partial_t - \alpha \right) \tilde{x}(t)+\lambda(t)\right].
The e^{ \sigma\int_{t} \tilde{x}(t)^2 } factor remains since it is the only term in the action independent of x(t).

The \delta-functional fixes \tilde{x}(t) in terms of \lambda:
\displaystyle \tilde{x}(t) = i \int_{t}^\infty \mathrm{d} s \,e^{\alpha(t-s)}\, \lambda(s) .

From this we get
\displaystyle \int_t \tilde{x}(t)^2 = - \int_t \int_{t}^\infty \mathrm{d} s_1 \,e^{\alpha(t-s_1)}\, \lambda(s_1) \int_{t}^\infty \mathrm{d} s_2 \,e^{\alpha(t-s_2)}\, \lambda(s_2) \\  =  - \frac{1}{2\alpha}\int_{-\infty}^\infty \mathrm{d} s_1 \int_{-\infty}^\infty \mathrm{d} s_2 \,e^{-\alpha|s_1-s_2|}\, \lambda(s_1) \lambda(s_2)

The normalization factor \mathcal{N} is seen to be 1 by imposing that for \lambda=0 we must have \left\langle e^{\int_t \lambda(t) x(t)} \right\rangle = 1.
Plugging this in gives the generating functional, as claimed above:
\displaystyle \left\langle e^{\int_t \lambda(t) x(t)} \right\rangle = \exp\left[ \frac{\sigma}{2\alpha}\int_{-\infty}^\infty \mathrm{d} s_1 \int_{-\infty}^\infty \mathrm{d} s_2 \,e^{-\alpha|s_1-s_2|}\, \lambda(s_1) \lambda(s_2) \right] .

One simple consequence is the exponential decay of the connected correlation function:
\displaystyle \left\langle x(t)x(t')\right\rangle^c = \frac{\delta^2}{\delta \lambda(t) \delta \lambda(t')}\big|_{\lambda=0} \ln \left\langle e^{\int_t \lambda(t) x(t)} \right\rangle = \frac{\sigma}{2\alpha}e^{-\alpha|s_1-s_2|}.

But of course, the full generating functional tells us much more – in some sense, it is the exact solution of the SDE!

Note added 01/12/2012: Of course, the generating functional can be written down immediately once one knows the two-point correlation function and the fact that the Ornstein-Uhlenbeck process is Gaussian. However, the latter fact is not immediately obvious (and was not used in our derivation), in fact one can see our MSR calculation as one way to prove Gaussianity for the OU process.

Derivation of the Martin-Siggia-Rose formula

Now that we’ve seen what the MSR path integral formulation is useful for, let us see how it can be derived. The average over an observable can be written as a path integral with the SDE enforced through a \delta-functional:
\displaystyle \left\langle \mathcal{O}[x(t)] \right\rangle = \left\langle\int \mathcal{D}[x]\,\mathcal{O}[x(t)]\,\delta\left[\partial_t x(t) - F(x(t),t) - \xi(x(t),t)\right]\right\rangle.
Now, we rewrite the \delta-functional using the formula \delta(q) \propto \int_{\tilde{x}} e^{i \tilde{x} q}. Since the \delta-functional is a product of a \delta-function at all points in space, effectively we are introducing an auxiliary field \tilde{x}(t):
\displaystyle \left\langle \mathcal{O}[x(t)] \right\rangle = \left\langle\int \mathcal{D}[x,\tilde{x}]\,\mathcal{O}[x(t)]\,\exp\left\{-\int_t i\tilde{x}(t)\left[\partial_t x(t) - F(x(t),t) - \xi(x(t),t)\right]\right\}\right\rangle.
Now, the only thing that depends on the noise \xi(x(t),t) is the factor e^{\int_t i\tilde{x}(t)\xi(x(t),t)}. Since \xi is Gaussian, the average of this factor can be simply expressed as
\displaystyle \left\langle e^{\int_t i\tilde{x}(t)\xi(x(t),t)} \right\rangle = \exp\left[-\frac{1}{2}\int_{t,t'}\tilde{x}(t)\tilde{x}(t')\left\langle\xi(x(t),t)\xi(x(t'),t')\right\rangle\right] =  \exp\left[-\int_{t,t'}\tilde{x}(t)\tilde{x}(t')G(x(t),t,x(t'),t') \right]
Altogether we obtain
\displaystyle \left\langle \mathcal{O}[x(t)] \right\rangle = \int \mathcal{D}[x,\tilde{x}]\,\mathcal{O}[x(t)]\,\exp\left\{-\int_t i\tilde{x}(t)\left[\partial_t x(t) - F(x(t),t)\right]-\int_{t,t'}\tilde{x}(t)\tilde{x}(t')G(x(t),t,x(t'),t') \right\}
which is just the MSR formula claimed above.

Amazingly, I did not yet find a clear derivation of this path integral representation on Wikipedia or anywhere else on the web… So I hope this will be useful for someone starting to work with the formalism!

If there is interest, I may write another blog post in the future with more examples and explanations e.g. on the physical role of the auxiliary field \tilde{x}.

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Written by inordinatum

September 27, 2012 at 11:03 pm