Grand Canonical Ensemble

The goal of this section is to obtain the density operator

$$\hat{\rho} = \sum_j p_j \vert\phi_j><\phi_j\vert,$$ (72)

with

$$1 = \sum_j p_j,$$ (73)

that maximizes the entropy

$$S= -k \sum_j p_j p_j,$$ (74)

subject to the constraints of fixed volume V, average internal energy

$$E = \sum_j p_j E_j,$$ (75)

and average number of particles

$$\bar{N} = \sum_j p_j N_j.$$ (76)

Such density operator describes the maximum entropy ensemble distribution for a grand canonical ensemble --i.e., a collection of replica systems in thermal equilibrium with a heat reservoir whose temperature is $T$ as well as in equilibrium with respect to exchange of particles with a ``particle'' reservoir where the temperature is $T$ and the chemical potential of the particles is $\mu$. This problem is solved in terms of the Method of Lagrange Multipliers by maximizing the function

$$f(p_1,p_2, ...) \equiv -k \sum_j p_j p_j + \gamma (E - \sum_j p_j E_j) + \gamma' (N - \sum_j p_j N_j) + \gamma''(1 - \sum_j p_j),$$ (77)

where $\gamma$, $\gamma'$ and $\gamma''$ are Lagrange Multipliers. Solving for $p_j$ from the following equation

$$\frac{\partial f}{\partial p_j} \Bigg )_V = -k ( p_j+1) - \gamma E_j - \gamma' N_j - \gamma'' p_j,$$ (78)

we obtain

$$p_j = (1 - \gamma'') (- \frac{\gamma}{k} E_j - \frac{\gamma'}{k} N_j).$$ (79)

Introducing the quantities $\beta \equiv \gamma/k$ and $\beta \mu \equiv - \gamma'/k$ we obtain the generalized Boltzmann distribution

$$p_j=\frac{e^{-\beta E_j + \beta \mu N_j}}{\sum_j e^{-\beta E_j + \beta \mu N_j}} = \Xi^{-1} e^{-\beta E_j + \beta \mu N_j },$$ (80)

where

$$\Xi \equiv \sum_j exp(-\beta E_j + \beta \mu N_j),$$ (81)

is the grand canonical partition function. The goal of the remaining of this section is to find the relation between the canonical and the grand canonical partition functions, $Z$ and $\Xi$, respectively. Substituting Eq. (80) into Eq. (74) we obtain

$$S = - k \sum_j p_j (-\beta E_j + \beta \mu N_j - \Xi),$$ (82)

and solving for ln$\Xi$ from Eq. (82) we obtain

$$\Xi = \beta (- E + S T + \mu \bar{N}).$$ (83)

Therefore,

$$\Xi = \beta (-A + \mu \bar{N}),$$ (84)

and

$$\boxed{\text{ln} \Xi = \text{ln} Z + \beta \mu \bar{N}}.$$ (85)