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I am currently studying "Critical Dynamics - A Field Theory Approach to Equilibrium and Non-Equilibrium Scaling Behavior", and came across an issue I can't solve.

If you know about functional derivatives and/or Martin-Siggia-Rose formalism, please skip the rather long paragraph meant to bring some context. I hope the notations are standard (I'm new to the field).

The context

If you have access to the book (no publicly available version unfortunately), starting from chapter 4, the Martin-Siggia-Rose formalism is set up for Langevin dynamics, with in this case the example :

Generalizing the Landau–Ginzburg–Wilson Hamiltonians for the Ising and Heisenberg models,[...], to n-component isotropic systems undergoing a second-order phase transition, we consider \begin{equation} H[S]=\int d^{d}x \sum_{\alpha} \left(\frac{r}{2}[S^{\alpha}(x)]^{2}+\frac{1}{2}[\nabla S^{\alpha}(x)]^{2}+\frac{u}{4!} \sum_{\beta}[S^{\alpha}(x)]^{2}[S^{\beta}(x)]^{2}-h^{\alpha}(x)S^{\alpha}(x) \right) \end{equation} With $S$ the n-component order parameter and h the conjugate external field.

The associated evolution is a Langevin equation, given by :

\begin{equation} \frac{\partial S^{\alpha}(x,t)}{\partial t}=-D(i\nabla)^{a} \frac{\delta H[S]}{\delta S^{\alpha}(x,t)}+\zeta^{\alpha}(x,t) \end{equation}

Where fluctuations are taken into account via $\zeta$, a zero mean diagonal white noise $\left< \zeta^{\alpha}(x,t) \zeta^{\beta}(x',t') \right> = 2 L\delta(x-x')\delta(t-t')\delta^{\alpha\beta}$, and $L=D(i\nabla)^{a}$.

From here is derived the generating functional for correlation functions, \begin{equation} Z[\tilde{j},j]=\left< exp \int d^{d}xdt \sum_{\alpha} \tilde{j^{\alpha}}(x,t) \tilde{S^{\alpha}}(x,t)+j^{\alpha}(x,t)S^{\alpha}(x,t)\right> \end{equation}

The $j$'s being the source currents and the tilde quantities the auxiliary MSR fields/source currents

By means of functional derivatives w.r.t. the currents, taken at vanishing currents, one obtains the correlation functions.

Taking the logarithm of Z gives you the generating functional of the connected correlation functions, and now performing a Legendre transform of the logarithm with the new variables,

\begin{align} \tilde{\Phi^{\alpha}}(x,t)=\frac{\delta \log Z[\tilde{j},j]}{\delta \tilde{j^{\alpha}}(x,t)} && \Phi^{\alpha}(x,t)=\frac{\delta \log Z[\tilde{j},j]}{\delta j^{\alpha}(x,t)} \end{align}

One ends up with the generating functional for vertex functions, \begin{equation} \Gamma[\tilde{\Phi^{\alpha}},\Phi^{\alpha}]=-\log Z[\tilde{j},j]+ \int d^{d}xdt \sum_{\alpha} \left( \tilde{j^{\alpha}}(x,t)\tilde{\Phi^{\alpha}}(x,t)+j^{\alpha}(x,t)\Phi^{\alpha}(x,t)\right) \end{equation}

All of this to map a perturbation expansion of the correlation functions in terms of Feynman diagrams. The connected correlation functions will correspond in perturbation theory to the contribution from connected graph, but as often in field theory one is interested in the "real" building blocks of the diagrammatic expansion, namely the 1-particle irreducible graphs whose contribution come from the functional derivatives of $\Gamma$, the vertex functions,

\begin{equation} \Gamma^{(\tilde{N},N)}_{\{\alpha_{i}\};\{\alpha_{k}\}}(\{x_{i},t_{i}\};\{x_{k},t_{k}\})=\prod^{\tilde{N}}_{i} \frac{\delta}{\delta \tilde{\Phi^{\alpha_{i}}}(x_{i},t_{i})} \prod^{N}_{k} \frac{\delta}{\delta \Phi^{\alpha_{k}}(x_{k},t_{k})} \Gamma[\tilde{\Phi^{\alpha}},\Phi^{\alpha}] \biggr\rvert_{j=\tilde{j}=0} \end{equation}

It can be shown, that $\Gamma^{(1,1)}_{\alpha;\beta}(x,t;x',t')$ is of the form $\Gamma^{(1,1)}(x-x',t-t')\delta^{\alpha\beta}$. In turn this form is interesting because we have, after performing a Fourier transform,

\begin{equation} \Gamma^{(1,1)}(q,\omega)=G(-q,-\omega)^{-1} \end{equation}

Thus, up to a sign in the variables, the loop diagrams contributing to $\Gamma^{(1,1)}(q,\omega)$ are nothing but the one-particle irreducible self-energy graphs.

Therefore, one can compute a perturbative expansion of the propagator by computing only the contribution from these graphs.


The issue itself

To compute the perturbative expansion of the propagator, it seems that you need to establish the Feynman rules for your theory first, and then make use of the results above to assert that you only need to compute the contributions from the 1-particle irreducible graphs.

But we also saw an explicit form of the propagator in terms of the vertex functions. Let's compute directly $\Gamma^{(1,1)}_{\alpha;\beta}(x,t;x',t')$ then !

\begin{equation} \frac{\delta \Gamma[\tilde{\Phi},\Phi]}{\delta \tilde{\Phi}^{\alpha}(x,t)} =\tilde{j}^{\alpha}(x,t) \end{equation}

and

\begin{equation} \frac{\delta^{2} \Gamma[\tilde{\Phi},\Phi]}{\delta \tilde{\Phi}^{\alpha}(x,t) \delta\Phi^{\beta}(x',t')} =\frac{\delta \tilde{j}^{\alpha}(x,t)}{\delta\Phi^{\beta}(x',t')} = ? \end{equation}

The first identity is rather easy to obtain. Though, I am not able to compute the derivative of the current at all. I tried to make use of the definition of $\tilde{\Phi}$ to make a change in variable but there it seems to be a functional change of variables. I tried the usual method, ie to take the multivariable case and let the number of variables go to infinity, changing sums into integrals, but nothing came out of it.

Is there any way to proceed like this ? Is this "functional change of variable" even a thing ? And if not, is it possible to compute the vertex functions in another way ?


Many thanks for reading up to here. I don't really expect a complete answer (unless Mr. Tauber is on this site and sees my post), but any reference, either to related content in QFT or SFT, or to a math text about this idea of functional change of variable will be greatly appreciated.

Please do not hesitate to ask any precision you'd like to see on the matter, I may have been too quick in the context (even though it is already too long).

Note: I could not attend functional analysis courses during my studies, that is why I had troubles finding anything from mathSE/math articles. Nevertheless, I'm ready to dive in if necessary.

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  • $\begingroup$ You would have the same problem in any QFT, so maybe you should have a look at a more standard $\phi^4$ theory. The thing is, you cannot compute the derivative you want, because for that you would need to know $\Gamma$, which is the thing you are trying to compute... So you need to write $\Gamma$ as a loop expansion, and compute its derivatives order by order. $\endgroup$ – Adam May 18 '18 at 13:19
  • $\begingroup$ Thanks for the comment ! Are you saying there may be a logic flaw here ? Actually, I have an full expression for $\Gamma$, because I have one for $Z$ (that's the full MSR formalism I somewhat skipped) althought it is still in a functional integral form. And I wanted to avoid that expansion beforehand, because otherwise I would rather compute the contributions to the propagator directly on my Fourier transformed Langevin equation, say. Still, I'll look more in depth at $\phi^{4}$. $\endgroup$ – Naptzer May 18 '18 at 13:28
  • $\begingroup$ "althought it is still in a functional integral form" : that's exactly what I mean by "you need to compute $\Gamma$". You have an implicit non-perturbative definition of it, which is not really helpful. Another way to phrase it is that if you don't have a useful right-hand side to the equation $\Gamma=...$ (that is, not its formal definition), you cannot compute $\Gamma^{(1,1)}$. Otherwise, you would have access to the exact effective action without performing any explicit calculation... $\endgroup$ – Adam May 18 '18 at 13:32
  • $\begingroup$ @Adam Now that I've dug more into this i definitely realize your last comment is the answer to my question. I'll accept it as the answer if you also point out that ofc, what matters is using $\Gamma^{(1,1)}=G^{-1}$ with an ansatz for $\Gamma$. I could do it myself but I want to give you credit, for being right before I could understand you were. $\endgroup$ – Naptzer Jul 30 '18 at 18:14
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There are two issues at stake here.

First, concerning the question itself :

By definition, $\Gamma$ is the Legendre transform of $W=\ln Z$, $$ \Gamma[\phi]=-W[J]+\phi.J,\\ \frac{\delta \Gamma}{\delta\phi_A}=J_A, $$ where I use condensed notations $J_A$, $\phi_A$, $J.\phi=J_A \phi_A$, etc. with $A$ collecting space-time coordinates, indices, and so on.

We also know that $\frac{\delta W}{\delta J_A}=\phi_A$ and $\frac{\delta^2 W}{\delta J_A\delta J_B}=\frac{\delta \phi_A}{\delta J_B}=G_{AB}$.

Thus, we find that $$ \frac{\delta^2 \Gamma}{\delta\phi_A\delta \phi_B}=\frac{\delta J_A}{\delta\phi_B}=\left(\frac{\delta \phi_B}{\delta J_A}\right)^{-1}=(G^{-1})_{AB}. $$

Up to the Fourier transform, this is the equation given in the OP. In fact, this equation, and the one one can obtain by taking more functional derivatives, just give us the relationship between the connected correlation functions, and the vertex function, nothing more.

This leads us to the second issue : using the formal definition of $\Gamma$ does not help to compute neither $\Gamma^{(2)}$ nor $G$ (unless we know $W$ of course). We thus need another way to compute $\Gamma$, and then use this to compute $G$.

One possibility is to compute $\Gamma$ as a loop expansion (it is better to compute $\Gamma$ than $W$ because there are less diagrams (only the 1PI) and resummations of the self-energies are already performed explicitly).

There are other kind of approximations, using Ansatz of $\Gamma$ and using some RG equation to find the coefficients (in the context of out-of-equilibrium dynamics, see for instance Phys. Rev. Lett. 92, 195703 (2004))

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The trick to compute $\Gamma^{(1,1)}$ is to compute the cross derivative

$$ \frac {\delta^2 \Gamma}{\delta\Phi^\alpha(x,t)\delta j^\beta(x',t')}. $$ This can be done in two ways. On one hand, you know the result is $\delta_{\alpha\beta} \delta(x-x') \delta(t-t')$, because this is simply the derivative of $j^\alpha(x,t)$. But you can also compute it using the chain rule to express the derivatives of $\Gamma$ wrt $j^\beta(x,t)$ in terms of derivatives of $\Gamma$ wrt $\Phi^\alpha(x,t)$, which is the natural variable of $\Gamma$. You obtain

$$ \int\!\!d^dx''\int\!\!dt''\, \Gamma_{\alpha,\beta}^{(1,1)}(x,t;x'',t'') G(x''-x',t''-t') =\delta(x-x')\delta(t-t')\delta_{\alpha,\beta} $$ And you invert the convolution in Fourier space to obtain $$ \Gamma^{(1,1)}(q,\omega) = G^{-1}(-q,-\omega). $$

This is done with more detail in sec. 4.4.2 in the book by Tauber.

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  • $\begingroup$ Thanks a lot for the answer @TomasG, though I knew this at the time of the question. My issue was that I did not see that what is needed is to have an ansatz for the theory under study, and then make use of the identity you mentioned. $\endgroup$ – Naptzer Jul 30 '18 at 20:04
  • $\begingroup$ Sorry, I followed your instructions and skipped the "context" part since I knew what you were talking about, and I missed that part. I misunderstood your question. $\endgroup$ – TomasG Jul 31 '18 at 13:51

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