How Does the Grand Partition Function Apply to Electron Occupancy in Defects?

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jmz34
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The example which I'll use to illustrate my problem is not a homework question but something I've found in a book and already know the answer to.

The grand partition function, G, is defined as SUM(over i)[exp(-B(Ei-yNi))] where B=1/kT, y is the chemical potential and Ei is the energy of the state i. I'm fine with this definition.

But the total grand partition of a system is also equal to the product of the individual grand partition functions of each energy level: G=G1*G2*...*GN . I thought I understood this as well until I came across the following example:

A point defect in a solid may be occupied by 0 ,1 (spin up or spin down) or 2 electrons, and the solid provides a reservoir of electrons at chemical potential y. The energy for occupation by a single electron is E, and that for 2 electrons is 2E+U, where U is the Coulomb repulsion energy between the two electrons. Obtain an expression for the average occupancy of the defect.

Now I tried doing this with the first definition to start with and that gave me the correct answer. So I said, G=1+exp(-B(E-y))+exp(-B(2E+U-2y)) and then used the grand potential etc.

However the second method didn't give me the correct answer. I identified the 4 independent energy levels as : 0, E, E, 2E+U. Then for each level I worked out the grand partition function:

G1=1
G2=1+exp(-B(E-y))
G3=G2
G4=1+exp(-B(2E+U-y))

since the energy level 4 can only exist if it's occupied by two electrons I grouped those two electrons together as effectively one particle with an associated energy of occupation 2E+U.

G=G1G2G3G4 does not give me the correct answer. Why is this?

Thanks.
 
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