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I am reading a paper on Neutron Degeneracy (http://www.physics.drexel.edu/~bob/Term_Reports/John_Timlin.pdf) and it is discussing the Pauli Exclusion Principle. There has to be trillions and trillions of neutrons packed together. Does the PEP mean than none of these trillions and trillions of neutrons are at the same energy level? How is that possible?

Does the PEP stop after a certain point or does it cover the entire universe?

Rick
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3 Answers3

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The PEP forbids indistinguishable fermions from occupying the same quantum state. They can have the same energy.

Clearly, particles localised at totally different positions do not share the same wavefunction or quantum state.

Cold neutrons confined in a gravitational potential well simply occupy all the available quantum states per unit volume, starting from zero (kinetic) energy upwards. There are a finite number of quantum states because the spatial part of the neutron wavefunction is constrained to be zero at the edges of the potential well.

As a handwaving rule, the PEP is only going to come into play when particles are within a de Broglie wavelength of each other. i.e. $\lambda = h/p$, where a typical neutron momentum might be of order $p \sim 0.1 m_n c$ and so $\lambda \sim 10^{-14}$ m.

ProfRob
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  • "the PEP is only going to come into play when particles are within a de Broglie wavelength of each other", can you please elaborate on that? – Árpád Szendrei Jan 16 '21 at 20:35
  • @ArpadSzendrei Each momentum eigenstate takes up $h^3$ of phase space, so the states will be getting full as $(\Delta x \Delta p)^3 \sim h^3$. i.e. when $\Delta x \sim h/\Delta p$. But the typical momentum difference between adjacent particles will be like the RMS value of $p$. – ProfRob Jan 17 '21 at 00:15
  • Thanks.. This makes sense. – Rick Jan 17 '21 at 12:52
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The PEP is a direct result of the fermionic statistics of the neutrons. The fact that they are fermions mean that if we write a multiparticle wave function it must satisfy $$\psi(x_1,...,x_i,...,x_j,...x_n)=-\psi(x_1,...,x_j,...,x_i,...x_n)$$ Note that this means that the function is zero for $x_i=x_j$: two identical fermions cannot be at the same position, or in general in the same state. This is a very basic law of QM and is always relevant (whenever all particles in the system are fermions. Photons, for example, are bosons, and can exist in the same state).

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The Pauli principle applies to indistinguishable fermions. There’s quite a bit of literature on what it means for particles to be indistinguishable, v.g.

Bach, A., 1988. The concept of indistinguishable particles in classical and quantum physics. Foundations of physics, 18(6), pp.639-649,

Kaplan, I.G., 1975. The exclusion principle and indistinguishability of identical particles in quantum mechanics. Soviet Physics Uspekhi, 18(12), p.988,

Kaplan, I.G., 2020. The Pauli Exclusion Principle and the Problems of Its Experimental Verification. Symmetry, 12(2), p.320,

but a good rule of thumb is that the wavefunctions describing those fermions must have significant overlap, i.e. they should not be “independent” particles.

This is rarely the case for two fermions that are well separated in space, as their (localized) wavepackets have very small overlap, although it is conceivable that entangled fermions could be well separated and still satisfy the Pauli principle. However, entanglement is “fragile” and so it is usually the case that widely separated particles are taken as distinguishable.

Note that a more sophisticated statement is that the state vector describing $n$ indistinguishable particles must transform - for bosons and fermions respectively - by the trivial or the alternating representations of the symmetric group ${S}_n$.

ZeroTheHero
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