# Negative amount of particles in a grand canonical ensemble

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 $N_{actual}<0$?

• $N_G$ 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). – Sebastian Riese Oct 1 '15 at 16:47

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).