Example: Ensemble Averages

Consider a system with two spins with magnetic moments $ \mu_1$ and $ \mu_2$, respectively, at thermal equilibrium with a bath at temperature T. There is an external magnetic field B that interacts with each spin according to $ E(i, \pm) = \pm \mu_i B$. Assuming that the spins do not interact with each other, compute: (A) The average value of the total internal energy of the system. (B) The mean square fluctuation of the total energy of the system. (C) The entropy of the system. (D) The Helmholtz free energy. (E) Assuming that the contributions to the total magnetization M are $ m_i(\pm) = \pm \mu_i$, compute the average value of the magnetization of the system at temperature T. Note: The spins can only be "up"(+) or "down" (-) relative to the magnetic field. When a spin is "up" its interation with the external field is $ E(i, +) = \mu_i B$ and when the spin is "down" its interation with the external field is $ E(i, -) =-\mu_i H$.





Solution (A) In order to compute the internal energy as an ensemble average we first compute the partition function

$\displaystyle Z=e^{-\beta B (-\mu_1 -\mu_2 )} + e^{-\beta B (-\mu_1 +\mu_2 )} + e^{-\beta B (\mu_1 -\mu_2 )} +e^{-\beta B (\mu_1 +\mu_2 )},$ (113)

which is

$\displaystyle Z=4$   cosh$\displaystyle (\beta B \mu_2)$   cosh$\displaystyle (\beta B \mu_1).$ (114)

According to Eq. (35),

$\displaystyle E= - \frac{\partial \text{ln} Z}{\partial \beta} = -\frac{1}{Z} \... ...(\beta B \mu_1) + \mu_1 \text{cosh}(\beta B \mu_2) \text{sinh}(\beta B \mu_1)),$ (115)

which gives

$\displaystyle E = -B \mu_2$   tanh$\displaystyle (\beta B \mu_2) - B \mu_1$   tanh$\displaystyle (\beta B \mu_1).$ (116)

(B) According to Eq. (67),

$\displaystyle \overline{(\delta E)^2} = -\frac{\partial \bar{E}}{\partial \beta... ...text{cosh}^2(\beta B \mu_2)} -\frac{(B \mu_1)^2}{\text{cosh}^2(\beta B \mu_1)}.$ (117)

(C) According to Eq. (37),

$\displaystyle S = k \beta \bar{E} + k$   ln$\displaystyle Z.$ (118)

Therefore,

$\displaystyle S = -k \beta B \mu_2$   tanh$\displaystyle (\beta B \mu_2) - k \beta B \mu_1$   tanh$\displaystyle (\beta B \mu_1) + k$   ln$\displaystyle (4$cosh$\displaystyle (\beta B \mu_2)$   cosh$\displaystyle (\beta B \mu_1)).$ (119)

(D) According to Eq. (37),

$\displaystyle A = - k T$   ln$\displaystyle Z = -k T$   ln$\displaystyle (4$   cosh$\displaystyle (\beta B \mu_2)$cosh$\displaystyle (\beta B \mu_1)),$ (120)

(D) The ensemble average magnetization is

$\displaystyle \overline{M} = \sum_j p_j m_j,$ (121)

where, according to Eq. (44),

$\displaystyle p_j = Z^{-1} exp(-\beta E_j) = exp(-\beta(E_j-A)).$ (122)

Therefore,
\begin{displaymath}\begin{array}{ll} \overline{M} & =\Bigg ( (-\mu_1-\mu_2) e^{-... ...xt{cosh}(\beta B \mu_2)\text{cosh}(\beta B \mu_1)). \end{array}\end{displaymath}     (123)