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.

Improve this question
$\endgroup$
1
  • 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, 2019 at 18:19

0

Reset to default

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.