Re: splitting wave energy into KE and PE

[David Bowman <David_Bowman@GEORGETOWNCOLLEGE.EDU>]

Regarding John Denker's comments:

> At 09:31 AM 3/18/00 -0500, David Bowman wrote:
> > [at low temperature] each term in the Hamiltonian is of the form:
> >     h*f*[a^dagger]*[a]
> > (where the quantities in the brackets are creation/destruction operators
>
> I agree that's a perfectly reasonable way to write the Hamiltonian.
> OTOH it is not the only reasonable way.

What is a reasonable representation depends on just which purpose the
Hamiltonian is to be used for.

> Suppose we choose
>         Q = (a^dagger + a)/2
> and call it our canonical coordinate.  The dynamically conjugate variable is
>         P = (a^dagger - a)/2i
> which we can think of as momentum.

True.  In fact, the canonical quantization procedure for the EM
Hamiltonian arrives at the representation in terms of field creation and
destruction operators *via* its classical representation in terms of
canonical coordinates and their conjugate momenta.  The system is
quantized by imposing the canonical commutation relations between these
coordinates and momenta.  But such a representation is not very useful
for doing statistical mechanics because it is not diagonal.  The needed
traces are best found in a diagonal representation (i.e. in terms of the
appropriate photon number operators that are made of the [a^dagger]*[a]
products).  Summing over the eigenvalues of these operators involves just
summing over the nonnegative integers (boson occupation numbers).  This
representation does not have a potential/kinetic energy split since
neither the kinetic energy nor the potential energy are compatible with
(commute with) the photon number operators so that states of well-defined
energy and photon numbers do not have well-defined separate kinetic nor
potential energy.

> Notes:
>  1a) Physically, for electromagnetic waves, Q is the electric field
> strength in some appropriate units, but formally we don't even need to know
> that.

I suppose that a canonical transformation of the dynamical variables
could be made so that the electric field amplitude would correspond
such dynamical coordinate variables.  But, typically, we would consider
the tranverse part of the electric field as the dynamical *momenta*
because it is made from the time derivative of the vector potential A
(where the various Fourier amplitudes for A make the usual dynamical
coordinates) and it is obtained by differentiating the Lagrangian density
w.r.t the time derivative of A (or its Fourier amplitudes).

>  1b) For transverse waves on a slinky, Q is the displacement;  the
> equations are formally identical.

I'd agree with 'similar' or 'analogous' rather than 'identical'.

>  2a) Physically, for EM waves, P is essentially the magnetic field
> strength, but formally we don't need to know that.

Since B = curl(A) we have in Fourier (wave number) q space:
|B(q)|^2 = |q(cross)A(q)|^2 = (q^2)*|A(q)|^2 (since q is perpendicular
to the transverse A(q) potential amplitude).  Thus the |B|^2 term in
the Hamiltonian generates the *potential* energy term and the magnetic
field strength amplitude is essentially the wave number q times the
canonical coordinate A(q).

>  2b) For transverse waves on a slinky, P is the transverse momentum.
>  **) In any case, all we really need to know is that P is dynamically
> conjugate to Q.  When in doubt, look at the Lagrangian.  It knows.

Yes, it does.

> We can write the Hamitonian as
>         H = [sum over modes] hbar omega (a^dagger a + 1/2)
> or equivalently as
>         H = [sum over modes] hbar omega (Q^2 + P^2)
>
> (Just grind out the expansion for P and Q to show the equivalence.)

True, the canonical representation is equivalent, but it is not as
useful as the number operator representation, and it is somewhat
backwards since the diagonal [a^dagger]*[a] representation comes from
the canonical Q^2 + P^2 representation which is not diagonal in either
a Q nor P basis, and these two terms do not mutually commute (i.e. have
simultaneous values) in *any* basis.

> Now ISTM that one could quite reasonably call the Q^2 piece a potential
     ^^^^
What does this abbrev. mean?  Is it "I suppose that means"?

> energy, and the P^2 piece a kinetic energy.  Averaged over a full cycle
> these two terms are equal in magnitude.

These terms do not even have simultaneous values at all, let alone
equal such values.  However, in a state of definite photon occupation
numbers then the *expectation* of each of these terms is equal by the
virial theorem.

> According to this way of looking
> at things, one could say the field energy is half potential and half kinetic.

Or, more properly, the expectation of the potential energy is half of the
total energy and the expectation of the kinetic energy is also half of
the total energy.

> (Note the validity of this viewpoint does not depend on any assumptions
> about temperature;  the aforementioned average does not need to be a
> thermal average.)

The reason that we were discussing the stat mech of the EM field is
because the discussion topic concerned the partition of *thermal energy*
into potential and kinetic energy in the cases of *dissipation* and
*latent heat*.  Also, you wrote:

> To generalize the idea...  One must decide what to say about the common and
> important case of _radiative_ cooling.  Is photon energy kinetic or
> potential or both or neither? ^^^^^^^

Thus, the topic being discussed concerned thermal issues.  But you are
correct because the virial theorem will always guarantee equality in the
expectations of the the kinetic and potential energies--even for a
nonthermal density matrix--as long as the density matrix commutes with
the Hamiltonian.

> David also wrote:
> > E = |p|*c.... by this way of looking at things, all the photon
> > energy is kinetic.
>
> I don't find this argument very convincing.

You don't?  Remember, in this section I was discussing the assembly of
photons as a classical assembly of massless particles that did not
interact with each other.  Noninteracting classical massless particles
(such as photons) must have only kinetic energy since they don't have
any rest energy and their lack of interaction prevents any potential
energy from arising.  All that's left is just their kinetic energy.

> One could make an analogous
> statement about an ordinary mass-on-a-spring oscillator:

I don't think so.  A mass on a spring is not a classical massless
particle.

> The mean energy
> is proportional to the mean-square displacement, so by this way of looking
> at things, all the oscillator energy is potential.

Huh?  I don't follow this apparent non sequitur.

>  The fallacy is that the
> average PE is proportional to the average KE, so an expression for one can
> always be dressed up to look like an expression for the other.  (And to
> complete the analogy, I think E=|p|c only holds when averaged over a cycle.)

No, E =|p|*c holds in general for any classical massless particle which
doesn't interact with anything else. (Remember, for now we are keeping
any charged sources which could interact with the photons strictly
segregated from the region where the photons are propagating.  We banish
any such interactions to the exterior of that region).  Classically
moving particles--even ultrarelativistic ones--don't even have a 'cycle'
to average over.

> ===========
> Bottom line:
>
> 1) AFAIK, it is always safe to think of the wave energy as half kinetic and
> half potential.  If anybody has an example of a situation where this leads
> to trouble, please explain.

This is the case where a wave equation is linear and dispersionless
(with the group velocity and phase velocity for every wave number both
being equal to a single *constant*).  Of course, if we treat the photons
as classical massless particles then any underlying wave equation, whose
quanta of excitation are the photons, is not part of our particle model.

> 2) In some situations the PE/KE split doesn't matter, in which case the
> question of how to do the split need not be asked and need not be answered.

True.  It is also possible to perform a canonical transformation in phase
space which rotates (by 90 deg) a system of SHOs' dynamical coordinates
and momenta in such a way that the old momenta become the new coordinates
and the old coordinates become the negatives of the new momenta.  After
performing such a transformation then the kinetic and potential energy
terms in the Hamiltonian switch their identities.  So just what we call
the generalized kinetic energy and the generalized potential energy for
a system of SHOs is to some extent a subjective matter of definition.

David Bowman
David_Bowman@georgetowncollege.edu