Fermi-Dirac distribution normalization

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The Fermi-Dirac distribution is not a true probability distribution, as it represents the probability of occupancy of quantum states rather than a normalized probability density function. At absolute zero (T = 0), the integral yields the chemical potential, allowing for normalization by 1/μ. However, at higher temperatures (T >> 0), the distribution does not require normalization in the conventional sense. Instead, the values of the distribution function f(E) range between 0 and 1 for each energy state. Understanding this distinction is crucial for applying the Fermi-Dirac statistics correctly.
Davide82
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Hi!

I have a little question which is puzzling me.
Maybe it is a very simple question.

It is my understanding that the Fermi-Dirac distribution is a probability density function and, as such, its integral between 0 and infinite should be 1.
When T = 0, the integral gives the chemical potential and so the distribution can be normalized by 1 / \mu.
But if I calculate the integral while T >> 0 I don't understand which could be the normalization factor. Do you have an answer?

Thank you
 
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No, the Fermi-Dirac distribution is not a true probability distribution. It tells you the probability of occupation of a state at energy E given the chemical potential, which means that for every energy value the value of f(E) has to range between 0 and 1. It does not get normalized in the way that you're thinking.
 
Thank you!
 
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