# Partition function to find expected occupancy of a lattice defect

1. Apr 25, 2012

### jncarter

1. The problem statement, all variables and given/known data
An impurity can be occupied by 0, 1 or 2 electrons. The impurity orbital in non-degenerate, except for the choice of electron spin. The energy of the impurity level is $\epsilon$, but to place the second electron on the site requires an additional energy $\delta \epsilon$.

Calculate the expected number of electrons <N> at the site as a function of $\epsilon, \delta \epsilon$ and the temperature T and the chemical potential $\mu$

2. Relevant equations
The expected number of electrons is given by:
<N> = $\Sigma n_{i}*p_{i}$
Where i iterates the possible states and pi is the probability of the ith state.
$p_{i} = \frac{e^{-\beta(\epsilon_{i}-\mu)*n_{i}}}{Z}$
Where Z is the partition function.
Alternatively, the expected number of electrons may be calculated by:
<N>= $\frac{1}{\beta} \frac{\partial}{\partial \mu} ln(Z)$

3. The attempt at a solution
I'm unsure of the best way to approach forming the partition function. I am also unsure of the second method to calculate the expected occupancy. I've seen it done that way as well as:
$-\frac{1}{\beta} \frac{\partial}{\partial \epsilon} ln(Z)$
My latest attempt was to write the partition function as follows:
$Z = 1 +e^{-\beta(\epsilon_{1} -\mu)} + e^{-\beta(\epsilon_{2} -\mu)2}$
Where $\epsilon_{1} = \epsilon$ and $\epsilon_{2} = \epsilon +\delta \epsilon$
And then use the first method given to find the expected number of electrons.
<N> = $\frac{e^{-\beta(\epsilon -\mu)}+2e^{-\beta(\epsilon +\delta \epsilon -\mu)2}}{1+e^{-\beta(\epsilon_{1} -\mu)} + e^{-\beta(\epsilon_{2} -\mu)2}}$

This is all well and good, but I took a look at the limiting behavior and my result does not match my expectations. I would expect that as the temperature increased N would approach two and as it decreased, <N> would approach zero. Yet, the limit as $\beta$ approaches zero (increasing temperature) is one rather than two.