Re: ideal Bose gas

[John Denker <jsd@MONMOUTH.COM>]

At 03:17 PM 5/12/99 -0400, Samuel Held wrote:
>        I thought that part of the definition of an ideal gas was one
> that is free of inter particle interactions.

That's basically right, but see below.

> I would think that
> idenitical particle interactions would cause the presence of a potential
> (binding energy) that would change the state of the system.

Yes and no.

 (a) Strictly speaking, the identical-particle effects do not cause a
potential energy.  Potential energy is a contribution to the Hamiltonian.
Exchange is something that happens as a *consequence* of various things
including the Hamiltonian.
 (b) Metaphorically speaking, the identical-particle effects can be
modelled by a "molecular field" term artificially added to the Hamiltonian.

In particular, consider a bottle of spin-aligned hydrogen at (say)
atmospheric density.  I think we would all consider that a reasonably ideal
gas at room temperature.  Now cool the gas to a few milliKelvin,
maintaining constant density.  The potential-energy terms in the
Hamiltonian remain just as negligible as they were at room temperature.
The kinetic-energy terms retain the same form: p^2/2m.

At these temperatures, the atoms are the size of bacteria.  They are not
little hard balls that bounce off each other;  they are little fluffy
clouds that diffract through each other.

Now if you calculate the scattering of a particular atom, you will find
that the scattering rates do depend on the presence of the other atoms.

 (a) Strictly speaking, this should be calculated using the ideal-gas
Hamiltonian, and the identical-particle effects emerge from the algebra of
the creation operators.

 (b) Metaphorically speaking, once you know the right answer (from the
strict calculation) you can toss a molecular-field term into the
Hamiltonian and forget about creation operators.  You can pretend the atom
you are watching is a classical atom precessing around the molecular field.

OK?

-- jsd