CHEM-GA 2651: Statistical Mechanics. Final Exam. Instructions: This is a ... other
textbooks, or journal articles may be used to answer the problems. Do not work ...
CHEM-GA 2651: Statistical Mechanics Final Exam
Instructions: This is a take-home exam. However, you may only use the book and your lecture notes. No Web sites, other textbooks, or journal articles may be used to answer the problems. Do not work together. The work you turn in must be your own work. Partial credit will be given, so be sure to show all your work and to present it in a neat an organized fashion. The exam is due Friday morning by 10:00 am. Maximum point value: 120
1. (20 points) a. (5 points) Consider an anharmonic oscillator with coordinate x, momentum p, mass m, and natural frequency ω, having a potential energy of the form U (x) =
1 mω 2 x2 − gx3 + f x4 2
Working in the classical canonical ensemble, assuming that g and f are small quantities, find expressions for the average energy, heat capacity, and entropy that are valid to first order in g and f . b. (10 points) Suppose now that g = 0 but f 6= 0, and suppose that we can no longer assume f is small. Derive an expression for the classical canonical partition function that is an infinite series in powers of f . All integrals must be explicitly carried out! c. (5 points) Using your expression in part (b), derive expressions the average energy and heat capacity.
2. (30 points) Derive Eqns. (14.6.8) - (14.6.10) for the quantum time correlation function GAB (t), and generalize these formulas for N particles in three dimensions obeying Boltzmann statistics. 3. (20 points) A one-dimensional quantum system has energy levels given by the expression 1 κ, n = 0, 1, 2, ..., ∞ En = n + 2 However, the energy levels are degenerate with a degeneracy factor gn given by the recursion gn = gn−1 + (n + 1)
n = 1, 2, 3, ....
and g0 = 1. Calculate the canonical partition function, average energy, and heat capacity of this oscillator. How do your results differ from the non-degenerate case?
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4. (30 points) In this problem, we will illustrate how a simple change of integration variables in the partition function can be used to create an enhanced sampling method. Consider a classical particle with mass m, coordinate, x, and momentum p moving in the double-well potential U (x) =
2 U0 2 x − a2 . 4 a
Z(β) =
Z
The configurational partition function is dx e−βU(x) .
a. (5 points) Consider the change of variables q = f (x). Assume that the inverse x = f −1 (q) ≡ g(q) exists. Show that the partition function can be expressed as an integral of the form Z Z(β) = dq e−βφ(q) , and give an explicit form for the potential φ(q). b. (5 points) Now consider the transformation q = f (x) =
Z
x
˜
dy e−β U(y)
−a
˜ for −a ≤ x ≤ a and q = x for |x| > a and U(x) a continuous potential energy function. Show that f (x) is a monotonically increasing function of x and, therefore, that f −1 (q) exists. Write down the partition function that results from this transformation. c. (5 points) Would you expect this partition function to generate the correct free energy? Carefully explain why it would or would not. d. (5 points) ˜ (x) is chosen to be U ˜ (x) = U (x) for −a ≤ x ≤ a and U ˜ (x) = 0 for |x| > a, then the If the function U function φ(x) is a single-well potential energy function. Sketch a plot of q vs. x, and compare the shape of φ(q) to the original potential U (x). How do the two functions differ? e. (5 points) Argue, therefore, that a Monte Carlo calculation carried out based on φ(q), or molecular dynamics calculation performed using the Hamiltonian H(q, p) = p2 /2m + φ(q), leads to an enhanced sampling algorithm for high barriers. f. (5 points) Suppose the Hamiltonian in part (e) were used to construct a microcanonical partition function. Would you expect this partition function to generate the same thermodynamic and equilibrium properties as that generated by the Hamiltonian H = p2 /2m + U (x)? ˆ 0 having eigenvalues Ek and eigenvectors |Ek i is subject to a 5. Suppose a quantum system with Hamiltonian H perturbation of the form ˆ 1 (t) = VF ˆ (ω) eiωt + e−iωt e−ǫ|t| H
representing a monochromatic field with frequency ω. Here, Vˆ is a Hermitian operator, F (ω) is an amplitude ˆ 1 (−∞) = H ˆ 1 (∞) = 0. Derive the resulting transition rate Rf i (t) function, and ǫ is a parameter ensuring that H ˆ 0 to a final state Ef and show that in the limit ǫ → 0, for transitions from a given initial energy level Ei of H the resulting transition rate is given by Fermi’s golden rule.
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