# Is Landau free energy really the free energy?

I recall reading in Negele&Orland's book that Landau free energy function is not really the free energy that one obtains from the partition function $$F = -\frac{1}{\beta}\log Z.$$ Indeed, the real free energy is already averaged over all possible states and its expansion will never produce a nice shape with multiple minima, used in Landau's analysis of phase transitions.

Question
Is Landau energy just a useful theoretical conjecture or is there a principled way of constructing it from a partition function? Examples will be appreciated.

• There is a good discussion in Section 5.6.2 of Goldenfeld's book. Apr 3 '20 at 12:14

Given the density-of-states $$\Omega(E)$$, we have (taking $$k_B=1$$) $$S(E) = \log \Omega(E),\\ \frac{1}{T}=\frac{\partial S(E)}{\partial E},\\ F(E)=E-TS(E)$$
$$Z=\sum_n e^{-\frac{E_n}{T}}=\int dE \Omega(E)e^{-\frac{E}{T}}= \int dE e^{-\frac{E-T\log\Omega(E)}{T}}=\\\int dE e^{-\frac{F(E)}{T}}=e^{-\frac{F_0}{T}}$$ When taking the thermodynamic limit, the integral can be evaluated using the steepest descent method, which gives $$F_0=\min_{E}F(E)$$