Homework #12 Solutions6 problems

. 1. A particular (made-up) system is in thermal and diffusive equilibrium with a reservoir with temperature Tand chemical potential µ; the particles that make up the system and reservoir are indistinguishable. The systemcan contain at most two particles, and the state of the system is simply determined by the number of particles itcontains: n = 0, 1, 2; there is no degeneracy. The energy of the system is E(n) = an2, where a > 0 is a constant.(a) Find the grand partition function.(b) Find the average number of particles 〈n〉 this box contains, as a function of T , µ, and a.(c) If µ is negative (as it usually is), find the limit of 〈n〉 at high temperatures (T → ∞) and low temperatures(T → 0). Explain why these limits make sense.

/ (a) There are three states: zero, one, or two particles in the system.

Z =∑n

e−β(ε−µn) =

2∑n=0

e−β(an2−µn) = 1 + eβ(µ−a) + e2β(µ−2a)

(b) The probability of state n is the ratio of its Gibbs function to the grand partition function: P (n) =1Z e−β(an2−µn). And the average number of particles in the state is

〈n〉 =∑n

nP (n)

=1

Z∑n

ne−β(an2−µn)

=eβ(µ−a) + 2e2β(µ−2a)

1 + eβ(µ−a) + e2β(µ−2a)

(c) As the temperature approaches infinity, β approaches zero, and all of the exponentials approach one;thus the average number of particles in the system approaches

〈n〉 → 1 + 2

1 + 1 + 1=

3

3= 1

At very high temperatures, the particles can basically have any energy they want, and so all three statesare equally accessible: there will be equal numbers of times when the system has 0 particles, 1 particle, or 2particles, and the average will indeed be 1.

As the temperature approaches zero, β approaches infinity, and all the exponentials approach e−∞ = 0

(because µ− a < 0 etc). Thus the average number of particles is 〈n〉 → 0

1= 0.

Having no particles in the system is the lowest energy state here, and when the temperature is very low thesystem can’t get out of its ground state.

. 2. Consider a system consisting of a single hydrogen atom/ion, which has two possible states: unoccupied (noelectron present) and occupied (one electron present, in the ground state). The hydrogen is in thermal anddiffusive equilibrium with a sea of electrons at temperature T . It takes energy I = 13.6 eV to ionize a hydrogenatom.(a) Assume that the electrons are a monatomic ideal gas, and ignore the internal degrees of freedom of electrons(like, for example, their spin states). Find the chemical potential µ of the ideal gas; you can keep your answer interms of the quantum volume vQ introduced in Chapter 6.(b) Calculate the ratio of the probabilities of the excited and ground states; this is the Saha equation.

/ (a) If the electrons are a monatomic ideal gas with no internal degrees of freedom (so that Zint = 1), thentheir chemical potential is the chemical potential of an ideal gas:

µ = −kT ln

(V ZintNgasvq

)= −kT ln

(V

NvQ

)If we include the spin states, then the chemical potential is reduced by −kT ln 2 (assuming no magneticfield).

(b) The two states are unoccupied (N = 0, ε = 0) and occupied (N = 1, ε = −13.6 eV ≡ −I) where I is theionization energy. The Gibbs factors of these two states are e−β(ε−µN) = 1 and e−β(−I−µN) = eβ(I+µ). Theratio of the probabilities of these two states is the ratio of the Gibbs free energies:

PunoccPocc

=1

eβ(I+µ)=e−βI

eβµ

Given the chemical potential above,

βµ = − ln

(V

NvQ

)=⇒ e−βµ =

V

NvQ

and soPunoccPocc

=V

NvQe−I/kT

Another way of writing this is in terms of the pressure of the electron gas, which is given by the ideal gaslaw PeV = NkT , so V

N = kTPe

. This gives us the usual form of something called the Saha equation

PunoccPocc

=kT

PevQe−I/kT

. 3. Suppose you have a “box” in which each particle may occupy any of 10 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?(b) What is the partition function of this system if the box contains two distinguishable particles?(c) What is the partition function of this system if the box contains two identical bosons?(d) What is the partition function of this system if the box contains two identical fermions?(e) Calculate (7.16) Z = 1

N !ZN1 to find the partition function for two indistinguishable, non-interacting particles.

(f) What is the probability of finding both particles in the same single-particle state, for the three cases ofdistinguishable particles, identical bosons, and identical fermions?

/ Since all energy states are energy zero, all Boltzmann factors are e−β(0) = 1, and so Z =∑s e−βEs is

simply the number of available states, or Ω.

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(a) One particle can be in any of the ten energy levels, so Z = Ω = 10.

(b) The two distinguishable particles can each choose from 10 energy levels, so Z = Ω = 102 = 100.

(c) For two identical bosons, there are 10 ways to put both in the same energy level, and there are(102

)= 45

ways of putting them in different energy levels, so Z = 10 + 45 = 55.

(d) For two identical fermions, the result is the same as (c) except for the 10 states with both in the sameenergy level; thus Z = 45.

(e) Using the formula with N = 2, and Z1 = 10 (Z1 is the number of states for the single particle), we have

Z =1

N !ZN1 =

1

2!102 = 50

(f) For distinguishable particles, 10 (out of 100) of the system states have both particles in the same state,so the probability is 10%. For identical bosons, there are still 10 such system states, but only 55 systemstates total, so the probability is higher, 10/55 = 18%. For identical fermions, you can’t have two particlesin the same single-probability state, so the probability is zero.

. 4. For a system of fermions at room temperature, compute the probability of a single-particle state beingoccupied if its energy is 0.01 eV less than µ.

/ The probability of a state being occupied (which is equal to the average occupancy in the case of fermions)

is given by the Fermi-Dirac distribution function,1

eβ(ε−µ) + 1. At room temperature, kT = 0.026 eV. If the

energy of the state is 0.01 eV less than µ, then ε− µ = −0.01 eV, and so the probability is

P =1

e(−0.01 eV)/(0.026 eV) + 1=

1

1.68= 59%

. 5. For a system of bosons at room temperature, compute the average occupancy n of a single-particle state, andthe probability that the state contains 0, 1, 2, and 3 bosons, if the energy of the state is 0.001 eV greater than µ.

/ This is similar to the previous problem. The energy of the state is ε = µ + 0.001 eV, so (ε − µ)/kT =0.001 eV

0.026 eV= 0.0385. The average occupancy is given by the Bose-Einstein distribution:

n =1

e(ε−µ)/kT − 1=

1

e0.0385 − 1=

1

0.0392= 25.5

The probability that the state is occupied by n particles is given by Eq. (7.20):

P (n) =1

Ze−n(ε−µ)/kT = (1− e−(ε−µ)/kT )e−n(ε−µ)/kT

(where I used Equation 7.24, the grand partition function of bosons. Since e−(ε−µ)/kT = e−0.0385 = 0.962,and so

P (n) = (1− 0.962)(0.962)n = 0.0377(0.962)n

and soP (0) = 3.77% and P (1) = 3.63% and P (2) = 3.49% and P (3) = 3.36%

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. 6. Imagine that there exists a third type of particle, which can share a single-particle state with one other particleof the same type, but no more than one. Thus the number of these particles in any state can be 0, 1, or 2. Derivethe distribution function for the average occupancy n of a state by particules of this type, and plot the occupancyas a function of the state’s energy, for three different temperatures.

/ If up to two particles of a given type can occupy a state of energy ε, then the grand partition function is

Z = e0 + e−(ε−µ)/kT + e−2(ε−µ)/kT = 1 + e−x + e−2x

where x = (ε− µ)/kT . Therefore the average number of particles in the state is

n = − 1

Z∂Z∂x

= − 1

1 + e−x + e−2x∂

∂x(1 + e−x + e−2x) =

e−x + 2e−2x

1 + e−x + e−2x

Not as pretty as bosons and fermions; maybe it’s lucky that these particles don’t exist. Here’s a graph of nfor three different choices of T ; we see the same basic shape that we had for fermions, only maxing out at 2for low energy levels instead of at 1.

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