An introduction to the Martin-Siggia-Rose formalism

Posted on September 27, 2012February 20, 2022 by inordinatum

Note: I’ve posted an example of how the below can be applied in a non-trivial example in a recent article on solving the Heston model with the MSR formalism

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

$\begin{array}{rl}\displaystyle \int_t \tilde{x}(t)^2 & \displaystyle = - \int_t \int_{t}^\infty \mathrm{d} s_1 \,e^{\alpha(t-s_1)}\, \lambda(s_1) \int_{t}^\infty \mathrm{d} s_2 \,e^{\alpha(t-s_2)}\, \lambda(s_2) \\[1cm] & \displaystyle = - \frac{1}{2\alpha}\int_{-\infty}^\infty \mathrm{d} s_1 \int_{-\infty}^\infty \mathrm{d} s_2 \,e^{-\alpha|s_1-s_2|}\, \lambda(s_1) \lambda(s_2) \end{array}$

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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25 thoughts on “An introduction to the Martin-Siggia-Rose formalism”

The Alessandro-Beatrice-Bertotti-Montorsi model « inordinatum October 11, 20128:37 pm Reply

[…] obtain information on avalanches in the ABBM model is mapping the SDE above to a path integral (the Martin-Siggia-Rose formalism). This is done e.g. in references [3] and [4] below, where the resulting path integral is […]
Sourabh Lahiri January 20, 20143:56 am Reply

I am a postdoctoral fellow in nonequilibrium statistical mechanics, and have recently got interested in the MSR formalism outlined by you in an elegant fashion. I am very interested in knowing the physical implications of the auxiliary field, through a few examples, and would greatly appreciate if you could briefly explain this. In particular, what is the significance of the auxiliary field in Langevin dynamics for a Brownian particle?
1. inordinatum January 21, 201410:31 pm Reply
  
  Hi Sourabh,
  thank you for reading my blog, and for your interesting question!
  
  In a “handwaving” way, the response field generates an infinitesimal external driving which acts on the particle trajectory. Computing the average of an observable $\mathcal{O}[x]$ multiplied by $\tilde{x}(t_0)$ is the same as taking the response of $\mathcal{O}[x]$ to a tiny “kick” at $t_0$ .
  
  To see this more formally, redo the calculation adding an additional driving $D\delta(t-t_0)$ to the original SDE. Then you get that
  $\left\langle \mathcal{O}[x] \right\rangle_D = \int \mathcal{D}[x,\tilde{x}]\,\mathcal{O}[x] \,e^{D\tilde{x}(t_0)-S[x,\tilde{x}]}$ , where by $\left\langle \cdot \right\rangle_D$ on the left I mean an average over all trajectories of the SDE with additional driving $D\delta(t-t_0)$ . This shows that
  $\left\langle \tilde{x}(t_0) \mathcal{O}[x] \right\rangle = \partial_D\big|_{D=0} \left\langle \mathcal{O}[x] \right\rangle_D$ ,
  i.e. inserting an $\tilde{x}(t_0)$ into an observable amounts to taking its change (formally the derivative) by an additional “push” of the particle at $t=t_0$ through some external driving — hence the name “response field”.
  
  Let me know if this clarifies things!
  1. Sourabh Lahiri January 22, 20141:40 am
    
    Dear inordinatum,
    Thank you very much for your reply. It clears the significance of the auxiliary field.
    
    Best Regards, Sourabh
Ryan Daly August 13, 201410:46 pm Reply

What are the analogous variables in quantum field theory? I am assuming that this is a generalization of QFT. What is the auxiliary field’s counterpart in QFT?
1. inordinatum August 18, 20148:50 pm Reply
  
  Hi Ryan,
  good question!
  
  The situation here is really more similar to standard quantum mechanics in a path integral approach (see here for details), than to a QFT: We have a single degree of freedom $x(t)$ which only depends on $t$ , in a QFT we would consider a field $\phi(x,t)$ with one degree of freedom at every point in space.
  Another difference is the starting point: Here we derive the path integral action assuming a given stochastic differential equation $\partial_t x=F(x)+\xi$ . One wouldn’t do that for QM: I don’t know of a natural way to add “quantum” noise to an equation of motion.
  
  Nevertheless, mathematically, the auxiliary field we introduce here is very similar to momentum in quantum mechanics. In particular, as the discussion with Sourabh in the comments above shows, $\tilde{x}$ acts as a derivative with respect to $x$ . This is just the property known from basic quantum mechanics for the momentum operator.
  
  Hope this helps!
  Alexander
  1. Ryan Daly August 23, 20143:24 pm
    
    Alexander,
    
    Thanks for the reply. I’m familiar with path formulations, but maybe I was generalizing too much. There is a theory which mimics QFT called the time-dependent Ginzburg-Landau theory which does exactly this: add Gaussian noise to a field at every point. It comes out with a diagrammatic theory similar (but still quite different from) the standard Feynman formulation of QFT perturbation theory.
    
    I guess this Ginzburg-Landau theory was what I was thinking of when I asked about QFT. I see that the main difference between your formulation here and QFT is that here x is a dependent variable and in QFT/etc. x and t are both independent variables. I just wonder if there is a way to extend MSR to the Ginzburg-Landau theory to get some sort of new formulation.
    
    The TDGL equation looks like this (can I use LaTeX?):
    
    $\frac{\partial\psi}{\partial t} = \gamma(x) \frac{\delta}{\delta \psi(x,t)} H + \eta(x,t)$
    
    And you’re right… I have been reading for years in this area and not once have I come upon MSR! I was kind of shocked at this. I hope I didn’t miss anything too important. Thanks for posting this article!
Ryan Daly August 23, 20143:29 pm Reply

I guess I can’t use latex. Here is the generated image of the equation in the previous post of mine:

https://drive.google.com/file/d/0B-QC89qDrwx_bVh0NlZyUnpTclU/edit?usp=sharing
1. inordinatum August 30, 20142:50 pm Reply
  
  Dear Ryan,
  
  indeed, the TDGL equation you are interested in can also be transformed into a field theory using the MSR formalism. In this case, you need to multiply it by a conjugate field $\bar{\psi}(x,t)$ , and obtain (following the same steps as in this article) a field theory action like
  
  $S[\psi,\bar{\psi}] = \bar{\psi}\left(\partial_t \psi - \gamma(x)\frac{\delta}{\delta \psi} H \right) + \bar{\psi}^2$ .
  
  Here you have a close analogy to QFT: the action looks very similar to the Dirac action for electrons (up to the spinor/symmetry structure which is not present here), and $\bar{\psi}$ can be identified with the antiparticle of $\psi$ . Also the diagrammatic expansion of the field theory is similar: In both cases, the fields $\psi$ and $\bar{\psi}$ correspond to incoming and outgoing legs in an interaction vertex. The propagators have a time direction (due to the first-order dynamics) and link an outgoing leg $\psi$ to an incoming leg $\bar{\psi}$ .
  
  May I ask what applications of this formalism and the TDGL you’re working on an interested in?
  
  Alexander
  
  P.S. I edited your post above so that the latex shows up, in fact you need to put $ latex …formula… $ (without the first space) for it to do so on wordpress…
Ryan Daly August 31, 20149:23 pm Reply

HI Alexander,

I am trying to look at a type of glass transition (or locking) which may occur in ionic liquids. I am just becoming familiar with this application of a classical field theory similar to QFT. QFT I have had experience with through reading and working out details on numerous legal pads (lots of money spent on those terrible things).

It may turn out that this isn’t very enlightening, but I want to try every avenue. My current understanding of glass transitions is via a type of generalized kinetics which has shown up in literature over the years. A lot of it has been brought into existence by people like Gene Mazenko, Peter Wolynes, and Hans Andersen. These people know what they are doing and I’m trying to catch up.

MSR could enhance my comprehension of this type of field theory I think.

I appreciate it!

Ryan
Ryan Daly September 8, 20147:20 am Reply

The paper the MSR methodology is originally found in is Phys. Rev. A 8, 423 (1973). It turns out that this is exactly what I am looking for. This is more than a reformulation of classical dynamical systems with a conjugate “field.” This paper demonstrates how to extend the beautiful formality of quantum field theory (the LSZ reduction, correlators as multi-state propagators, etc.) to classical mechanics. The result is a method for creating renormalized perturbation theories for dynamical problems.

More to the point, MSR can help me develop mode-coupling theories for glassy liquids (particularly ionic liquids)!

Thanks again!
Ryan Daly September 8, 20147:45 pm Reply

Alexander,

If you have any more insight into the physical role of the auxiliary field variable, it would definitely be helpful. I need to visualize the importance of the new variable. The math all makes sense, but artificially introducing another field “conjugate” to the original field variable which creates “excitations” in the initial state is not obvious to me.

I’m reading, but the MSR paper seems to be somewhat vague.
Braden March 19, 20155:06 am Reply

For the interested who stumble across this, there is a formalism for quantum many-body non-equilibrium problems, called the “Schwinger-Keldysh” or “closed time path” formalism. This formalism reduces to MSR in the \hbar -> 0 limit. The correspondence between fields in MSR and the quantum analogues is made clear by taking this limit, though I don’t remember the precise statement off the top of my head.

Kleinert (http://www.amazon.com/dp/9814273562/ ) had a nice intro to the method, and this stackoverflow page has several links to material on it: http://physics.stackexchange.com/questions/13997/good-reading-on-the-keldysh-formalism
1. inordinatum March 19, 20155:28 pm Reply
  
  Thanks, this is interesting, indeed!
Ivar November 27, 20162:43 pm Reply

Hi Alexander, stumbled upon your blog when searching for MSR. Wonderful explanation — thank you! I am in the process of building a public wiki-type collaborative resource, knowen.org. It already has some physics content, but none of the topics you cover. Would you be interested in contributing? thanks!
1. inordinatum June 30, 20178:21 am Reply
  
  Hi Ivar,
  thanks for your interest – glad to hear that! knowen.org sounds like a very good idea – I’ll check it out and get back to you 🙂
Leo June 29, 20176:31 pm Reply

Excellent post. May I ask how the following integral is derived?
$\displaystyle \int_t \tilde{x}(t)^2 = - \int_t \int_{t}^\infty \mathrm{d} s_1 \,e^{\alpha(t-s_1)}\, \lambda(s_1) \int_{t}^\infty \mathrm{d} s_2 \,e^{\alpha(t-s_2)}\, \lambda(s_2) \\ = - \frac{1}{2\alpha}\int_{-\infty}^\infty \mathrm{d} s_1 \int_{-\infty}^\infty \mathrm{d} s_2 \,e^{-\alpha|s_1-s_2|}\, \lambda(s_1) \lambda(s_2)$

Any help is appreciated!
1. inordinatum June 30, 20178:15 am Reply
  
  Hi Leo,
  glad you like it!
  The first equality follows from the definition of $\tilde{x}$ in the line directly above this equation. To see the second equality, consider:
  $- \int_t \int_{t}^\infty \mathrm{d} s_1 \,e^{\alpha(t-s_1)}\, \lambda(s_1) \int_{t}^\infty \mathrm{d} s_2 \,e^{\alpha(t-s_2)}\, \lambda(s_2) \\ = - \int_t \int_{-\infty}^\infty \mathrm{d} s_1 \int_{-\infty}^\infty \mathrm{d} s_2 \,e^{\alpha(t-s_1)}\, e^{\alpha(t-s_2)} \,\theta(s_1-t)\theta(s_2-t)\, \lambda(s_1)\,\lambda(s_2) \\ = - \int_{-\infty}^{\mathrm{min}(s_1,s_2)}\mathrm{d}t\, \int_{-\infty}^{\infty} \mathrm{d} s_1 \int_{-\infty}^\infty \mathrm{d} s_2 \,e^{2\alpha\,t-\alpha\,s_1-\alpha\,s_2} \, \lambda(s_1)\,\lambda(s_2) \\ = - \frac{1}{2\alpha}\int_{-\infty}^\infty \mathrm{d} s_1 \int_{-\infty}^\infty \mathrm{d} s_2 \,e^{-\alpha(s_1+s_2-2\mathrm{min}(s_1,s_2))}\, \lambda(s_1) \lambda(s_2) \\ = - \frac{1}{2\alpha}\int_{-\infty}^\infty \mathrm{d} s_1 \int_{-\infty}^\infty \mathrm{d} s_2 \,e^{-\alpha|s_1-s_2|}\, \lambda(s_1) \lambda(s_2)$
  1. Leo July 3, 20179:17 am
    
    I also found it out. Thanks a lot!
Fluctuations of the time-weighted average price | inordinatum June 4, 20185:03 pm Reply

[…] is another way to obtain the full distribution of the TWAP directly, via the Martin-Siggia-Rose formalism. Using the MSR approach, we know that the generating function of any linear functional of the […]
Solving the Heston model with the Martin-Siggia-Rose formalism | inordinatum May 1, 202012:03 pm Reply

[…] In the Heston model, one can calculate closed-form expressions for many observables. Let us focus here on the probability distribution of the log-returns of the asset price, i.e. . It has been calculated via the Fokker-Planck formalism e.g. in this paper by Drǎgulescu and Yakovenko [1]. Here I will show how the same result can be obtained via the Martin-Siggia-Rose formalism explained in an earlier post of mine. […]
H.K. Janssen June 12, 20206:14 am Reply

The MSR formalism is an operator formalism, not a path integral formalism. More correctly referred as MSRJD formalism since the path integral was introduced by Janssen (Z.Phys. B 23, 377 (1976), Z.Phys. B 24, 176 (1976)) and DeDominicis (J.Phys. C 37, 247 (1976), Phys.Rev. B 18, 353 (1978)). See for instance the book of Taeuber: Critical Dynamics – A field theory approach to equilibrium and non-equilibrium scaling behavior (2014).
philosophicalmath June 23, 20206:16 pm Reply

Hello, I have a quick questions about a part of the MSR derivation. How do you show that $\left\langle e^{i\int_t \tilde{x}(t)\xi(x(t), t)}\right\rangle = \exp\left( \int_{t, t'} \tilde{x}(t) \tilde{x}(t') \langle \xi(x(t), t) \xi(x(t'), t') \rangle \right)$ ?
1. inordinatum June 24, 20208:59 am Reply
  
  Thanks for your good question!
  This equality is the moment generating function of a Gaussian process. For a single Gaussian random variable $\xi(0)$ (i.e. in the special case where $\tilde{x}(t) = a\, \delta(t)$ ), you obtain it by integrating over the Gaussian distribution and completing the square in the exponent (see e.g. these lecture notes). For multiple jointly Gaussian variables $\xi(t_0), \xi(t_1), ... \xi(t_N)$ you follow exactly the same steps with an $N$ -dimensional Gaussian integral. Then you just convince yourself that the continuum limit works 🙂 Hope this helped!
Ivan September 7, 20208:20 am Reply

Thanks for the clear explanation. However, as you said before, the hypothesis that the process is gaussian is necessary in the derivation of the Martin-Siggia-Rose. Hence, MSR formalism does not show that the Ornstein-Uhlenbeck process is gaussian, instead this is an hypothesis implicitely done writing down the SDE (xi is GAUSSIAN white noise). Do you agree?