0
$\begingroup$

Knowing that, in the $\mu$-canonical (or micro-canonical) and canonical ensembles, the number of particles is held constant and usually reflects the actual number of particles, in which case $N_{MC} \simeq N_{actual}$, and also $N_C \simeq N_{actual}$. Or otherwise

$\text{sgn}(N_{MC}) = \text{sgn}(N_{actual})$

$\text{sgn}(N_C) = \text{sgn}(N_{actual})$

where $\text{sgn}(x)$ is the sign of x. But here, $N_{GC}=-\left(\frac{\partial\Omega}{\partial\mu} \right)_{VT}$, defined as a derivative and may not reflect the actual number of particles, and $N_{GC}$ can even be negative, given its definition.

For this reason, I would tend to believe that a system where

$\text{sgn}(N_{GC}) \neq \text{sgn}(N_{actual})$

could exist, that is, $N_{GC}<0$ while $N_{actual}\geq 0$, and hence negative numbers of particles are a physical reality, albeit only in a grand canonical ensemble. Or would even a grand canonical ensemble necessarily imply

$\text{sgn}(N_{GC}) = \text{sgn}(N_{actual})$

i.e., $N_{GC}<0$ also imply $N_{actual}<0$?

$\endgroup$
  • $\begingroup$ $N_G$ cannot be smaller than zero as it is defined as $N_G = \frac 1 {Z_G} \mathrm{Tr}(\rho_G N)$ and $N$ is a positive operator. To be smaller than zero would mean for $\rho$ to be not positive, but then $\rho$ is not a valid density matrix (i.e. does not describe a state as it assigns negative probability to some configurations). $\endgroup$ – Sebastian Riese Oct 1 '15 at 16:47
1
$\begingroup$

No, this cannot happen, simply because in general $N_G \ge 0$, which can easily be proven in a quantum-statistical setting.

$N_G$ is defined as $N_G = \frac 1 {Z_G} \mathrm{Tr}(\rho_G N)$, now consider the following expansion of the trace in terms of a complete set of states $\left|n\alpha\right>$ which are eigenstates of the number operator with the eigenvalue $n$: \begin{align*} N_G &= \frac 1 {Z_G} \mathrm{Tr}(\rho_G N) = \frac 1 {Z_G} \sum_{n\alpha, m\beta} \left<n\alpha \right| \rho_G \left|m\beta\right>\left<m\beta\right| N \left|n\alpha\right> \\ &= \frac 1 {Z_G} \sum_{n\alpha, m\beta} \left<n\alpha \right| \rho_G \left|m\beta\right>n\delta_{nm}\delta_{\alpha\beta} \\ &= \frac 1 {Z_G} \sum_{n\alpha} n \left<n\alpha\right| \rho_G \left| n\alpha \right> \ge 0. \end{align*} In the last step the inequality follows trivially as a density matrix is a positive operator ($\left<a\right|\rho\left|a\right> \ge 0$ for all $\left|a\right>$). If it were not a positive operator, it would not represent a mixed state as it would then assign negative probability to some pure state (which is in contradiction to the axioms of probability).

Also note, that $N_\text{actual}$ is not a valid concept for a system in the grand canonical equilibrium, in equilibrium there is no actual particle number, the grand canonical equilibrium is a mixed state. So again, a concept $N_\text{actual}$ cannot be well defined and is only confusing, not helpful, especially as we usually use the grand canonical ensemble for large systems where we fix $\mu$ by requiring a fix particle number/density (which will never be negative as this has no physical meaning).

$\endgroup$

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.