Exam 1, 02/03/03

Exam 1 CHEM 430b/530b
Statistical Methods and Thermodynamics
02/03/03 Exercise 1

(20 points) Item (1.1): Explain the fundamental postulates of Statistical Mechanics.

(20 points) Item (1.2): Show that the definitions of entropy given by Gibbs and Boltzmann can be obtained from the Von Neumann definition of entropy.

(20 points) Item (1.3): Prove that

$$S=-k \beta \frac{\partial \text{ln} Z}{\partial \beta} \Bigg )_{V,N} + k \text{ln} Z.$$ (124)

Exercise 2

Consider a system of $N$ non-interacting spins in a magnetic field $B$, in thermal equilibrium with a bath at temperature $T=1/(k \beta)$. Each spin has a magnetic moment of size $\mu$ and can point either parallel or anti-parallel to the field.
(10 points) Item (2.1): Determine the internal energy of the system as a function of $\beta$, B and N.
(10 points) Item (2.2): Determine the entropy of the system as a function of $\beta$, B and N.
(10 points) Item (2.3): Determine the average total magnetization of the system as a function of $\beta$, B and N.
(10 points) Item (2.4): Determine the average square fluctuation of the total magnetization of the system $\overline{(\delta M)^2}$ as a function of $\beta$, B and N.

Solution:

Item (1.1):

First Postulate: The experimental result of a measurement of an observable in a macroscopic system is the ensemble average of such observable. Second Postulate: Any macroscopic system at equilibrium is described by the maximum entropy ensemble, subject to constraints that define the macroscopic system. The first postulate is needed to equate the ensemble average to a more elementary description of what is begin observed. The second postulate is needed to connect the attributes of a quantum state to its probability. We found that the maximum entropy postulate established the connection between $p_j$ and the attributes of $j$ as follows. For a canonical ensemble $p_j$ is determined by the Boltzmann distribution; for a microcanonical ensemble $p_j$ is independent of $j$ and is determined by inverse of the total number of states and for the grand canonical ensemble $p_j$ is determined by the generalized Boltzmann distribution. Therefore, the second postulate established that all quantum states with the same energy and the same number of particles are equally probable.

Item (1.2): According to Eq. (22), the Von Neumann definition of entropy is:

$$S \equiv -k Tr\{\hat{\rho} \hat{\rho} \},$$ (125)

which according to the definition $\hat{\rho} = \sum_k p_k \vert\phi_k><\phi_k\vert$ becomes,

$$S \equiv -k \sum_j <\phi_j\vert \sum_k p_k \vert\phi_k><\phi_k\vert (\sum_k p_k \vert\phi_k><\phi_k\vert) \vert \phi_j >.$$ (126)

Expanding ln$\hat{\rho} = 0 + (\hat{\rho}-1) - (\hat{\rho}-1)^2/2! + ...$ we obtain

$$S \equiv -k \sum_j p_j p_j,$$ (127)

which is the Gibbs definition of entropy. According to Eq. (45), $p_j$ = 1/$\Omega$ for a microcanonical ensemble. Therefore,

$$S = k \Omega,$$ (128)

which is the Boltzmann definition of entropy.

Item (1.3): According to Eq. (44), the probability $p_j$ of observing a system in quantum state $\vert j>$ is

$$p_j = Z^{-1} exp(-\beta E_j) = exp(-\beta(E_j-A)),$$ (129)

where $Z=\sum_j exp(-\beta E_j)$. Substituting this expression in the Gibbs definition of entropy, introduced in Item (1.2), we obtain

$$S = k \beta Z^{-1} \sum_j E_j exp(-\beta E_j)+ k Z = -k \beta \frac{\partial \text{ln} Z}{\partial \beta} \Bigg )_{V,N} + k \text{ln} Z.$$ (130)

Item (2.1): According to Eq. (59), the canonical partition function for a system of N two-level particles $j$ with energies $E_j= \pm \mu B$ is

$$Z = \prod_{j=1}^N (e^{-\beta \mu B}+e^{\beta \mu B}) = (2 (\beta \mu B))^N.$$ (131)

Therefore,

$$E = - \frac{\partial \text{ln}Z}{\partial \beta} \Bigg )_{N,V} = - \mu B N \text{tanh}(\beta \mu B).$$ (132)

Item (2.2): According to Eq. (44),

$$S = k Z + \frac{E}{T} = k N (2 (\beta \mu B)) - \beta k \mu B N (\beta \mu B).$$ (133)

Item (2.3): According to Eq. (121), the ensemble average magnetization is

$$\overline{M} = \sum_j p_j m_j,$$ (134)

where $m_j= \pm \mu$ and

$$p_j = Z^{-1} exp(-\beta B m_j) = exp(-\beta(B m_j-A)).$$ (135)

Therefore,

$$\overline{M} = \frac{\partial \text{ln} Z}{\partial (B \beta)} \Bigg )_{N,V} = N \mu \text{tanh}(\beta B \mu).$$ (136)

Item (2.4):

$$\overline{(\delta M)^2} = \frac{\partial^2 \text{ln} Z}{\partial (B \beta)^2} =N \mu^2 \text{sech}^2(\beta B \mu).$$ (137)