Two energy level system calculate average occupancy

[Bananen]

Hi!

I need help with the following question:
A system has two energy levels, ε and ε1 that can be occupied by fermions (spin=1/2) that are non-interacting from a reservoir at temperature T and chemical potential μ. Compute the avarage occupation number of the state with energy ε.

I have written down the grand canonical partition function and it is (1+e^(-β(ε-μ))^2)((1+e^(-β(ε1-μ)))^2)

and I know the formula for avarage occupation number for fermions is
ΣnP(n) where n goes from 0 to 1.

But then I'm stuck. The hint I got was that there can be up to four particles and there are different possibilities to put these into the energy levels.

Very thankful for any help!

 

[member 553083]

The answer to this question is as follows:The average occupation number of the state with energy ε is given by:N_ε = 1/Z * (1+e^(-β(ε-μ)))^2 where Z is the grand canonical partition function. This can be calculated by considering the different possibilities for the four particles to occupy the two energy levels: 1 particle in ε and 3 in ε1; 2 particles in ε and 2 in ε1; 3 particles in ε and 1 in ε1; 4 particles in ε and 0 in ε1.The probability of each case is given by the ratio of the corresponding partition function to the total partition function, which is the grand canonical partition function. For example, the probability of having 1 particle in ε and 3 in ε1 is given by: P_1ε_3ε1 = (1+e^(-β(ε-μ)))^2/(1+e^(-β(ε-μ)))^2 + (1+e^(-β(ε1-μ)))^4Calculating the probability of all four cases and summing them up gives the desired result: N_ε = 1/Z * (1+e^(-β(ε-μ)))^2 + (1+e^(-β(ε1-μ)))^2 * [ (1+e^(-β(ε-μ)))^2 + (1+e^(-β(ε1-μ)))^2 ]