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Re: Accessible States

At 02:14 AM 5/12/99 -0600, Jim Green wrote:

Consider a box of volume V which contains N atoms of an ideal gas which
have a total energy E.

OK.

You didn't say that the gas was in thermal equilibrium but I'll assume
that's what you had in mind.

On the other hand, since you are doing a quantum-mechanical calculation, I
will *not* assume that the gas is classical. That is, we will consider the
possibility that the gas is cold enough and dense enough that
identical-particle effects must be taken into account. Examples include
spin-aligned hydrogen, and the degenerate electron gas in a metal.

You didn't say the gas was pure, but for simplicity I'll assume that's what
you meant (as opposed to, say, a mixture of hydrogen isotopes).

Now I start to worry: Assume that each atom is contained in one of a
number of 6-dimensional cells of

V(6) = DxDyDzDpxDpyDpz = h^3 (eq 1)

OK. That's not the only way to go, but it's OK. The choice of cell size
(h) has consequences which will be discussed below.

The total number of states in the box for each atom is

Omega(1) = (total 6-d volume)/cell volume
= V(real)V(momentum)/DxDyDzDpxDpyDpz
= VrVp/h^3 (eq 2)

OK.

Now Vp = (4/3)pi x p(max)^3 & p(max) = sqrt(2mE) (eq 3)

This introduces another approximation. A thermal distribution doesn't have
a "max" p. Perhaps "typical" p would be a better concept.

So Omega(1) = (4/3)pi(2m)^(3/2) x V E^(3/2) (eq 4)

Now Omega(N) =Prod[Omega(i)] (eq 5)

This assumes that we can distinguish which cell the atom is in. That is
true, because we chose just the right cell size back in equation (1). See
also below.

So Omega(N) = Const x V^N x E^(3N/2) Cf big Rief Eqn 2.5.19

The result seems OK (assuming I typed it correctly), but I have problems:

1) Is it ok to say that the atom fills a Heisenberg volume? ie that the
possible 6-d positions of the atoms are this size?

Interesting question. Heisenberg tells us the atom must fill an area in
phase space (delta P delta X) greater than roughly hbar/2. But why should
we assume it is on the order of hbar/2, and not vastly larger?

Well, for a nonthermal situation, it could perhaps be vastly larger. But
for this thermal situation, it's not. In fact, for a gas that's almost
classical (not *too* degenerate) we know delta P and delta X separately:
delta Px is on the order of sqrt(m k T) (eq A)
delta X is on the order of hbar / sqrt (m k T) (eq B)

I calculated equation (B) using ideas from Feynman and Hibbs, _Quantum
Mechanics and Path Integrals_.

The basic ideas are as follows:
1) Quantum mechanical exchange is a *physical* process. It is not a
mathematical magic wand that we wave over the equations. If there is a
physical barrier, such as a potential energy barrier, the exchange will not
occur.
2) In a thermal situation, the particle has a time hbar / kT to complete
the exchange. That's the rule.

Now as a consequence of that rule, we have in this case a *kinetic* energy
barrier. The temperature limits the available kinetic energy, which in
turn limits the velocity, so the atom simply cannot get very far in the
time hbar / kT. Atoms separated more widely than this are distinguishable
by their position. Atoms separated less widely than this are busy
exchanging with each other, so they are indistinguishable (unless they are
distinguishable by isotopic identity, spin state, or something like that).

So the issue here is not just accessible states, but accessible
*distinguishable* states.

-----------------

Pedagogical points:

1) Feynman&Hibbs is *not* suitable for high-school use. I know folks who
have Nobel prizes who are terrified of that book.

2) On the other hand, the idea that exchange is a *physical* process is
simple and beautiful.

3) Similarly, the idea that under some (indeed many) conditions, particles
are distinguishable by their position is simple and beautiful.

4) Bohr taught us the importance of the correspondence limit. We believe
the quantum-mechanical description is correct, and we believe our every-day
classical observations are correct, so we had darn well better be able to
cough up the classical result as a consequence of, not an exception to, the
quantum description. The notion that atoms are distinguishable by their
position unless they get too cold and/or too close together gives a nice
physical picture of the gentle crossover from classical to fully quantum
behavior.

Cheers --- jsd