Fermion
Definition
- subatomic particle
- half integer spin (in units of ħ)
- only one per state
- follows the Fermi-Dirac statistical distribution
e. g. electrons, protons, neutrons He 3 atoms
Multiplicity of Fermions
Derivation
- there is only one or zero per state
- imagine n particles in m levels as n particles in m boxes
- simplify by replacing the boxes by partitions
- m different boxes can be arranged in m! ways but there are n identical states with one particle, i.e. swapping two particles doesn’t produce a different distribution so divide by n!
- the (m-n) states without particles are also identical so swapping them doesn’t produce a different distribution either, divide by (n-m)!
Multiplicity of Fermions
Equation
Ω = m! / n!(m-n)!
m = no. of states n = no. of particles
Fermi-Dirac Distribution Function
ni / ωi = fi = 1 / [ exp(α + Ei/kT) + 1 ]
Fermi-Dirac Function
f (E) = 1 / [ exp(-(Ef-E)/kT) + 1 ]
- where Ef is the Fermi energy
- f is the Fermi-dirac function
Fermi Dirac Distribution
Graph
- step curve for 0K (horizontal from probability of occupancy 1 on the y axis, 90 degree angle with vertical line down to the x axis, horizontal along x axis (y=0) )
- as temperature increases, the 90 degree angle becomes more rounded
Fermi Energy
Definition
- the Fermi energy is the energy boundary between the occupied and empty states at T=0
- since at T=0 all states from the lowest state up will be filled first so there will be a boundary state above which there are no particles the energy of this boundary state is the Fermi energy
Density of States for the Electron Gas
g(ε) = V/2π² [2m/ħ²]^(3/2) ε^(1/2)
Electron Density
- the Fermi energy is a function of electron density
- an intensive variable
Fermi Tempertaure
Tf = Ef / k ≈ few eV ≈ few 10^4 K
Approximating the Fermi-Dirac Distribution Near Ef With a Linear Function
- for T>0K, the Fermi-Dirac distribution is not a step function, there is a slope, the middle of the slope is at E=Ef
- in that sloped region are the states that are partially occupied
- the function in that region can be approximated by a straight line with gradient equal to the differential of the Fermi-Dirac distribution at Ef
- the gradient is 1//4kT
Heat Capacity of a Cold Fermi Gas
-Fermi-Dirac statistics explains the (low) heat capacity of a metal
-excitation only occurs within a narrow range ~kT around Ef
-the fraction of electrons we transfer to higher energies is ~kT/Ef
-the energy increase for these electrons is ~kT
-thus the increase in internal energy with temperature is proportional to:
N * kT/Ef * kT ~ N(kT)²/Ef
Cv = (∂U(T)/∂T)v ∝ N(kT)²/Ef
-which is smaller than Cv = 3/2 * Nk for an ideal gas
Why is the Fermi gas heat capacity much smaller than that of a classical ideal gas?
- the Fermi gas heat capacity is much smaller than that of a classical ideal gas with the same energy and pressure
- this is because only a small fraction KT/Ef of the electrons are excited out of the ground state
- as required by the third law, the electronic heat capacity in metals goes to 0 as T->0
Why is the Helmholtz free energy equal to the internal energy at T=0 ?
because entropy=0 at T=0
Fermi Pressure in Metals
- for a typical metal, p=2/5 nEf = 5x10^10 Pa
- in metals this enormous pressure is counteracted by the Coulomb attraction of the electrons to the positive ions
- this pressure balancing the Coulomb attraction is what allows electrons in metals to delocalise
Fermi Pressure
p = - (∂F/∂V)|N
-at T=0 , F=E
p = - (∂E/∂V)|N = 2/5 N/V Ef
-this is a non-zero pressure at T=0 and does not depend on T at T<
Internal Energy in Terms of Fermi Energy
E = 3/5 N Ef
What is the entropy of a Fermi gas at T=0 ?
-the entropy of a Fermi gas is zero at T=0 , it is a single state
