David Bowman and I have been slowly converging on agreement about what to
say about the KE and PE of a wave.
Let me see if I can get it more-or-less right this time:
1) Each mode of the electromagnetic field is a harmonic oscillator. The
following statements apply to harmonic oscillators in general. They also
apply to other fields such as small-amplitude waves on a slinky. For
anharmonic oscillators (including large-amplitude waves on a slinky) the
story is different and beyond the scope of this discussion.
2) The quantum analysis is correct in all cases. As a subset of those
cases, sometimes the classical analysis is also correct, i.e. an excellent
and convenient approximation.
3) In the classical limit, for mechanical oscillators, there are standard
definitions of PE (which depends on the "coordinate" variable) and KE
(which depends on the "momentum" variable). On the other hand, for
non-mechanical oscillators we can choose the coordinate arbitrarily. For
an RC oscillator, we could take charge to be the coordinate and flux to be
the momentum, or vice versa.
No matter what coordinates we choose, for any harmonic oscillator, in the
classical limit it is clear that KE averaged over a cycle is equal to PE
averaged over a cycle. We conclude that (averaged over a cycle) half the
energy is kinetic and half is potential. (Proof: whatever is KE now will
be PE a quarter-cycle later, and vice versa.)
4) In the general case, we know how to write the quantum operators for
position Q and momentum P. We can write the Hamiltonian as
H = [sum over modes] hbar omega (Q^2 + P^2)
But in general we cannot simultaneously measure both P and Q. We cannot
average either of them over a cycle, because we cannot measure either of
them at more than one point in the cycle. Suppose we make an accurate
measurement of P at some phase which (without loss of generality) we call
the zero-degree phase. Then we can measure Q at the 90 degree phase and
get the same number; we can measure P again at the 180 degree phase and
get the same number; and we can measure Q at the 270 degree phase and get
the same number. But there is only one measurable number, and any
measurement thereof scrambles the dynamically conjugate variable, rendering
meaningless any subsequent attempt to measure P or Q except at the four
special phases just mentioned.
5) On the other hand, I still contend that the KE and PE of the oscillator
are equal in the following sense: Let's use a gambler's estimate. We will
set up a gambling game and sell two kinds of tickets, one of which pays off
in proportion to a measurement of P^2 at a random phase, and the other of
which pays off in proportion to a measurement of Q^2 at a random
phase. You buy a ticket, and then we perform the appropriate measurement
on a photon drawn IID from some ensemble. So the question is, would you
pay more for a P^2 ticket or a Q^2 ticket? I claim that a rational gambler
would pay the same. If I can find somebody who bets otherwise, I can
systematically win money from him.
(This is quite a bit more subtle and cautious than my previous
contention; I thank David for pointing out the bugs in the previous versions.)
6) Note the classical limit requires the energy to be large compared to
hbar omega for each mode of interest; that is, the occupation number must
be large compared to unity. For blackbody radiation in 3D, this is
generally not a good assumption, because the peak of the Planck spectrum is
made up of a large number of sparsely-occupied modes. On the other hand,
there are lots of non-black bodies in the world, such as macroscopic damped
harmonic oscillators at ordinary frequencies and temperatures. The
classical description works fine for such things. There are also lots of
non-thermal distributions, such as the 10,000 watts of electromagnetism
that comes from the local radio station at 880 kHz. You would have no
trouble measuring the P^2 and the Q^2 of that.
7) I renew my assertion that "E=pc" is not an effective
counterargument. For one thing, although E=pc is valid if you take a
suitable average, and in certain other special cases, it is not valid in
the general unaveraged case. (Proof: just superpose a DC electric
field; you get big E in one place and big p in another place.) And if we
do average over a cycle, <E>=<p>c is perfectly consistent with everything
said above.
8) Why does any of this matter? If all you care about is the Hamiltonian,
I'm not sure it does matter very much. The Hamiltonian only cares about
the sum of KE+PE, and you could arbitrarily rename part of the KE to be PE
and it wouldn't affect the sum.
But remember the Lagrangian! The Lagrangian is where the action is
(literally, if you'll pardon the pun). When in doubt, run to the safety of
the Lagrangian. It knows things the Hamiltonian doesn't. The Lagrangian
density is relativistically invariant, which the Hamiltonian obviously is
not. As mentioned above, you can arbitrarily choose a generalized
coordinate; having done so, the only way you can identify the dynamically
conjugate momentum is to ask the Lagrangian. That means that given only
the Lagrangian, you can construct the Hamiltonian, but not vice versa.
The Lagrangian most definitely cares about the difference between KE and
PE. So ISTM the question was worth asking.