My name is Alexander FufaeV and here I write about:
Fermi Distribution: The Occupation Probability of Fermions
Important Formula
$$P(W) ~=~ \frac{1}{\mathrm{e}^{ \frac{ W - \mu }{ k_{\text B} \, T}} ~+~ 1}$$
$$P(W) ~=~ \frac{1}{\mathrm{e}^{ \frac{ W - \mu }{ k_{\text B} \, T}} ~+~ 1}$$
$$W ~=~ k_{\text B} \, T \, \ln\left( \frac{1}{P(W)} ~-~ 1\right) ~+~ \mu$$
$$\mu ~=~ W ~-~ k_{\text B} \, T \, \ln\left( \frac{1}{P(W)} ~-~ 1\right)$$
$$T ~=~ \frac{ W ~-~ \mu }{ k_{\text B} \, \ln\left( \frac{1}{P(W)} ~-~ 1\right) }$$
What do the formula symbols mean?
Probability
$$ P(W) $$ Unit $$ - $$
The occupation probability indicates the probability \(P\) that a state with energy \( W \) is occupied at temperature \( T \). At absolute zero (\(T=0 \, \text{K}\)), the probability that the state with energy \( W \) is occupied is exactly 50%: \( P(W) ~=~ \frac{1}{2}\).
Energy
$$ W $$ Unit $$ \mathrm{J} = \mathrm{Nm} = \frac{ \mathrm{kg} \, \mathrm{m^2} }{ \mathrm{s}^2 } $$
Energy state which can be occupied by a fermion, for example by an electron.
Chemical potential
$$ \mu $$ Unit $$ \mathrm{J} $$
Chemical potential gives the change of the internal energy when the particle number of the Fermi gas (e.g. free electron gas) changes. At \( T=0 \, \text{K} \) the chemical potential correspons to the Fermi energy: \( \mu = W_{\text F} \).
Temperature
$$ T $$ Unit $$ \mathrm{K} $$
Absolute temperature of the Fermi gas, for example a free electron gas in a metal.
Boltzmann Constant
$$ k_{\text B} $$ Unit $$ \frac{\mathrm J}{\mathrm K} = \frac{\mathrm{kg} \,\mathrm{m}^2}{\mathrm{s}^2 \, \mathrm{K}} $$Boltzmann constant is a physical constant from many-particle physics and has the following exact value:
$$ k_{\text B} ~=~ 1.380 \, 649 ~\cdot~ 10^{-23} \, \frac{\mathrm{J}}{\mathrm{K}} $$
The Fermi distribution (also known as the Fermi-Dirac distribution) is particularly important in the description of electrons in solids, especially in metals.
The Fermi distribution indicates the occupation probability \( P(W) \) that a fermion in a system acquires the energy \( W \). The occupation probability depends on the temperature \( T \) of the system.
$$ \begin{align} P(W) ~=~ \frac{1}{\mathrm{e}^{ \frac{ W - \mu }{ k_{\text B} \, T}} ~+~ 1} \end{align} $$
In contrast to the Bose distribution, the Fermi distribution contains a positive term in the denominator, which ensures that the probability of occupying a state does not approach zero when the temperature approaches zero. This means that fermions cannot occupy the same quantum mechanical state according to the Pauli principle.