Electron Degeneracy Pressure is A phenomenon arising from the symmetry properties of wavefunctions of degenerate electrons, although the concept holds true for all degenerate fermions (particles with non-integral spin)such as neutrons. It is this electron degeneracy pressure that holds up a white dwarf star, e.g. Sirius B, against the force of its own gravity, and a more appropriate name for such a star is a degenerate star. The pressure arising from degenerate neutrons is what holds up a neutron star, although neutron-neutron interaction and beta decay make modelling such a star more difficult than for a white dwarf.

It turns out that fundamental particles such as electrons are indistinguisable and thus obey different statistical lawsthan macroscopic objects. Bosons, or particles with integral spin such as photons (spin 1), obey Bose-Einstein Statistics and tend to coalesce into the same states if conditions permit. Fermions, or particles with non-integral spin such as electrons, protons and neutrons (all with spin 1/2), obey Fermi-Dirac statistics and eschew one another and occupy different states. The Pauli-Exclusion principle states that no two fermions can occupy the same state, and this arises from the symmetry properties of their wavefunctions:

Consider two indistinguishable particles with positions **r**_{1} and **r**_{2}. These may also be states but we can take them to be positions for simplicity. The wavefunction that describes these two particles is ψ(**r**_{1}, **r**_{2}, t). However, all that we can measure about the particles have to do with |ψ|^{2} = ψ^{*}ψ where ψ^{*} is the complex conjugate of ψ. To see how symmetry enters the argument, consider exchanging the two particles. The new wavefunction is then ψ(**r**_{2}, **r**_{1}, t). Because the particles are indistinguishable, our measurements should not be able to discover this switch, so |ψ(**r**_{1}, **r**_{2}, t)| = |ψ(**r**_{2}, **r**_{1}, t)|, for all **r**_{1} and **r**_{2}. It turns out that this implies that ψ(**r**_{1}, **r**_{2}, t) = ± ψ(**r**_{2}, **r**_{1}, t). It also turns out that if ψ(**r**_{1}, **r**_{2}, t) = + ψ(**r**_{2},**r**_{1}, t) in one instant, then it must be true in all instants. The same is true for ψ(**r**_{1}, **r**_{2}, t) = - ψ(**r**_{2},**r**_{1}, t). Particles that obey the former are called Bosons while particles that obey the latter are called Fermions, so lets focus on them.

If the two particles where to come close to one another, or occupy nearly the same state (if we had chosen our position vectors to be state vectors instead), then we have that **r**_{1} ≅ **r**_{2} = **r**. We can then say that ψ(**r**,**r**, t) = - ψ(**r**,**r**, t), and ofcourse this implies that ψ = 0. The probability of observing the two particles approaches zero as they come close to one another, such that the probability of observing them in the same place or state is exactly zero. This property of the wavefunction causes a repulsion between particles, but one that involves no force! My professor likened this to two English gentlemen on poor terms with one another who simply arrange their own affairs so that they never meet.

Coming back to Electron degeneracy pressure, let us model a degenerate spin 1/2 fermi gas and formulate this pressure mathematically. Consider a number of such particles, electrons say, which occupy some finite volume from which they cannot escape. We'll assume the electrons to be non-interacting for simplicity, and this can easily be realized to some degree of accuracy if the electrons are submersed in an equal number of positive ions, such as in a metal or core of a degenerate star. Consider the electrons sitting in the bottom of an infinite 3D potential well. Because they are in bound states, each state will have a discrete allowable energy E_{i}. For the electron gas to be degenerate, we need all the particles to occupy the lowest possible states available to them (subject to the Pauli constraint), so that we may say T → 0. This is not entirely unreasonable because in the astronomical conditions that these phenomena occur, the temperature is relatively sufficiently cold for this assumption to be valid. The first two electrons will then occupy the lowest energy state, with the next two occupying the next, and so on, with no state being passed up. Remember that electrons can be spin +1/2 or -1/2 so that two electrons may occupy the same energy level, if their spins are antiparallel. As the temperature increases some electrons will gain enough energy to move into higher states, leaving intermediate states unoccupied, but this effect is minimal at the temperatures that we'll consider. We'll say that there are N such particles in N/2 states, and that the energy of the highest state is E_{f} (the Fermi energy).

First we must determine the energy levels that are available to the electrons. Lets assume that the well exists between x = 0 and x = L_{x}, y = 0 and y = L_{y}, z = 0 and z = L_{z}, so that the potential V = 0 here, and that V = &inf; outside. This is essentially your run of the mill particle in a box. The potential is time independent, so lets use the time-independent Schrödinger Equation:

-~~(h~~^{2}/2m) ∇^{2}Ψ + V(**r**)Ψ = E Ψ

Inside the box, V = 0, and we may use boundary conditions that Ψ = 0 outside to require that Ψ = 0 on the boundary of the box as well. We can use separation of variables and write Ψ(x,y,z) = A(x) B(y) C(z) to see if we can find a solution in this manner:

d^{2}A/dx^{2} BC + d^{2}B/dy^{2} AC +
d^{2}C/dz^{2} AB = - (2m E / ~~h~~^{2}) ABC

We can divide both sides by Ψ = ABC:

d^{2}A(x)/dx^{2} / A(x) + d^{2}B(y)/dy^{2} / B(x) +
d^{2}C(z)/dz^{2} / C(x) = - 2m E / ~~h~~^{2}

Now, since each term depends on a different variable, and since the equation must hold at all positions throughout the box, each term must separately be equal to a constant. We can solve the first to yield A(x) = A_{0} sin(k_{x} x) and similarily with the others. We want A(0) = 0 and A(L_{x}) = 0, so k_{x} = n_{x}π/L_{x}. We can now substitute the results back into our equation above, and get:

E_{nx}_{ny}_{nz} = (~~h~~^{2}π^{2} / 2m)( n_{x}^{2} / L_{x}^{2} +
n_{y}^{2} / L_{y}^{2} +
n_{z}^{2} / L_{z}^{2})

Now we can determine E_{f}. Here some cleverness is involved. We'll instead try to find the total number of particles in terms of the Fermi energy. We do this by adding up the amount of particles in each state (2), upto the highest occupied state, summing over all states. Lets give HTML a run for its money:

N = Σ_{E=0}^{Enxnynz ≤ Ef} (2)

Now, in our gas we may suppose that there are a great many electrons and that the levels are closely spaced, and transform the sum into an integral. Feeling masochistic yet?

N = ∫_{E=0}^{Enxnynz ≤ Ef} (2) dn_{x}dn_{y}dn_{z}

Where E_{nx}_{ny}_{nz} = (~~h~~^{2}π^{2} / 2m)( n_{x}^{2} / L_{x}^{2} +
n_{y}^{2} / L_{y}^{2} +
n_{z}^{2} / L_{z}^{2}). Lets make the substitution α = n_{x}/ L_{x}, β = n_{y}/ L_{y} and γ = n_{z}/ L_{z}.

N = L_{x}L_{y}L_{z}
∫_{E=0}^{Enxnynz ≤ Ef} (2) dα dβ dγ

Where E_{nx}_{ny}_{nz} = (~~h~~^{2}π^{2} / 2m)( α^{2} + β^{2} + γ^{2}). Now, the region of integration is 1/8^{th} of the sphere α^{2} + β^{2} + γ^{2} ≤ 2m E_{f} / (~~h~~^{2}π^{2}) = R^{2}, so letting V = L_{x}L_{y}L_{z}, we obtain:

N = (1/8) 2V ∫_{0}^{R} 4πr^{2} dr

N = (1/8) 2V (4π/3) R^{3}

N = (1/8) 2V (4π/3) (2m E_{f} / ~~h~~^{2}π^{2})^{3/2}

Putting it all together, we obtain

E_{f} = (~~h~~^{2}π^{2}/2m) (3/π N/V)^{2/3}

Since we're after the pressure created by the fermi gas, we'll start with a related value, the total energy contained within the volume. We simply sum the energy contained in each occupied state, so:

U = (1/8) 2V ∫_{0}^{R} E(n_{x},n_{y},n_{z}) 4πr^{2} dr

Now E(n_{x},n_{y},n_{z}) = (~~h~~^{2}π^{2}/2m) r^{2}, so

U = πV(~~h~~^{2}π^{2}/2m) ∫_{0}^{R}r^{4} dr

U = πV(~~h~~^{2}π^{2}/2m)(1/5)R^{5}.

U = πV(~~h~~^{2}π^{2}/2m)(1/5)(2m E_{f} / ~~h~~^{2}π^{2})^{5/2}.

But since E_{f} = (~~h~~^{2}π^{2}/2m) (3/π N/V)^{2/3}, we can write:

U = πV(~~h~~^{2}π^{2}/10m)(3/π N/V)^{5/3}.

Now we have the total energy contained within the potential well due to all the electrons present within it. This system is in equilibrium and we can use the thermodymic properties of the gas to obtain the pressure:

dU = T dS - P dV

Since the gas is degenerate and T is sufficiently small, we must have dS = 0 (any change in the system will go from one completely ordered state to another completely ordered state, so the change in entropy goes to zero). Therefore:

P = - ∂U/∂V

So,

P = (~~h~~^{2}π^{2}/5m)(3/π)^{2/3}(N/V)^{5/3}.

Now this is the pressure exerted by the gas due solely by the Pauli repulsion. We may write P ∝ ρ^{5/3} where ρ is the density of the electron gas. However, this result is only good for non-relativistic electrons, i.e. electrons travelling much slower than the speed of light. For a classical gas, P ∝ p v where p is the momentum and v is the velocity of the electron. Since v ∝ p, We have that P ∝ p^{2}. For relativistic electrons, v ∝ c but p remains unbounded, so P ∝ p c ∝ p. This difference, along with other effects, causes the pressure to depend on ρ^{4/3} rather than ρ^{5/3}.

This is an important result, since stars for which P ∝ ρ^{1+1/n} are known as Regular Polytropes. For such stars we can use the condition of hydrostatic equilibrium to solve the Lane-Emden equation, which allows us to find the pressure and density of the star as a function of the distance from the center. In some cases this can even be found in closed form. This is important in Stellar Astrophysics since in many cases we can use this information to determine things about the luminosity and power production of a star, at least to a first approximation.