Given a quantum mechanical density matrix $\rho$, the internal energy and entropy of the system is given by:

\begin{align} E &= \text{Tr}[H \rho] \\ S &= -k_B \, \text{Tr}[\rho \ln \rho] \end{align}

where $H$ is the quantum mechanical Hamiltonian.

Then, I can minimize the Free Energy $F = E - TS$ to deduce the form of $\rho$:

\begin{align} \delta F &= \text{Tr}(\delta \rho (H+ k_B T \ln\rho+k_BT)) = 0\\ &\Rightarrow \rho = \mathcal{N}e^{-H/k_B T} \end{align}

But then, when I insert this back into $F$, I don't get the usual formula

$$ F = -T \ln(\text{Tr}[e^{-H/k_B T}]) $$

Instead I get zero. Can someone help me out with what went wrong? Thanks

  • $\begingroup$ Did you get term $k_BT\rho \ln N$ there? This should not be cancelled by any other term. $\endgroup$ Aug 2, 2019 at 22:33

1 Answer 1


You've probably forgotten about $\cal{N}$ in $\ln \rho$.
$$ \rho = {\cal{N}} e^{- H /k_B T}\ \longrightarrow\ \ln\rho = -H/k_BT + \ln{\cal{N}}\ \longrightarrow\ TS = \mbox{Tr}[\rho H] -k_BT\ln{\cal N}\ \longrightarrow $$ $$ F = k_BT\ln{\cal N}. $$ Here ${\cal N}$ is the normalization constant $$ {\cal N} = \left(\mbox{Tr} e^{-H/k_BT} \right)^{-1}. $$ The usual formula follows from two last equations


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