The particle on a ring problem can be used to model electronic transitions in an

Question

The particle on a ring problem can be used to model electronic transitions in an aromatic hydrocarbon ring (benzene, naphthalene). For a particle on a ring, a particle of mass m is xed to move on a ring of radius a in the xy plane. The Hamiltonian of this system is H^ = ~ 2 2I d 2 d2 , I is the moment of inertia, I = ma 2 and  is the angle of the x-axis in the xy plane, which ranges from 0 to 2. a) Find the energy eigenstates for the system. b) Compute the normalization constant for these eigenstates. c) What is the boundary condition for this problem? d) What does this restrict the allowed energies of the system to be? e) What is the degeneracy of the energy levels?

Explanation / Answer

Number the green, gray, orange, and blue particles as particles 1, 2, 3, and 4, respectively. The energy of system A is the sum of the energies of the four particles, E = E1 + E2 + E3 + E4 = 0 + 3 + 6 + 6 = 15. Likewise, the energy of system B is the sum of the energies of its four particles, E = E1 + E2 + E3 + E4 = 0 + 4 + 4 + 8 = 16. As far as the ways the energy of each system can be distributed, I see some ambiguity. First of all, do the allowed levels keep going up and up (E_i = 0, 3, 6, 9, 12, 15, .... for system A, and E_i = 0, 4, 8, 12, 16, .... for system B), or are there only four allowed levels for system A (0, 3, 6, 9) and three allowed levels for system B (0, 4, 8)? Also, can any number of particles occupy one level, or is there a maximum of two particles per level? Furthermore, are the particles distinguishable (is particle 1 in level 3 and particle 2 in level 6 distinguishable from particle 2 in level 6 and particle 1 in level 3)? I will assume they are distinguishable, since they are different colors in the diagram. I will also assume that there are only four levels for system A and three levels for system B, and that any number of particles can occupy a single level. If this is not the case, you should be able to modify my answer to include the correct case. For system A, you want to find all the different ways you can distribute the particles such that E = E1 + E2 + E3 + E4 = 15. There are 12 ways of putting the four particles in levels that have energies 0, 3, 6, and 6 (one of which is shown in the diagram), 12 ways of putting the particles in levels that have energies 0, 0, 6, and 9, and 4 ways of putting the particles in levels that have energies 3, 3, 3, and 6. Therefore there are 12 + 12 + 4 = 28 ways of distributing the particles such that E = 15. (If the levels go up and up, you also need to consider the case where the four particles have energies of 0, 0, 3, and 12, and the case where the four particles have energies of 0, 0, 0, and 15. If only two particles are allowed per level, then leave out the 3, 3, 3, and 6 case and the 0, 0, 0, and 15 case.) I will enumerate the 28 ways of distributing the particles in a table below. E1 E2 E3 E4 ------------------- 0 3 6 6 0 6 3 6 0 6 6 3 3 0 6 6 6 0 3 6 6 0 6 3 3 6 0 6 6 3 0 6 6 6 0 3 3 6 6 0 6 3 6 0 6 6 3 0 0 0 6 9 0 0 9 6 0 6 9 0 0 9 6 0 0 6 0 9 0 9 0 6 6 9 0 0 9 6 0 0 6 0 9 0 9 0 6 0 6 0 0 9 9 0 0 6 3 3 3 6 3 3 6 3 3 6 3 3 6 3 3 3 Those are the 28 ways of distributing the 4 particles over the 4 levels in system A, assuming that any number of particles can occupy a given level. For system B, there are 12 ways of putting the four particles in levels that have energies 0, 4, 4, and 8 (one of which is shown in the diagram), 6 ways of putting the particles in levels that have energies 0, 0, 8, and 8, and 1 way of putting the particles in levels that have energies 4, 4, 4, and 4. All of these ways has a total energy of E = E1 + E2 + E3 + E4 = 16. Therefore there are 12 + 6 + 1 = 19 ways of distributing the energy in system B. Entropy is given by the equation S = kB ln[Omega(E)], where kB is the Boltzmann constant, ln(x) is the natural log of x, and Omega(E) is the number of ways you can distribute the energy in a system of total energy E. Therefore entropy is a monotonically increasing function of Omega. In this example, system A has Omega(E = 15) = 28, and system B has Omega(E = 16) = 19. Therefore system A has the greater entropy.

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