1
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ 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. $\endgroup$ – Nephente Dec 6 '19 at 18:19

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.