# Why/When does thermal Entropy equal von Neumann Entropy?

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 ?

• Give us the paper name and author. – 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 $$k_B$$.