Re: QUANTA

["John S. Denker" <jsd@MONMOUTH.COM>]

Hi Folks --

My previous posting was unclear, incomplete, and/or wrong (depending on how
generous you want to be).  Thanks to Paul Johnson for pointing this out.
Let me try again.

At 09:12 AM 3/7/99 -0500, David Abineri wrote:
> Does one look any differently at an EM wave generated by the falling of
> an electron from one energy level to another OR the EM wave generated in
> a radio antenna which, as I understand it, is generated by a continuous
> oscillation of electrons in the antenna?

They're not categorically different.

> Does the notion of a quantum of energy apply in both cases? If so, how
> does one think of it in the latter case?

The same laws apply in both cases, although simplifications that apply in
the atomic case don't apply in the radio case, and vice versa.

Consider the modes of a high-Q resonant cavity.  The different modes will
have different frequencies.  Because of the high Q, the frequency of each
mode will be sharp, definite, discrete.  This discreteness is sometimes
called "first quantization" and is completely classical.  You don't need to
know anything about hbar to understand it.

So let's see to what extent atoms are analogous to RF cavities.
  1) Atoms are really small.  They are too small to couple well to the
Maxwell field.  Therefore they are *always* high-Q oscillators.  A Q of 10^8
is typical.  (The same cannot be said for dye molecules, which are large
enough to couple rather better.)  In any case, if we match the Qs, the
RF/atomic analogy is pretty good so far.
  2) The atomic orbitals have discrete energy levels.  This is reasonably
analogous to the discrete modes in the RF resonant cavity.  In the atomic
case, the spacing between levels depends on hbar, but in some sense that's
because the size of the atom is determined by hbar.
  3) Atomic frequencies are high compared to kT/h (remember k/h is 20GHz
per Kelvin).  Therefore it is relatively easy to prepare atoms in
particular states.  In contrast, radio-frequency oscillators are typically
in a thermal mixture of states.  So the RF/atomic analogy breaks down
somewhat at this point.

We are still stuck on first quantization.  We have spoken only about what
frequencies of photons will be emitted.  We have not said anything about
how *many* photons will be emitted.

So let's move on to second quantization.

Go back to the RF cavity.  Pick one of the modes, and ask what is the
energy of the standing wave in that mode.  You will find that this energy
is quantized.   You can put one photon into that mode, or hundreds, or
billions.  This counting of photons in a given mode (second quantization)
is in some sense "perpendicular" to the enumeration of discrete modes
(first quantization).

So this leads us to additional points of comparison between atoms and RF
cavities:
  4) Atoms are typically quite anharmonic.  If you pump one photon into an
atom, you typically can't put in another one of the same frequency.  In
contrast, the cavity is remarkably linear -- the frequency of the mode is
unaffected by the number of photons in it (up to the point where the field
gets very intense and the electromagnetic stresses start warping the walls
of the cavity).

  5) This is to some extent a combination of items (3) and (4):  By the
time an RF cavity mode has an energy content that is easy to measure, it
contains a huuuuge number of photons.  In contrast, single photons at
optical frequencies are rather easy to count.

  6) In RF work, it is traditional to speak of voltage and not photon
number.  In atomic work, it is traditional to speak of photons and not
voltage.  But these are mere traditions and/or conveniences;  the
non-traditional choices give perfectly correct descriptions.  Indeed they
sometimes shed very interesting new light on old problems.

In particular, a lot of people know how to formulate the quantum-mechanical
operator that counts photons.  Far fewer people know what the
quantum-mechanical voltage operator looks like.  But it exists, and there
are things you can do with it that you can't do with the number operator.

Cheers --- jsd