How does the Heisenberg Uncertainty Principle relate to electron degeneracy pressure?
1 Answer
From the Heisenberg Uncertainty Principle, we know that
#DeltaxDeltap >= ℏ/2#
i.e. the momentum and position cannot be simultaneously observed to the same precision.
Now, imagine compressing two electrons into a confined space.
If a second electron is placed into a half-filled orbital, the electron already there necessarily has to move to a new, higher energy level. If one further contracts the orbital size, the electrons are going to have more defined positions,
That implies higher uncertainties in their momenta, i.e. on average they are moving at very high momentum. This gives rise to something called "degeneracy pressure":
#P_(deg) = ((3pi^2)^"2/3"ℏ^2)/(5m_e) rho_N^"5/3"# where:
#ℏ# has units of#"J"cdot"s"# , or#"kg"cdot"m"^2"/s"# .#m_e# is the mass of the electron in#"kg"# .#rho_N -= N/V# is the number of free electrons per unit volume, with#"m"^3# for the volume.#P_(deg)# is the degeneracy pressure in#"Pa"# .#P_(deg)# here is only weakly dependent on temperature, unlike the ordinary pressure in the gas laws.
Degeneracy pressure pertains to the amount of energy involved in condensing matter to make it degenerate, i.e. all the same energy levels.
Usually, degeneracy pressure is negligible compared to the kind of pressure we all are familiar with, i.e. at ordinary matter densities,
#P_(deg)# #"<<"# #[P -= (Nk_BT)/V]# where
#k_B# is the Boltzmann constant in#"J/K"# and#T# is temperature in#"K"# .
When electron density is extraordinarily packed, i.e.
#P_(t ot) = P_(deg) + P ~~ P_(deg)# .
When that is the case, the electrons in the space are considered degenerate, since energy at constant entropy (particle distributions in each energy level) is a function of the volume, i.e. smaller volumes leads to more-degenerate energy levels.
(Although this is easiest at
