Re: Negative energy particles, Mass Gap & 3rd Law

whopkins_at_csd.uwm.edu
Date: 03/01/05 

Date: Tue, 1 Mar 2005 20:14:50 +0000 (UTC)

Daryl McCullough wrote:
> whopkins@csd.uwm.edu (Mark Hopkins) says...
> >That already seems to be implicit in the remarks; particularly given
> >the tendency for a system at negative temperature to absorb energy
from
> >the environment (here: the universe), which is at positive
temperature.
> >
> >A negative energy fermion absorbs positive energy from the
environment
> >and jumps into a positive energy state. This shows up as the
creation
> >of a positive energy state, and the vacating of a negative energy
state
> >-- pair production.
>
> But in order for the *absence* of a negative energy electron to
> appear as the *presence* of a positively-charged particle, you
> need to assume that all negative energy levels are filled, don't
> you?
>
> Another way to ask the question: what, in your view, is the vacuum
> state?

To followup further on my previous reply: a thermal vacuum would be a
mixed state of the form:
                  P_0 = sum exp(-beta E_n) |n><n|.

Now, there is a general theorem which enables one to treat a mixed
state as being a pure state in a larger Hilbert space. This operation
doesn't come for free, however, since the original algebra of
observables embeds as an operator algebra in the larger space with a
non-trivial center -- i.e., you end up trading mixture for
superselection.

In this case, one thinks of the vacuum, itself (since, as a thermal
state, it's a heat bath) as having its own particle states. So a
one-particle Hamiltonian
                   H = sum E_n |n><n|
gets replicated into a part H+ for the particles and H- for the "holes"
(the vacuum states):
                  H = H+ - H-.

The corresponding state space is then the direct product:
                 {|n+>} x {|n->}.
In this representation, the vacuum is then a superposition of all the
0-energy combinations; i.e., all the |m,n> = |m+> x |n-> which have H
as a 0-eigenvalue (i.e., all the |n,n>). To get the desired
expectation values for operators on the particle space {|n>+} the
corresponding pure state would be:
                   |T> = sum exp(-beta E_n/2) |n,n> / sqrt(Z)
where
                   Z = sum exp(-beta E_n)
                   beta = 1/(kT), k = Boltzmann's constant.
For an operator A in the original space
                     A = sum |m>A^m_n<n|
one has
    <T|A|T> = 1/Z sum <p,p| |m+>A^m_n<n-| |q,q> exp(-beta(E_p+E_q)/2)
            = 1/Z sum <p-|q-> A^p_q exp(-beta(E_p+E_q)/2)
            = 1/Z sum A^p_p exp(-beta E_p)
which is what one is looking for. This also makes it clear the role
that superselection plays in generating the appearance of a mixture.
The |m> and |n> states decohere, in effect, when represented as |m,m>,
and |n,n> because of the action of the coefficient <m-|n->.

This basically is how "thermofield dynamics" works.

For a fermionic or bosonic degree of freedom, you can also begin to see
where the Escher-like swap between foreground and background also
starts to arise. The Hamiltonian, identifying (a, a') and (b, b') as
the operators on the particle and hole spaces, would be:
              H = h-bar omega/2 (a a' - a' a - b b' + b' b)
with states
               |0+>,|1+>,|0->,|1->
that combine to give you the 4 states:
                |V> = |0,0> = vacuum (= vacuum as T -> 0+)
                |P> = |1,1> = plenum (= vacuum as T -> 0-)
                |+> = |1,0> = particle
                |-> = |0,1> = hole
with
  |T> = sqrt(1/Z)
        |V> exp(beta h-bar omega/4) + |P> exp(-beta h-bar omega/4)
and
        Z = exp(beta h-bar omega/2) + exp(-beta h-bar omega/2);
so that
  |T> = (|V> + |P> exp(-beta h-bar omega/2))/sqrt(Z')
with
         Z' = 1 + exp(-beta h-bar omega)

Notice what happened to the zero-point energy!

Defining the thermionic angle Theta by
                 tan(Theta) = exp(-beta h-bar omega/2)
one can also write this, for positive temperatures T > 0, as
            |T> = |V> cos(Theta) + |P> sin(Theta)
For negative temperatures the roles of V and P are swapped.

This shows that the effect of temperature is to rotate the vacuum into
a plenum. It rotates all the way over, as one crosses to negative
temperature through positive infinity (note the resemblance to a spin
system).

But it is discontinuous across T = 0 -- which is, in effect, a
rendition of the 3rd law: no transitions to or through T = 0 are
possible.

The operators, whose actions on the larger state space become:
              (a,a') particle creation and annihilation
            a = |V><+| + |-><P|
            a' = |+><V| + |P><->
and
              (b,b') hole creation and annihilation
            b = |V><-| - |+><P|
            b' = |-><V| - |P><+|
yield two sets of mutually anti-commuting operators (just like for a
Dirac fermion) with
             a a' + a' a = b b' + b' b = 1.

For temperature T, the corresponding vacuum annihilators a(T), b(T)
then become
            a(T) = a cos(Theta) - b' sin(Theta)
            b(T) = b cos(Theta) + a' sin(Theta)
which effects the above-mentioned Escher swap between (anti-particle
create) <-> (hole annihilate).

Bosons work in a similar way, and I won't go through this in too much
detail here except in outline:
     |T> = sum (|n,n> (tanh Theta)^n/cosh(Theta))
     tanh(Theta) = exp(-beta h-bar omega/2).
The rotations now becoming hyperbolic
           a(T) = a cosh(Theta) + b' sinh(Theta)
           b(T) = b cosh(Theta) + a' sinh(Theta).

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