Why/When does thermal Entropy equal von Neumann Entropy?

Question

I'm reading a paper that claims that for thermal states both entropies are equal up to the Boltzmann-Faktor.

'for states in thermal equilibrium, i.e. states of the form ... it is known that the thermodynamic entropy equals the Von Neumann entropy. '

I've been trying to find out more about this but to no avail :/

Can anyone help ?

Answer 1

The von Neumann entropy reads $$ S(\rho)= -\mathrm{Tr}( \rho \log \rho),$$ where $\rho$ represents the state for which you want to compute the entropy. For thermal states we have that $$ \rho = \frac{e^{-\beta H}}{Z},$$ with $Z$ the partition function and $\beta=\frac{1}{k_B T}$. Now, if you plug the expression for $\rho$ into the definition of the von Neumann entropy you will find, after el little of algebra, the following expression $$ S(\rho) = \frac{1}{k_B T} \langle E \rangle + \log Z.$$ Now, from thermodynamics we know that $$F = U - T S.$$ Also, from statistical mechanics, we saw that for the canonical ensamble (this case) $F=-k_B T \log Z$. With these things in mind we can see that $$F=-k_B T \log Z= \langle E \rangle - T(k_B S(\rho_{thermal})).$$ Then, for thermal states the von Neumann entropy coincides with the thermodynamic entropy up to a factor equals to $k_B$.

Answer 2

For quantum states the von Neumann entropy quantifies the informational entropy. If a quantum system is weakly coupled to a bath at equilibrium we may associate a canonical ensemble and Gibbs entropy to it.

In this circumstance, the von Neumann entropy is equal to the Gibbs entropy giving an explicit connection between information and thermodynamics.

Consider the Gibbs entropy may be expressed in terms of the Gibbs Free Energy as $$ F = U - TS \, \implies \, S = -\beta\left(U - F\right)\\ S_G = -\beta\left(\langle \hat{H}\rangle - t\ln \mathcal{Z}\right)\\ = -\beta \text{Tr}\left\{\hat{\rho}\hat{H}\right\} + \ln \mathcal{Z} $$ and since this is the entropy of a Gibbs state we have $$ = -\beta \sum^{N}_{i=0} \frac{e^{-\beta E_i}}{\mathcal{Z}}E_i + \ln\mathcal{Z}. $$ For malleability, we may multiply the logarithm of the partition function by the trace of the density matrix, which is 1, giving $$ = -\beta \sum^{N}_{i=0} \frac{e^{-\beta E_i}}{\mathcal{Z}}E_i + \ln\mathcal{Z}\text{Tr}\{\hat{\rho}\}\\ = -\beta \sum^{N}_{i=0} \frac{e^{-\beta E_i}}{\mathcal{Z}}E_i + \ln\mathcal{Z}\sum^{N}_{i=0} \frac{e^{-\beta E_i}}{\mathcal{Z}}\\ = \sum^{N}_{i=0} \frac{e^{-\beta E_i}}{\mathcal{Z}} \left(-\beta E_i + \ln \mathcal{Z}\right) \\ = -\sum^{N}_{i=0} \frac{e^{-\beta E_i}}{\mathcal{Z}}\left(\ln\left(\frac{e^{-\beta \hat{H}}}{\mathcal{Z}}\right)\right)\\ = -\sum^{N}_{i=0}\lambda_i \ln \lambda_i =-\text{Tr}\left\{\hat{\rho}\ln\hat{\rho}\right\} = S_{vN} $$ where $S_{vN}$ is the von Neumann entropy.