# Work check and advice on a statistical mechanics problem

## Homework Statement:

I don't know if I am correct or not....

## Relevant Equations:

Probabilities

b)
Consider P_j(n) as a macrostate of the system,
Bosons: P_1(1) = P_2(1) = 1/2*1/2=1/4 ,P_1(2)=P_2(2)=1/2*1/2=1/4

Fermions: P_1(1)=P_2(1)=1 (Pauli exclusion principle), P_1(2)=P_2(2)=0

Different species: P_1(1)=P_2(1) = 2*1/2*1/2=1/2 (because there are two microstates with corresponding to one atom in each well, atom A in well 1, atom B in well 2 and atom B in well A, atom A in well B).

c)
Bosons: Suppose well 2 has an energy of E. 3 microstates as above
Z = 1 + e^(-2bE) + e^(-bE),
P_1(1)=P_2(1) = e^(-bE)/Z, P_1(2)=1/Z, P_2(2)=e^(-2bE)/Z

Fermions: P_1(1)=P_2(1)=1 (Pauli exclusion principle), P_1(2)=P_2(2)=0

Distinct species: Z = 1 + e^(-2bE) + 2e^(-bE)
P_1(1)=P_2(1) = 2e^(-bE)/Z, P_1(2)=1/Z, P_2(2)=e^(-2bE)/Z

d)
Bosons:
Z = e^(-bU) + e^(-2b(E+U)) + e^(-bE)
P_1(1)=P_2(1) = e^(-bE)/Z, P_1(2)=e^(-bU)/Z, P_2(2)=e^(-2b(E+U))/Z

Fermions: P_1(1)=P_2(1)=1 (Pauli exclusion principle), P_1(2)=P_2(2)=0

Distinct species: Z = e^(-bU) + e^(-2b(E+U)) + 2e^(-bE)
P_1(1)=P_2(1) = 2e^(-bE)/Z, P_1(2)=e^(-bU)/Z, P_2(2)=e^(-2b(E+U))/Z

For U->infinity, Z->e^(-bE) for bosons, 2e^(-bE) for distinct atoms,
P_1(1)=P_2(1) -> 1 for bosons and distinct atoms and P_2(2)=P_1(2)->0 as U->infinity.

I am not sure that whether this method is correct for distinct atoms and is there any more general methods that work for more than 2 wells? For example, I don't know what to do if the question ask: What is the probability of p_j(n) if there are N atoms and J wells....

DrClaude
Mentor
Looking at (b), your answers are not clear, and you have probabilities that do not sum to 1. State clearly what the different possible states are.

Clara Chung
Looking at (b), your answers are not clear, and you have probabilities that do not sum to 1. State clearly what the different possible states are.
b)
Consider P_j(n) as a macrostate of the system,
Bosons:
State 1: Two atoms in well 1
State 2: Two atoms in well 2
State 3: Each well is occupied by 1 atom
Can I use the apriori principle that each microstate is equally possible? In this case all states will have a probability of 1/3.

P_1(1)=P_2(1)=1/3+1/3=2/3
P_1(2)=P_2(2)=1/3

Fermions:
State 1: Each well is occupied by 1 atom, so the probability will be 1.

Distinct atoms:
State 1: Two atoms in well 1 p=1/4
State 2: Two atoms in well 2 p=1/4
State 3: Atom A in well 1 Atom B in well 2 p=1/4
State 4: Atom A in well 2 Atom B in well 1 p=1/4

P_1(1) = P_2(1) =1/4+1/4=1/2
P_1(2)=P_2(2) =1/4

I will try part c and d again...

c)
Z = 1 + e^(-2bE) + e^(-bE)
Bosons:
State 1: Two atoms in well 1 probability: 1/Z
State 2: Two atoms in well 2 p=e^(-2bE)/Z
State 3: Each well is occupied by 1 atom P=e^(-bE)/Z

P_1(1)=P_2(1)=e^(-bE)/Z
P_1(2)= 1/Z
P_2(2)= e^(-2bE)/Z

Fermions:
State 1: Each well is occupied by 1 atom, so the probability will be 1.

Distinct atoms:
Z = 1 + e^(-2bE) + 2e^(-bE)
State 1: Two atoms in well 1 p= 1/Z
State 2: Two atoms in well 2 p= e^(-2bE)/Z
State 3: Atom A in well 1 Atom B in well 2 p= e^(-bE)/Z
State 4: Atom A in well 2 Atom B in well 1 p= e^(-bE)/Z

P_1(1) =P_2(1) = 2e^(-bE)/Z
P_1(2) = 1/Z
P_2(2) = e^(-2bE)/Z

I think my original part c and d are right.. so I will save the time for writing out part d...