Maximum-Entropy Density Operator

The goal of this section is to obtain the density operator $\hat{\rho}$, with $Tr\{ \hat{\rho}\}=1$, that maximizes the entropy $S= -k$   Tr$\{ \hat{\rho}$   ln$\hat{\rho} \}$ of a system characterized by an ensemble average internal energy

$$E = \{\hat{\rho} \hat{H}\},$$ (25)

and fix extensive properties $X$ such as $X=(V,N)$ (i.e., canonical and microcanonical ensembles). This is accomplished by implementing the Method of Lagrange Multipliers to maximize the function

$$f(\hat{\rho}) \equiv -k \{\hat{\rho} \hat{\rho} \} + \gamma (E -Tr\{\hat{\rho} \hat{H}\}) + \gamma'(1-Tr\{\hat{\rho} \}),$$ (26)

where $\gamma$ and $\gamma'$ are Lagrange Multipliers. We, therefore, solve for $\hat{\rho}$ from the following equation

$$\frac{\partial f}{\partial \hat{\rho}} \Bigg )_X= 0,$$ (27)

and we obtain that the density operator that satisfies Eq. (27) must satisfy the following equation:

$$Tr \{ -k \hat{\rho} - k - \gamma \hat{H} - \gamma' \} =0.$$ (28)

Therefore,

$$- \hat{\rho} = 1 + \frac{\gamma}{k} \hat{H} + \frac{\gamma'}{k}.$$ (29)

Exponentiating both sides of Eq. (29) we obtain

$$\hat{\rho} = (-(1 + \frac{\gamma'}{k})) (-\frac{\gamma}{k} \hat{H}),$$ (30)

and, since Tr$\{\hat{\rho}\}$=1,

$$(-(1 + \frac{\gamma'}{k}))=\frac{1}{Z},$$ (31)

where $Z$ is the partition function

$$Z \equiv Tr \{ exp(-\beta \hat{H}) \},$$ (32)

with $\beta \equiv \gamma/k$. Substituting Eqs. (32) and (31) into Eq. (30), we obtain that the density operator that maximizes the entropy of the ensemble, subject to the contraint of average ensemble energy $\bar{E}$, is

$$\hat{\rho} = Z^{-1} (-\beta \hat{H}).$$ (33)

Note that

$$\frac{\partial \hat{\rho}}{\partial t} = 0,$$ (34)

when $\hat{\rho}$ is defined according to Eq. (33) and, therefore, the system is at equilibrium.

Exercise 6: Use Eqs. (17) and (33) to prove Eq. (34).