# Von Neumann entropy for non-interacting systems

I am trying to compute the von Neumann entropy for a system described by a non-interacting Hamiltonian:

$$H = \sum_\alpha\epsilon _\alpha N_\alpha$$

where N is the number operator $$N_\alpha = a^\dagger_\alpha a_\alpha$$

The system is placed in the grand-canonical ensemble:

$$\rho = e^{(-\beta\sum_\alpha(\epsilon_\alpha - \mu )N_\alpha)}/tr{(e^{-\beta\sum_\alpha(\epsilon_\alpha - \mu )N_\alpha})}$$

Since in the sum $$a^\dagger_\alpha a_\alpha$$ commute with $$a^\dagger_{\alpha'} a_{\alpha'}$$ for any $${\alpha'}$$, $${\alpha}$$, I can write the denominator as:

$$tr{(e^{-\beta\sum_\alpha(\epsilon_\alpha - \mu )N_\alpha})} = \prod_\alpha tr{(e^{-\beta\sum_\alpha(\epsilon_\alpha - \mu )N_\alpha})}.$$

For fermions, for example, this is $$\prod_\alpha tr{(e^{-\beta\sum_\alpha(\epsilon_\alpha - \mu )N_\alpha})} = \prod_\alpha \sum_{n = 0,1}e^{-\beta\sum_\alpha(\epsilon_\alpha - \mu )n})$$ = $$\prod _\alpha(1 + e^{-\beta(\epsilon _\alpha - \mu)})$$.

I am stuck in the direct calculation of the entropy:

$$S = - tr(\rho ln(\rho))$$

My atempt:

$$ln(\rho) = -\sum_\alpha [\beta(\epsilon_\alpha - \mu)N_\alpha + ln(1 + e^{-\beta(\epsilon_\alpha - \mu)})].$$

And $$\rho ln(\rho) = [\prod_\alpha {(e^{-\beta\sum_\alpha(\epsilon_\alpha - \mu )N_\alpha})}/(1 + e^{-\beta(\epsilon_\alpha - \mu)})] \times [-\sum_\alpha [\beta(\epsilon_\alpha - \mu)N_\alpha + ln(1 + e^{-\beta(\epsilon_\alpha - \mu)})]]$$

But I cannot go any further to take the trace of this thing.

• Possibly the simplest way to obtain the entropy is taking the derivative of the grand-canonical potential $S=-\partial_T\Phi$ where $\Phi = -T\ln Z$ is the logarithm of the partition sum which you already calculated. – Nephente Dec 6 '19 at 18:19