### Eigenvalues and Boltzmann statistics

#### Problem:

Due to the presence everywhere of the cosmic background radiation the minimum possible temperature of a gas in interstellar space is 2.7 K.  This implies that a significant fraction of molecules in space may be in low-level excited states.  Consider a hypothetical molecule with one possible excited state.  What would the excitation energy have to be for 20% of the molecules to be in the excited state?

Solution:

• Concepts:
Boltzmann statistics
• Reasoning:
Boltzmann statistics tells us that Nexcited/Nground = e-ΔE/kT, if the degeneracy of both states is 1.
• Details of the calculation:
0.2/0.8 = exp(-ΔE/(kT))
ln(0.25) = -ΔE/(kT)
ΔE = -ln(0.25)*(kT) = -ln(0.25)*8.617*10-5*2.7 eV = 3.2*10-4 eV = 0.32 meV

#### Problem:

Consider a particle in a simple two level system with eigenstates ψ1 and ψ2 such that
1 = E1ψ1 and Hψ2 = E2ψ2.  Derive an expression for the specific heat per particle cV = ∂<E>/∂T of the system as a function of temperature.

Solution:

• Concepts:
Boltzmann statistics
• Reasoning:
The system consisting of many particles is in a statistical mixture of states.
In thermodynamic equilibrium P(Ei) ∝ giexp(-Ei/(kT)
Here the degeneracy gi is 1 for all i.
• Details of the calculation:
Average energy:  <E> = (E1 exp(-E1/(kT)) + E2 exp(-E2/(kT)))/(exp(-E1/(kT)) + exp(-E2/(kT)))
<E> = (E1 exp(-βE1) + E2 exp(-βE2))/(exp(-βE1) + exp(-βE2))
= (E1 + E2 exp(-β∆))/(1 + exp(-β∆)),
with β = 1/(kT) and ∆ = E2 - E1.
cV = ∂<E>/∂T.  cV = (∂<E>/∂β)(∂β/∂T) = -(1/(kT2))∂<E>/∂β.
cV = [1/(kT2)][(E2∆exp(-β∆))/(1 + exp(-β∆)) - (E1 + E2exp(-β∆))(∆exp(-β∆)))/(1 + exp(-β∆))2].

#### Problem:

Consider a system of N particles with only 3 possible energy levels separated by ε.  Let the ground state energy be zero.  The system occupies a fixed volume V and is in thermal equilibrium with a reservoir at temperature T.  Ignore interactions between particles and assume Boltzmann statistics apply.

(a)  What is the partition function for a single particle in the system?
(b)  What is the average energy per particle?
(c)  What is the probability that the 2ε level is occupied in the high-temperature limit kBT >> ε?
(d)  What is the average energy per particle in the high-temperature limit kBT >> ε?
(e)  At what energy is the ground state 1.1 times as likely to be occupied as the 2ε level?
(f)  Find the heat capacity CV of the system, analyze the low-T (kBT >> ε) and high-T (kBT >> ε) limit, and sketch CV as a function of T.

Solution:

• Concepts:
Boltzmann statistics
• Reasoning:
In thermodynamic equilibrium P(Ei) ∝ giexp(-Ei/(kT).
Here the degeneracy gi is 1 for all i.
• Details of the calculation:
(a)  P(E) = Cexp(-E/(kBT)).
ΣjP(Ej) = CΣjexp(-Ej/(kBT)) = 1, C = 1/Σjexp(-Ej/(kBT)) = 1/Z.
Let β = 1/(kBT)).
Z = 1 + exp(-βε) + exp(-2βε).
(b)  <ε> = (ε exp(-βε) + 2ε exp(-2βε))/Z
= ε(exp(-βε) + 2exp(-2βε))/(1 + exp(-βε) + exp(-2βε)).
(c)  P =  exp(-2βε)/(1 + exp(-βε) + exp(-2βε)) ~ (1 - 2βε)/( 1 + 1 - βε + 1 - 2βε) = 1/3.
In the high energy limit the probability is the same for all three levels.
(d)  <ε> = 0/3 + ε/3 + 2ε/3 = ε.
(e)  exp(-2βε) = 1/1.1,  2βε = ln(1.1),  T = 2ε/(kB*ln(1.1)).
(f)  CV = dU/dT = Nd<ε>/dT = Nd<ε>/dβ)(dβ/dT).
CV = (Nε2/(kBT2))[ exp(-βε) + 4 exp(-2βε) + exp(-3βε)]/[1 + exp(-βε) + exp(-2βε)]2.
Low temperature limit:  CV = (Nε2/(kBT2)) exp(-ε/(kBT)).  (CV --> 0 as T --> 0)
High temperature limit:  CV = (2/3)(Nε2/(kBT2)).  ( CV --> 0 as T --> ∞)