Re: Entropy, Objectivity, and Timescales

[David Bowman <dbowman@tiger.gtc.georgetown.ky.us>]

Dan wrote:

> > What James means is that we express our microstate as a combination of
> > single system states (e.g. particles in a box of gas) and allow them to
> > interact. The propagation of the interaction then proceeds at some
> > finite speed near sound speed.
>
> So instead of a basis of exact stationary states for the whole system, 
> you'd prefer to use a basis of products of single-particle states.
> To me it's not at all obvious what happens in this basis, that is,
> whether all 10^(10^23) possible states will eventually get mixed into
> the wavefunction (according to the Schrodinger equation).  That's why
> I chose to work in a basis where the answer is obvious.

I'm not exactly sure what Leigh's interpretation of James above means, but 
if Dan's interpretation is correct about using a basis of (I assume properly
symmetrized or antisymmetrized to properly account for quantum statistics)
products of single-particle states then the results obtained will be the
same as in any other basis, including Dan's.  Every orthonormal basis in
Hilbert space is connected to any other one by a unitary operator and all
expectations of all quantum physical observables are invariant under
unitary transformations.  The Schroedinger equation will not allow the
system's microstate to develop a nonzero incoherent probability for all
10^(10^23) microstates if there were not this many microstates with a
nonzero incoherent probability to begin with.  This is a consequence of
the Schroedinger evolution operator being unitary, and the result is true
in *any* basis.  To the (unrealistic) extent that the system can be
described as a wave function (i.e. pure state), and to the (unrealistic)
extent that the Schroedinger equation actually does describe the system's
microscopic dynamics, then to that extent the system has exactly 1 possible
(micro)state at all times.

> Do you know a way to prove that all 10^(10^23) states *will* get
> mixed into the final wavefunction, using your basis?  This isn't 
> obvious to me at all.  And even if it's true, can you tell me why 
> your basis is better than mine?  (If entropy depends on our choice
> of basis, doesn't that make it subjective?)

It's not true, and Leigh's basis is not intrinsically any "better" or
"worse" than Dan's or anyone else's.  One basis may be more convenient than
another for doing some calculation, but it is not any more or less correct.
Also, the entropy (w.r.t. the incoherent probabilities of stat. mech, but
not w.r.t. the coherent probabilities of quantum measurements) does *not*
depend on our choice of basis.  The most convenient basis to evaluate the
entropy in is one for which the density matrix is diagonal.

David Bowman
dbowman@gtc.georgetown.ky.us