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 ?

  • $\begingroup$ Give us the paper name and author. $\endgroup$ – DanielC Nov 23 '18 at 23:12

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 $k_B$.


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