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Don't many Hamiltonians include the 'exchange interaction' for this effect?
Or is my memory fading?
At 2:21 PM -0600 on 1/27/02, Jack Uretsky wrote
Hi all-
I take it that the question is the one that is posed in the
second paragraph.
For the sophisticated reader, degeneracy pressure of a
non-interacing Fermi gas is calculated, e.g., in Chapter 2 of Fetter and
Walecka. Of course the gas is assumed to be at equilibrium, and one might
ask how equilibrium can be reached without interactions. So it is a bit
easier to talk about a weakly interacting Fermi gas.
The pressure arises from the nature of the statistical
distribution (Fermi-Dirac), there is no force term in the Hamiltonian that
gives rise to the pressure. So, depending on what one means by "force",
the answer to the question is "No."
More precisely, as is usual with such innocent-sounding questions,
the calculations that one does are unambiguous. The words used to
describe the calculations, however, can be made as ambiguous as a
sufficiently fertile mind desires. As we lawyers learned in law school,
it is always possible to make an argument on any side of any question
(including the one implied by this sentence). So I suppose that I can
foreclose some lines of argument by answering the question this way:
If, by force, you are asking whether "degeneracy pressure" can
be made to correspond to a force term in a Hamiltonian, then the answer is
"No" as far as anyone has determined to date (in the published literature
as far as I know).
Regards,
Jack
On Sun, 27 Jan 2002, John Mallinckrodt wrote:
A sophisticated layperson friend of mine asked me a question which>
I had to confess had never really occurred to me before. I told
him I'd consult a prestigious panel of friends. So here goes:
Why should the Pauli principle *not* be considered itself to be
the equivalent of a "fifth force"?
Most sources explain the origin of degeneracy pressure along
something like these lines: Fermions cannot occupy the same
quantum state. Thus, as the spatial extent of their wavefunctions
become more and more similar, they are forced to occupy higher
energy levels. This is equivalent to the effects of a repulsive
force.
Fair enough, but the effect is at least *seemingly* independent of
whether or not the particles interact in any other way--i.e., via
the strong, electroweak, or gravitational mechanisms. Now it is
true that all fermions *do* interact by at least one of these
mechanisms, but *is* this a requirement for fermions? If so, why
exactly? And if not, how would we explain the existence of
degeneracy pressure in the case of two noninteracting fermions?
> John Mallinckrodt mailto:ajm@csupomona.edu
> Cal Poly Pomona http://www.csupomona.edu/~ajm
>
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