This question concerns an interacting system's thermodynamic free energy $\Omega$. Generally speaking, The action $S$ for an interacting system has the following form:
\begin{equation} S(\phi,\psi) = S_{\mathrm{free}}(\phi) + S_{\mathrm{free}}(\psi) + S_{\mathrm{int}}(\phi,\psi) \end{equation} Where $\phi$ and $\psi$ are two generic fields.
The thermodynamic potential of the system is then given by $$\Omega = -T \ln Z$$ Where the partition function is given by $$ Z = \int d\phi d\psi ~e^{-\beta S(\phi,\psi)} = \int d\phi d\psi ~e^{-\beta \Bigl(S_{\mathrm{free}}(\phi) + S_{\mathrm{free}}(\psi) + S_{\mathrm{int}}(\phi,\psi)\Bigr)} $$
The problem is to evaluate the integral $$ \Omega = -T \ln~ \left( \int d\phi d\psi ~e^{-\beta \Bigl(S_{\mathrm{free}}(\phi) + S_{\mathrm{free}}(\psi) + S_{\mathrm{int}}(\phi,\psi)\Bigr)} \right) $$
In general, it is quite hard to evaluate the above integral. Landau invented a trick to evaluate this, which has been discussed in his book. Later AGD extended Landau's work in field theory. There exists a whole simulation industry dedicated to evaluating this integral numerically. The way I understand the problem goes as follows:
Let's first introduce a dimensionless "adiabatic parameter" $\lambda$, such that when $\lambda = 0$, the system is "free," and we know how to evaluate $\Omega$. Now, when $\lambda = 1$, the system is fully interacting, and we are interested to know the corresponding $\Omega$. This suggests that the change in $\Omega$ can be written as: $$ \Delta\Omega = \int_{0}^{1} d\lambda \frac{\partial\Omega(\lambda)}{\partial\lambda} $$
In terms of $\lambda$ the $Z$ can be written as
$$ Z(\lambda) = \int d\phi d\psi ~e^{-\beta S(\lambda;\phi,\psi)} = \int d\phi d\psi ~e^{-\beta \Bigl(S_{\mathrm{free}}(\phi) + S_{\mathrm{free}}(\psi) + \lambda S_{\mathrm{int}}(\phi,\psi)\Bigr)} $$
Then $$ \Delta\Omega = -T\int_{0}^{1} d\lambda ~ \frac{1}{Z(\lambda)} \frac{\partial Z(\lambda)}{\partial\lambda} $$
Then $$ \Delta\Omega = T\int_{0}^{1} d\lambda ~ \frac{1}{Z(\lambda)} \left[\int d\phi d\psi~ S_{\mathrm{int}}(\psi,\phi) e^{-\beta \Bigl(S_{\mathrm{free}}(\phi) + S_{\mathrm{free}}(\psi) + \lambda S_{\mathrm{int}}(\phi,\psi)\Bigr)} \right] $$
Which can be written as
$$ \Delta\Omega = T\int_{0}^{1} d\lambda ~ \frac{\langle S_{\mathrm{int}}(\psi,\psi)\rangle_{\lambda}}{Z(\lambda)} $$ Where, $$ \langle S_{\mathrm{int}}(\psi,\psi)\rangle_{\lambda} = \int d\phi d\psi~ S_{\mathrm{int}}(\psi,\phi) e^{-\beta \Bigl(S_{\mathrm{free}}(\phi) + S_{\mathrm{free}}(\psi) + \lambda S_{\mathrm{int}}(\phi,\psi)\Bigr)} $$
We have been able to get rid of the log with the price of introducing the integration over $\lambda$! A further complication arises since both the numerator and the denominator of $\Delta\Omega$ depend on $\lambda$. To my knowledge, a variant of the above expression was first given by John G. Kirkwood.
Question: The way one can evaluate this expression is to evaluate the averages of interaction terms for different orders in $\lambda$. Then, pull the denominator up, expand it to a power series in $\lambda$, and simplify and integrate over $\lambda$ at the end. Does this proposition sound reasonable? Hopefully, PSE people can inject their input.
