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Mayer expansions and the Hamilton–Jacobi equation by D. Brydges and T. Kennedy begins mentioning that many problems in statistical mechanics and QFT center on the analysis of integrals of the form: \begin{equation} \int d\mu_{C}(\varphi)e^{-V(\varphi+\varphi')} \tag{1} \label{1} \end{equation} where $d\mu_{C}$ is a Gaussian measure with mean zero and covariance $C$. I'm primarily interested in statistical mechanics so I'd like to understand these integrals in this context. I don't know much about QFT but (correct me if I'm wrong) a common object of study which satisfies (\ref{1}) is the so-called effective action $\mathcal{G}_{\lambda, C}$. But this effective action is also useful in statistical mechanics. To be more specific, we define it as: \begin{equation} \mathcal{G}_{\lambda,C}(\psi) \mathrel{:=} \log \int d\mu_{C}(\varphi)e^{-\lambda V(\psi+\varphi)} \tag{2}\label{2} \end{equation} where $V$ is an interaction function (I'm assuming here that (\ref{2}) is well-defined). Note that (\ref{2}) has the exact same form of (\ref{1}). Now, let us consider: \begin{equation} Z_{\lambda}(\psi) \mathrel{:=} \int d\mu_{C}(\varphi) e^{-\lambda V(\varphi)+\langle C^{-1}\psi, \varphi\rangle} \tag{3}\label{3} \end{equation} where $\langle \cdot, \cdot \rangle$ is a given inner product. This is called the partition function of the system with source term $C^{-1}\psi$. The partition function (and also its logarithm) is a powerful tool to calculate correlation functions. With a little algebra, we can prove that: \begin{equation} \mathcal{G}_{\lambda,C}(\psi) = -\frac{1}{2}\langle \psi, C^{-1}\psi\rangle + \log Z_{\lambda}(\psi) \tag{4}\label{4} \end{equation} and thus we can also use the effective action to study correlation functions. It is often the case that the covariance $C$ can be written as a sum of covariances $C= C_{1}+\dotsb + C_{n}$. Taking $n=2$ for simplicity, plugging it into $\mathcal{G}_{\lambda V,C}$ and using the property that a sum of Gaussian random variables is again a Gaussian random we get: \begin{equation} \mathcal{G}_{\lambda V,C_{1}+C_{2}}(\psi) = \mathcal{G}_{\mathcal{G}_{\lambda V, C_{1}}, C_{2}}(\psi) \tag{5}\label{5} \end{equation} which is the semi-group property of the effective action. Recursively, this also hold for $n>2$. Thus, it is sufficient to study the map: \begin{equation} -\log \int d\mu_{C_{i}}(\varphi)e^{-\lambda V_{i}(\varphi+\psi)} \tag{6}\label{6} \end{equation} where $V_{i}$ is properly chosen.

The point that is not clear to me is the following. In statistical mechanics, one is primarily interested in evaluating the partition function, even though other quantities (such as the ones defined above) are also relevant. I'd like to better understand the relation between the above objects and the actual partition function of the system (not the one with source terms) and what is the difference between approaching a problem using the effective action or the partition function itself, is there is any.

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These correspondences might be helpful: we have the partition function $Z(\beta)$, the effective action $W(\beta)$, the classical action $I_\beta[\phi]$ for physical variables $\phi(x,\tau)$ in Euclidean time $\tau$ with period $\beta$ equal to inverse temperature. These are related by $$ e^{-W(\beta)}=Z(\beta)=\int D\phi\, e^{-I_\beta[\phi]}$$ $$\qquad\qquad = \int D\phi \exp\left[-\int_0^\beta d\tau\,\int d^3 x\,\bigl(\tfrac{1}{2}(\nabla\phi)^2+V(\phi)\bigr)\right].$$ The thermodynamic free energy is $F(\beta)=\beta^{-1}W(\beta)$, the entropy \begin{equation} S(\beta)=\beta^2 \frac{dF(\beta)}{d\beta}=\left(\beta\frac{d}{d\beta}-1\right)W(\beta). \end{equation}

My answer to the question in the OP, "what is the difference between approaching a problem using the effective action or the partition function itself" would be --- there is no difference, one is the logarithm of the other.

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  • $\begingroup$ Thanks for the answer. So, if I understood it correctely, the partition function is my effective action (\ref{2}) at $\psi=0$? $\endgroup$ – IamWill May 3 at 20:00
  • $\begingroup$ (2) is (minus) the effective action, to obtain the partition function you would exponentiate (2) --- indeed at $\psi=0$. $\endgroup$ – Carlo Beenakker May 3 at 20:34

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