Fermi-Dirac Distribution and Fermi Energy

The previous topic ended with a diagnosis: electrons are not a classical gas. This topic supplies the statistics that replace the classical assumption, and the failures resolve themselves.

The exclusion principle changes the counting

Electrons are fermions. No two of them may occupy the same quantum state. So when you put $N$ electrons into a metal, they cannot all sit in the lowest energy state — they fill states from the bottom upwards, one pair per level (spin up and spin down), until the electrons run out.

At absolute zero the filling is sharp: every state below a certain energy is occupied, every state above it is empty. That dividing energy is the Fermi energy $E_F$.

The distribution function

At any temperature $T$, the probability that a state of energy $E$ is occupied is

This is the Fermi-Dirac distribution function. Note what it is and is not: it is a probability of occupancy for a state that exists, not a count of electrons. To get the number of electrons you multiply it by the density of states.

Reading the function

Three cases tell you everything.

At $T = 0$. The exponent is $\pm\infty$ depending on the sign of $E - E_F$:

A perfect step. Every state below $E_F$ full, every state above empty.

At $E = E_F$, any $T > 0$. The exponent is zero, so $e^0 = 1$:

This is the definition worth memorising. The Fermi level is the energy at which the occupation probability is exactly one half, at any non-zero temperature. It is the cleanest way to state what $E_F$ means, and it is the form most exam questions want.

At $T > 0$, away from $E_F$. The step softens. States a little below $E_F$ lose some occupancy; states a little above gain some. The softening extends over an energy range of roughly $k_B T$ on either side — and that width is the crux of the whole module.

The width is tiny, and that is the point

At room temperature $k_B T = 0.0259\ \text{eV}$. For copper $E_F = 7.05\ \text{eV}$. The ratio:

Under half a percent. Heating a metal to room temperature disturbs the occupancy of only the electrons within about 0.4% of the top of the filled sea. Everything deeper is locked: an electron 1 eV below $E_F$ cannot absorb $0.026\ \text{eV}$ of thermal energy, because every state it could move to is already occupied. The exclusion principle forbids the transition.

That single fact resolves the specific heat failure. Only a fraction of order $k_B T / E_F$ of the electrons can absorb heat at all, so the electronic specific heat is smaller than the classical prediction by roughly that factor — a couple of hundred, which is the observed order of magnitude. The classical model let every electron take part; the exclusion principle lets almost none.

It also explains conduction. An electric field can only accelerate an electron into an empty state. Deep in the sea there are none, so those electrons carry no current no matter how large the field. Conduction is done entirely by electrons within $\sim k_B T$ of $E_F$, moving into the empty states just above.

Fermi energy, velocity and temperature

For a free electron gas of density $n$:

For copper this gives 7.05 eV. Two consequences are worth stating because they are so unlike classical expectations:

• Fermi velocity, from $E_F = \tfrac12 m v_F^2$, is $1.57 \times 10^{6}\ \text{m s}^{-1}$. Electrons at the top of the sea move at over a million metres per second at absolute zero, where a classical gas would be at rest. Their energy is not thermal; it is the cost of the exclusion principle.
• Fermi temperature, $T_F = E_F/k_B$, is about $8.2 \times 10^{4}\ \text{K}$ — far above copper's melting point. A metal at room temperature is, in the statistical sense, extremely cold. This is why the distribution stays so nearly a step.