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Do electrons in an electron beam (cathode ray) follow the Pauli exclusion principle? Or in other words, does the Pauli exclusion principle apply for the beam of electrons?

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  • $\begingroup$ Please note that in a beam the electrons are not bound and are spatially separated $\endgroup$
    – anna v
    Apr 16, 2015 at 15:09
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    $\begingroup$ Contemplate this as well: for a 1nA, 20keV electron beam, the electrons are more than a centimeter apart, on average. $\endgroup$
    – Jon Custer
    Apr 16, 2015 at 15:23
  • $\begingroup$ @JonCuster That is what I thought, but by checking learned differently. Please have a look at my answer where a whole new world of electron beams is opened in the link $\endgroup$
    – anna v
    Apr 17, 2015 at 6:32

3 Answers 3

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A cathode ray tube where the electrons are randomly ejected and are not constrained in space the Pauli exclusion seems irrelevant..

In structured beams though, things may be different:

The Pauli Exclusion Principle places a fundamental limit on the brightness of an electron beam. Developing a cathode which can reach this limit is useful for achieving maximum operation in current applications of electron beams, but also opens new areas of physics to be explored. When the phase space of the electron beam is filled to the maximum density, the electrons will experience a degeneracy
pressure, similar to that which keeps a neutron star from collapsing. One promising source or a quantum degenerate beam is field emission from adsorbates on carbon nanotubes. Adsorbates have been shown to provide several orders of magnitude enhancement to emission brightness, which approaches the degeneracy limit. We have developed experiments to test various adsorbates, in order to find those which bind tightest and provide the largest enhancement in brightness. Continuing work to discover better adsorbates should soon allow for the generation of a quantum degenerate electron beam .

So it seems to be an aim, to reach the degeneracy limit given by the Pauli exclusion principle.

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If by the exclusion principle you mean that the total wave function is anti-symmetric under particle exchange, then yes.

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Yes, it must do.

The available number of quantum states per unit volume for an electron is given by $$ g(p)\ dp = \frac{8 \pi p^2}{h^3}\ dp\ ,$$ where $p$ is the particle momentum.

If we have a beam of particles with a spread of momentum $\Delta p$ around a mean value $p$, then the maximum number density that those particles can have is $$ n_{\rm max} \simeq \frac{8\pi p^2}{h^3}\ \Delta p$$

To increase the number density of particles you must either allow them to have a broader spread of momentum or increase the average momentum.

The limit on number density will in turn lead to a limit on the current density that can be carried by the beam. If for simplicity we consider non-relativistic electrons with $p \ll m_e c$, then $J = nep/m_e$ and $$J_{\rm max} = \frac{16 \pi e m_e}{h^3} E\ \Delta E,$$ where $E$ is the electron kinetic energy.

For example, imagine a 100 keV electron beam, collimated so that all the electrons have energies within 1 eV. The maximum current density would be about $6 \times 10^{19}$ A/m$^2$ and the electron number density would be $ \sim 10^{30}$ m$^{-3}$ !

All this of course ignores the vast Coulomb repulsion you would have to overcome in assembling such a beam.

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