Work check and advice on a statistical mechanics problem

  • #1
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Homework Statement:

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

Relevant Equations:

Probabilities
241555

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....
 

Answers and Replies

  • #2
DrClaude
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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.
 
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  • #3
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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...
 
  • #4
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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...
 

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