# Solutions for the Homework 5

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1 Solutions for the Homework 5 Problem 78: Suppose you have a box in which each particle may occupy any of 0 single-particle states For simplicity, assume that each of these states has energy zero (a) What is the partition function of this system if the box contains only one particle? Solution: Since the states has energy zero, each Boltzmann factor equals to and it means that the partition function Z is same as the total number of states Ω So, in this case, Z = 0 (b) What is the partition function of this system if the box contains two distinguishable particles? Solution: There are two independent particles, so Z 2 = Z 2 = 00 (c) What is the partition function if the box contains two identical bosons? Solution: There are two cases, one is that the particles are different states and another is that the particles are same state The number of possible states are 45 for the first case and 0 for the second case Then, Z 3 = 55 (d) What is the partition function if the box contains two identical fermions? Solution: Similar to the above boson case, but the fermion can have only the first case because of Pauli exclusion principle So, Z 4 = 45 (e) What would be the partition function of this system according to equation 76? Solution: If there are two particles, Z = Z 2 /2 = 50 (f) What is the probability of finding both particles in the same single-particle state, for the three cases of distinguishable particles, identical bosons, and identical fermions? Solution: Consider the case that the both particles in the same state, there are 0 possible states for distinguishable particles, also 0 possible states for bosons, and 0 state for fermions Then, the probability to find both particle in same state is /0 for distinguishable particles, 2/ for identical bosons and 0 for identical fermions Problem 70: Consider a system of five particles, inside a container where the allowed energy levels are nondegenerate and evenly spaced For instance, the particles could be trapped in a onedimensional harmonic oscillator potential In this problem you will consider the allowed states for this system, depending on whether the particles are identical fermions, identical bosons, or distinguishable particles (a) Describe the ground state of this system, for each of these three cases Solution: Suppose that there are 5 states, s to s 5, where the number indicate energy levels starting from the lowest level to 5 Then, the ground state of each cases are like below

2 Particles s s 2 s 3 s 4 s 5 Degeneracy Fermions Bosons Distinguishable (b) Suppose that the system has one unit of energy (above the ground state) Describe the allowed states of the system, for each of the three cases How many possible system states are there in each case? Solution: In this case, one of that particles is excited because of there is one unit of energy Similar to the above problem, Particles s s 2 s 3 s 4 s 5 s 6 Degeneracy Fermions 0 Bosons Distinguishable (c) Repeat part (b) for two units of energy and for three units of energy Solution: For two units and three units of energy, Particles s s 2 s 3 s 4 s 5 s 6 s 7 Degeneracy Fermions Bosons Distinguishable Particles s s 2 s 3 s 4 s 5 s 6 s 7 s 8 Degeneracy Fermions Bosons Distinguishable (d) Suppose that the temperature of this system is low, so that the total energy is low (though not necessarily zero) In what way will the behavior of the bosonic system differ from that of the system of distinguishable particles? Discuss Solution: Since the degeneracy of distinguishable particle states is much larger than the bosonic particle state, small energy unit leads to excite the particles easily at the distinguishable particles So, we can find the particles in the ground state for bosonic case easily 2

3 Problem 72: Consider two single-particle states, A and B, in a system of fermions, where ϵ A = µ x and ϵ B = µ + x; that is, level A lies below µ by the same amount that level B lies above µ Prove that the probability of level B being occupied is the same as the probability of level A being unoccupied In other words, the Fermi-Dirac distribution is symmetrical about the point where ϵ = µ Solution: From the Fermi-Dirac distribution, the probability of the state B being occupied is P B occupied = e xβ + () Now, the the probability of the state A being unoccupied is given below and that s the same with above P A unoccupied = e xβ + = e xβ e xβ + = e xβ + (2) So, the probability that the state A is unoccupied is same as the probability that the state B is occupied Problem 76: Consider an isolated system of N identical fermions, inside a container where the allowed energy levels are nondegenerate and evenly spaced For instance, the fermions could be trapped in a one-dimensional harmonic oscillator potential For simplicity, neglect the fact that fermions can have multiple spin orientations (or assume that they are all forced to have the same spin orientation) Then each energy level is either occupied or unoccupied, and any allowed system state can be represented by a column of dots, with a filled dot representing an occupied level and a hollow dot representing an unoccupied level The lowest-energy system state has all levels below a certain point occupied, and all levels above that point unoccupied Let η be the spacing between energy levels, and let q be the number of energy units (each of size η) in excess of the ground-state energy Assume that q < N Figure 78 shows all system states up to q = 3 (a) Draw dot diagrams, as in the figure, for all allowed system states with q = 4, q = 5, and q = 6 Solution: Similar to the given Figure 78, let s draw q = 4 q = 5 q = 6 3

4 (b) According to the fundamental assumption, all allowed system states with a given value of q are equally probable Compute the probability of each energy level being occupied, for q = 6 Draw a graph of this probability as a function of the energy of the level Solution: There are cases for q = 6, so each of cases have a probability / From the bottom to top, occupancies at the each energy levels are, 0 0, 9, 8, 7, 6, 5, 4, 3, 2, 0, (3) There are so many states that have the occupancies / and 0/ I will cut this as 5 states only 0 n Ε Η (c) In the thermodynamic limit where q is large, the probability of a level being occupied should be given by the Fermi-Dirac distribution Even though 6 is not a large number, estimate the values of µ and T that you would have to plug into he Fermi-Dirac distribution to best fit the graph you drew in part (b) Solution: When the energy is same as µ, the probability to find the particle in that state is /2 So, µ = 5η from the plot data of (b) And also from the fitting process of (b), kt 29η (d) Calculate the entropy of this system for each value of q from 0 to 6, and draw a graph of entropy vs energy Make a rough estimate of the slope of this graph near q = 6, to obtain another estimate of the temperature of this system at that point Check that it is in rough agreement with your answer to part (c) Solution: For each q, the total number of cases Ω tell us that the entropy of each system state, S = k ln Ω q Ω S/k Since U = qη, the slope of the below graph is η/kt because of T = U/ S From the fitting process, kt 233η, which is similar to the result of (c) The difference may caused by small q 4

5 S k q Problem 77: In analogy with the previous problem, consider a system of identical spin-0 bosons trapped in a region where the energy levels are evenly spaced Assume that N is a large number, and again let q be the number of energy units (a) Draw diagrams representing all allowed states from q = 0 up to q = 6 Instead of using dots as in the previous problem, use numbers to indicate the number of bosons occupying each level Solution: Let s draw diagrams similar to the above fermion cases q = 0 q = q = 2 q = 3 q = 4 q = N N- N- N-2 N- N-2 N-3 N- N-2 N-2 N-3 N-4 N- N-2 N-2 q = 5 q = N-3 N-3 N-4 N-5 N- N-2 N-2 N-3 N-2 N-3 N-4 N-3 N-4 N-5 N-6 5

6 (b) Compute the occupancy of each energy level, for q = 6 Draw a graph of the occupancy as a function of the energy of the level Solution: There are cases for q = 6, so each of cases have a probability / similar to the fermion case From the bottom to top, occupancies at the each energy levels are ( N 35 ) 9,, 8, 4, 2, 0, (4) There are so many states that have the occupancies 0/ I will cut this as 5 states only n Ε Η (c) Estimate the values of µ and T that you would have to plug into the Bose-Einstein distribution to best fit the graph of part (b) Solution: When the energy is same as µ, the probability to find the particle in that state is Infinite Let the number of particle N is sufficiently large Then we can treat this number as approximately infinite So, the chemical potential is µ = η And from the fitting process of (b), kt 220η (d) As in part (d) of the previous problem, draw a graph of entropy vs energy and estimate the temperature at q = 6 from this graph Solution: This is exactly same as the previous problem for the fermion case So, from the fitting except the q = 0 case, kt 233η, which is similar to the result of (c) Also the difference may caused by small q 6

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