Partition function to find expected occupancy of a lattice defect

[jncarter]

Homework Statement

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 $\mu$

Homework Equations

The expected number of electrons is given by:
<N> = $\Sigma n_{i}*p_{i}$
Where i iterates the possible states and p_i 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>= $\frac{1}{\beta} \frac{\partial}{\partial \mu} ln(Z)$

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.

All advice would be appreciated.

 

[mark726]

Dear fellow scientist,

Thank you for your post. Your approach to finding the expected number of electrons is correct, but there are a few errors in your calculations. First, the partition function should be written as:

Z = 1 + e^{-\beta(\epsilon - \mu)} + e^{-2\beta(\epsilon + \delta \epsilon - \mu)}

This is because the second electron can only occupy the impurity orbital if the first electron is already present, so the probability should be multiplied by an additional factor of e^{-\beta(\epsilon + \delta \epsilon - \mu)}.

Next, the expected number of electrons should be written as:

<N> = \frac{1}{Z} \left[e^{-\beta(\epsilon - \mu)} + 2e^{-2\beta(\epsilon + \delta \epsilon - \mu)} \right]

This is because the probability of each state should be divided by the partition function.

Finally, when taking the limit as \beta approaches zero, you should get <N> = 2 as expected. This can be seen by expanding the exponential terms and taking the limit.

I hope this helps. Keep up the good work!