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Homework Help: Partition function to find expected occupancy of a lattice defect

  1. Apr 25, 2012 #1
    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 [itex]\epsilon[/itex], but to place the second electron on the site requires an additional energy [itex]\delta \epsilon[/itex].

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

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

    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:
    [itex]-\frac{1}{\beta} \frac{\partial}{\partial \epsilon} ln(Z)[/itex]
    My latest attempt was to write the partition function as follows:
    [itex]Z = 1 +e^{-\beta(\epsilon_{1} -\mu)} + e^{-\beta(\epsilon_{2} -\mu)2}[/itex]
    Where [itex]\epsilon_{1} = \epsilon[/itex] and [itex]\epsilon_{2} = \epsilon +\delta \epsilon [/itex]
    And then use the first method given to find the expected number of electrons.
    <N> = [itex]\frac{e^{-\beta(\epsilon -\mu)}+2e^{-\beta(\epsilon +\delta \epsilon -\mu)2}}{1+e^{-\beta(\epsilon_{1} -\mu)} + e^{-\beta(\epsilon_{2} -\mu)2}} [/itex]

    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 [itex]\beta[/itex] approaches zero (increasing temperature) is one rather than two.

    All advice would be appreciated.
  2. jcsd
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