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Fermi statistics questions

  1. Jan 20, 2006 #1
    Okay, so I've been asked to calculate the Fermi energy of n 2D free particles of mass m, given that the density of states does not depend on energy i.e N(e) = D for E>0.

    Now I know that the general recipe for this is:

    [tex] n = \int{N(E).dE} = \int{D.dE} = DE [/tex]

    So that the energy is [tex] E = \frac {n}{D} [/tex] and this is the right answer for the Fermi energy. I'm just not sure why I know that this energy is the Fermi energy. I know it has something to do with the fact that at T = 0 the max energy is the fermi energy, but I'm not sure how to link that to this problem?

    Also does anyone know how I get from the above result to the total energy of the Fermi gas, and how this stuff relates to thermal velocity in metals? :biggrin:
     
  2. jcsd
  3. Jan 20, 2006 #2

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    To answer your first question, as you add particles at absolute zero, they settle into the lowest unoccupied states. In this case, there are n states with energy less than n/D, so the n particles will occupy these n lowest states. The Fermi energy is the energy of the highest occupied state (or, put another way, the incremental energy added to the system if you were to add another particle, explaining the connection between the Fermi energy and the chemical potential).

    The total energy of the system is the sum over the energies of each state. This can be converted from a sum over states to an integral in the usual way:

    [tex] \sum_k f(\epsilon_k) = \int D(\epsilon) f(\epsilon) d\epsilon[/tex]

    The thermal velocity can be found by getting the mean energy and using E=1/2 mv^2.
     
  4. Jan 21, 2006 #3
    I'm sorry I'm still a bit confused:blushing:

    Although I kept on thinking about it, and is theres something wrong with saying that since the F-D distribution says that at T=0, the Fermi energy is the maximum possible energy. I can then do the integral

    [tex] n = \int_{0}^{\epsilon_F} {N(E).dE} [/tex]

    Which would get me the right answer? Does that make sense or am I making that up.

    Also in the equation you wrote is [tex] f(\epsilon_k) [/tex] the distribution function? If so I'm still confused, I though summing that would give the number of particles with energy k. Should I then sum that over all k?
     
  5. Jan 21, 2006 #4

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    First of all, the equation I gave at the end is for an arbitrary function f. It just shows how to turn a sum over states into an integral over energy (a similar equation works for wave-vector).

    Now, the Fermi-Dirac distribution is (note that I'm using f as before, but now I'm referring only to the FD distribution):

    [tex]f(\epsilon) = \frac{1}{e^{(\epsilon-\mu)/k_B T}+1}[/tex]

    where [itex]\epsilon[/itex] is the energy and [itex]\mu[/itex] is the chemical potential, ie, the energy added to the system upon the addition of one more particle. This distribution gives the probability that a single (non-degenerate) state of energy [itex]\epsilon[/itex] will be occupied at a temperature T. [itex]\mu[/itex] is a function of temperature, and is found by:

    [tex]n=\sum_k d_k f(\epsilon_k)=\int d\epsilon f(\epsilon) D(\epsilon)[/tex]

    (where dk is the degeneracy of the kth state) and solving for [itex]\mu[/itex]. It reduces to the Fermi energy in the limit as T goes to zero. In this same limit, the distribution approaches one for energies less than the Fermi energy and zero for energies above it.

    So to answer your question, yes, the FD distribution says that as T approaches zero, all states below the chemical potential are occupied and all above it are not. So, using the general technique for solving for the chemical potential that I showed above, and using the fact that at T=0 we call the chemical potential the Fermi energy, you see that the method you used in your first post is the correct one.
     
    Last edited: Jan 21, 2006
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