Fermi distribution interpretation

  • #1
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Hello!
Let [itex]E_1, E_2, \ldots, E_n[/itex] be [itex]n[/itex] allowed energy levels for a system of electrons. This system can be described by the Fermi-Dirac distribution [itex]f(E)[/itex].
Each of those levels can be occupied by two electrons if they have opposite spins.
Suppose that [itex]E_1, E_2, \ldots, E_n[/itex] are such that

[itex]\displaystyle 2 \sum_{k = 1}^n f(E_k) = 1[/itex]

where the [itex]2[/itex] is due to the degeneracy of states (two electrons allowed for each state). So, can it be stated that in such a system is actually present one electron, that is the result of the sum?
If someone could even explain why, it would be very appreciated.
In any case, thank you for having read.

Emily
 

Answers and Replies

  • #2
Charles Link
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If I understand it correctly, the Fermi factor is the mean number of electrons in a state in k-space. You sum over all the k-space states and it adds up to the number of electrons N in your system. Unless you are very selective with the k-space states that you sum over, you won't get the sum of the mean numbers to add up to 1. (if you pick a bunch of very highly energetic states, then maybe on the average only one of these states will be occupied.)
 
  • #3
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If I understand it correctly, the Fermi factor is the mean number of electrons in a state in k-space.
Did you deduce this from my post or by your own?

Unless you are very selective with the k-space states that you sum over, you won't get the sum of the mean numbers to add up to 1. (if you pick a bunch of very highly energetic states, then maybe on the average only one of these states will be occupied.)
The first post was maybe not clear about this. Don't consider my problem as real and don't take its numbers as absolute: it was just an example to show how the Fermi distribution is used to count the electrons in a band of energies.
The strange fact about Fermi distribution is that we start talking that [itex]f(E)[/itex] is the probability that the quantum state at level E is occupied; then we talk about the number of electrons that populate the conduction band of a semiconductor by doing

[itex]\displaystyle \int_{E_c}^{+\infty} g_c(E) f(E) dE[/itex]

([itex]g_c(E)[/itex] is the density of states in the conduction band). So, how can we speak almost interchangeably about probability and actual presence of a particle?
 
  • #4
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So, how can we speak almost interchangeably about probability and actual presence of a particle?
What do you mean by actual presence of a particle?
 
  • #5
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What do you mean by actual presence of a particle?
I know that, in Quantum Mechanics, talking about "actual" quantities is inappropriate.
Anyway, it is common to compute the number of electrons in conduction band, for example, in a semiconductor (e.g. here, penultimate formula). During the computation, Fermi distribution and density of states are used. The resulting number [itex]n[/itex] is considered as if it were the number of actual particles populating that band.

Hoping to have clarified your doubt, this is what I meant.
 
  • #6
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Until observation shows the model to be found significantly deficient, and/or until there is a better one, this is how n is estimated.
 
  • #7
f95toli
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How many electrons are "actually" present is not really a meaningful question in this case. The Fermi distribution tells you something about the statistical properties of a system; it can't be used to predict e.g. exactly how many electrons are going to be in a specific state at a given time.

Note that this is not really a practical problem, in most cases we are dealing with either so many electrons or such long times that Fermi distribution will give you a very accurate estimate of the occupancy as in the case of the formula for the semiconductor.

There are system where we are really dealing with a few electrons (e.g. quantum dots) and then you have to be more careful about really interpreting the results as an probability.
 
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  • #8
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@EmilyRuck. What do you mean y actual presence ?
 

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