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

AI Thread Summary
The discussion revolves around the calculation of the grand partition function for a point defect in a solid that can be occupied by 0, 1, or 2 electrons. The correct expression for the average occupancy was derived using the grand partition function defined as G=1+exp(-B(E-y))+exp(-B(2E+U-2y). However, an alternative method using the product of individual grand partition functions for each energy level did not yield the correct result. The confusion arises from the treatment of energy levels, particularly the need to account for the occupancy of two electrons correctly. The participant questions whether the second term should include a factor of 2, indicating a potential oversight in their calculations.
jmz34
Messages
26
Reaction score
0
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.
 
Physics news on Phys.org
So I said, G=1+exp(-B(E-y))+exp(-B(2E+U-2y)) and then used the grand potential etc.
Shouldn't the second term have a factor of 2?
 
Thread 'Gauss' law seems to imply instantaneous electric field'
Imagine a charged sphere at the origin connected through an open switch to a vertical grounded wire. We wish to find an expression for the horizontal component of the electric field at a distance ##\mathbf{r}## from the sphere as it discharges. By using the Lorenz gauge condition: $$\nabla \cdot \mathbf{A} + \frac{1}{c^2}\frac{\partial \phi}{\partial t}=0\tag{1}$$ we find the following retarded solutions to the Maxwell equations If we assume that...
Maxwell’s equations imply the following wave equation for the electric field $$\nabla^2\mathbf{E}-\frac{1}{c^2}\frac{\partial^2\mathbf{E}}{\partial t^2} = \frac{1}{\varepsilon_0}\nabla\rho+\mu_0\frac{\partial\mathbf J}{\partial t}.\tag{1}$$ I wonder if eqn.##(1)## can be split into the following transverse part $$\nabla^2\mathbf{E}_T-\frac{1}{c^2}\frac{\partial^2\mathbf{E}_T}{\partial t^2} = \mu_0\frac{\partial\mathbf{J}_T}{\partial t}\tag{2}$$ and longitudinal part...
Back
Top