I Minimize grand potential functional for density matrix

[dRic2]

TL;DR Summary

$\rho$ is the density matrix
$\Omega$ is the grand potential
$\text{Tr}$ stands for 'trace'

I'd like to show that, by minimizing this functional
$$\Omega[\hat \rho] = \text{Tr} \hat \rho \left[ \hat H - \mu \hat N + \frac 1 {\beta} \log \hat \rho \right]$$
I get the well known expression
$$\Omega[\hat \rho_0] = - \frac 1 {\beta} \log \text{Tr} e^{-\beta (\hat H - \mu \hat N )}$$

I'm familiar with minimizing a functional of the form $F[g] = \int dx f(g(x))$, but this notations for operators eludes me.

Thanks,
Ric

 

[dRic2]

I've found the original paper (PhysRev.137.A1441) where Mermin introduced this functional. You can check the proof in the Appendix (If I'm not mistaken it was first proved by von Neumann).

To sum up the idea behind the proof is to define the operator $\rho_{\lambda} = \frac {e^{H-\mu N + \lambda \Delta}} {\text{Tr}e^{H-\mu N + \lambda \Delta}}$ with $\Delta = -H + \mu N - \frac 1 {\beta} \log \rho$. You can see that because of $\text{Tr} \rho =1$ for $\lambda = 1$ I get $\rho_1 = \rho$ and for $\lambda = 0$ you get $\rho = \rho_0$ (the equilibrium value). You can then proceed to study the following relation $\Omega[\rho] - \Omega[\rho_0] = \Omega[\rho_{\lambda = 1}] - \Omega[\rho_{\lambda = 0}] = \int_0^1 \frac {\partial \Omega[\rho_{\lambda}]} {\partial \lambda} d \lambda$ and check that is always greater than zero and zero only if $\lambda = 0$ thus proving that $\rho_0$ minimizes the functional