In quantum mechanics, why in a fully degeneracy fermion state pressure increases when momentum also increase? Heisenberg uncertainty principle, ΔpΔx ≥ ħ/2, where Δp is the uncertainty in the particle's momentum and Δx is the uncertainty in position. This implies that the momentum of a highly compressed particle is extremely uncertain, since the particles are located in a very confined space. Why is that the case? Why pressure would increase? and Why this relationship between momentum and position. I am a beginner and would appreciate a more intuitive answer, like Richard Feynman would ;O)
 A: For the simplest example consider the particle trapped within a box. Beginning with the ground state (n=1 lowest energy), there are an infinite number of states the particle can be in:

Because fermions, aren't allowed to share the same state with one another, if you add more particles to the box, they each must lie in a different state. A higher energy state means the particle has more kinetic energy/ momentum.
Say for simplicity, we have four particles in the box. If they were bosons, in the lowest energy state for the system, all four would have n=1. For four fermions however, they would (at the lowest energy) occupy states 1 through 4. This means that the fermions would have a higher average momentum than would four bosonic particles of the same mass. Interestingly as you add more fermions to the box, (even at the lowest energy state) the average momentum for the particles increases (which isn't necessarily the case for the bosonic particles)
Because pressure is simply (to use a classical analogue) the average momentum times the average number of particles hitting some cross sectional area of the wall, this immediately corresponds to a higher pressure for the fermions than the bosons (to put it in reverse order of your question).
In essence, I believe your question (as stated in the title) boils down to the Pauli exclusion principle.
