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Unformatted text preview: Physics 112 Fall 2010 Professor William Holzapfel Homework 7 Solutions Problem 1, Kittel 6.3: Double Occupancy Statistics (a) In this problem were considering a new particle which can at most doubly occupy an orbital (as opposed to single occupancy for fermions or unlimited occupancy for bosons). The Gibbs sum for a single orbital is then = 2 X N =0 e ( N- N ) / = 1 + e ( - ) / + e (2 - 2 ) / . The average occupancy is given by the weighted sum: h N i = 1 2 X N =0 N N e- N / = 1 1 + 1 e- / + 2 2 e- 2 / = e- / + 2 2 e- 2 / 1 + e- / + 2 e- 2 / . (b) This question is poorly phrased, but by usual quantum mechanics they mean ordinary fermions. Were asked to consider ordinary fermions occupying two degenerate orbitals. Notice that in part (a) there was one orbital that could be doubly occupied, whereas in part (b) there are two orbitals that each can be at most singly occupied. This gives rise to four total system states: one with N = 0, two with N = 1, and one with N = 2. Using the ASN formalism from Kittel, we have = 2 X N =0 X s N e ( N- s N ) / = 2 X N =0 X g ( ) e ( N- ) / = 1 + 2 e- / + 2 e- 2 / = 1 + e- / 2 . Notice that the Gibbs sum for a single orbital would be 1 = 1 + e- / , and so we have for the two orbital system 2 = ( 1 ) 2 , just as we did for the partition function in the canonical ensemble. (By way of terminology, the setting of the Gibbs sum, in which the system is allowed to trade energy and particles with the reservoir, is known as the Grand canonical ensemble .) We then have h N i = 1 1 + 2 e- / + 2 2 e- 2 / = 2 e- / + 2 2 e- 2 / 1 + 2 e- / + 2 e- 2 / = 2 e (- ) / + 1 which is just twice the Fermi-Dirac distribution, which gives the occupancy of one of the orbitals. Problem 2, Kittel 6.4: Energy of Gas of Extreme Relativistic Particles 1 For a particle in a box we have k x = n x L = p x = n x ~ L and so E = | p | c = q n 2 x + n 2 y + n 2 z L ~ c = n L ~ c where n q n 2 x + n 2 y + n 2 z . So the partition function for a single particle is: Z 1 = Z dn x Z dn y Z dn z e-| p | c/ = 1 8 Z 4 n 2 dne- ~ c L n . Let x = ~ c L n . Then Z 1 = V 4 3 (2 ~ ) 3 c 3 Z dxx 2 e- x 3 and so U = 2 ln Z 1 = 3 2 ln = 3 . Problem 3: The Relation Between Pressure and Energy Density Revisited (a) Kittel 6.7 Since the system is in thermal contact with a heat reservoir we are to assume that is constant....
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