Probability of occupation of a state of energy E

EDerkatch

Could someone please explain why it is f(E) for a gas of electrons, where f(E)=1/(Exp[B(E-(mu)]+1).

Thanks! :smile:

Reply 1

Morbo

That's not a simple explanation - a textbook devotes many pages to showing that. It's called the Fermi-Dirac distribution function and relies on a knowledge of the Pauli exclusion principle, and statistical mechanics.

However, I can explain to you why it works.

f(E) = \frac{1}{\mathrm{e}^{\beta(E-\mu)}+1}

Due to the Pauli exclusion principle, only 1 electron is allowed in any quantum state. Hence, at absolute zero, electrons stack up in energy - with one per state - up to the maximum energy they need, called the Fermi energy. Above the Fermi energy, there are no electrons occupying states. The Fermi-Dirac distribution function models how the electrons fall into these states.

The Fermi energy

E_F = \mu

at absolute zero, hence the distribution function becomes:

f(E) = \frac{1}{\mathrm{e}^{\frac{(E-E_F)}{kT}}+1}

Now, consider when T~0. At energy E<E_F i.e. for energies below the Fermi energy, the exponential function = 0, and so f(E) = 1 i.e. one electron per state.

When E>E_F, i.e. for energies above the Fermi energy, the exponential function = infinty, thus f(E) = 0 i.e. no electrons per state.

Thus, the results we need at T=0 are produced. As T increases, the distribution doesn't become a simple "1 per state then 0 per state", but takes account of the thermal excitations which mean some electrons will be above the Fermi energy. In fact as T gets big enough, f(E) tends towards the fully thermal Boltzmann distribution.

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