Re: [Phys-L] zero point motion and E-M emission

[John Denker <jsd@av8n.com>]

*** Background idea #1:

Suppose we have an ordinary classical coin.  The probability of
the H state is equal to the probability of the T state.  The 
probability distribution is symmetrical.  However, if we talk 
about any particular coin (rather than the distribution), that
coin is in only one state.  There is nothing symmetrical about
the state of that particular coin.

There is a verrrry deep principle here, which we can summarize
by saying:
	There is no such thing as a random number.
	If it's random, it's not a number.
	If it's a number, it's not random

Also:
	There is no such thing as a random number.
	If you have a random distribution of numbers,
	the randomness is in the distribution, not in
	any particular number drawn from the distribution.

The same applies to distributions over non-numbers, including
distributions over coins, and distributions over quantum states.

For an introduction to the principles of probability from a
modern (post 1933) viewpoint, see
  http://www.av8n.com/physics/probability-intro.htm

*** Background idea #2:

There is more than one vector basis.  A vector that equal to one 
of the basis vectors in basis "A" will be a linear combination
of multiple basis vectors in basis "B".

This applies directly to quantum mechanics.  A photon that is
polarized in the X state in the XY basis will be in a linear
combination of states in the circular polarization basis.

I mention all this because on 06/11/2013 07:16 AM, Rauber, Joel wrote:
> The ground state of the atom is spherically symmetric and presumably
> has zero dipole moment.  I recall showing that EM radiation requires
> a time varying dipole moment (non-zero 2nd time derivative); which
> implies that a spherically symmetric charge distribution can not
> radiate.

I disagree with that on several points, including the conclusion.

If you use the SPDF spectroscopic basis i.e. spherical harmonics,
then the ground state is the S state.  However, if you use the 
position basis, the ground state is a superposition of many many 
positions.  Infinitely many.  We are talking about a change of
basis in an infinite-dimensional vector space.  This is an 
application of idea #2.

The probabilistic /distribution/ of positions is spherically symmetric;
however, the randomness is in the ensemble, not in any particular atom
drawn from the ensemble.  The electron is still a particle, and if you
measure the position, you will get a position.  You can choose any 
basis you like.  I can choose any basis I like, not necessarily the
same as your basis.  This is an application of idea #1.

Here's another line of reasoning that leads to the same conclusion:
To a very good approximation, there are only two terms in the atomic
Hamiltonian:  The potential energy and the kinetic energy.  To a 
decent approximation, the kinetic energy is given by the non-relativistic
formula:  It's goes like the  square of the velocity.  So ... in order
for atoms to exist at all, the electron needs to have a velocity.  Indeed
it's easy to accurately estimate the RMS velocity.  The calculation fits
on on the back of a post-it note.  Hint:  Virial theorem.

If you're going to have a nonzero velocity and still have the atom
not fly apart, the velocity has to be changing.  This is what we
call an acceleration.  An accelerated charge radiates.

Again:  You can project that atomic state onto any basis you like.
 -- If you measure the state in the spectroscopic basis, you'll
  get a spectroscopic state (S, P, D, F et cetera).
 -- If you measure the position you'll get a position.
 -- If you measure the velocity you'll get a velocity.
 -- If you measure the acceleration you'll get an acceleration.

Among other things, this tells us that the whole idea of "wave-particle
duality" is a misconception.  I try not to use the term at all, except 
on rare occasions, and then only within scare quotes.  Duality implies
two possibilities, but as you can see from the previous paragraph, in 
reality there are more than two possibilities.  Infinitely more.

> The ground state of the atom is spherically symmetric 

-- The S-wave wavefunction is spherically symmetric.
-- The /distribution/ over positions is spherically symmetric.
++ However, he actual position in an actual atom drawn from the
  ensemble is not symmetric.  It is either on the +X side of 
  the nucleus or the -X side of the nucleus, not both.

You can measure the instantaneous position of the electron within an
atom using X-ray scattering, or electron-electron scattering, et cetera.

> and presumably has zero dipole moment.

The dipole moment of any particular atom is nonzero.  The van der Waals
force between atoms depends on this.
 -- The mean dipole moment (averaged over the ensemble) is zero.
 -- The RMS dipole moment (averaged over the same ensemble) is nonzero.

> I recall showing that EM radiation requires a time varying dipole moment 
> (non-zero 2nd time derivative);

Sure.

> which implies that a spherically symmetric charge distribution can not radiate.

Be careful with the word "distribution".
 -- A spherically symmetric /arrangement/ of charge cannot radiate, so
  long as it stays spherically symmetric, and so long as the center of
  the arrangement is not accelerating.
 -- A spherically symmetric /probability distribution/ is something else
  entirely.  It may be that an individual element drawn from the distribution
  has far lower symmetry than the distribution itself.  See idea #1 above.
  A probability distribution is one kind of thing;  an element drawn from
  the distribution is something else entirely.

==========================

I stand by my previous answer.  You can analyze the atom and the EM field
in the position basis.  This is of course not the only possible approach,
or even the conventional approach, but it is entirely doable ... and it is
exceedingly informative.  Details can be found at
  Bernard Yurke and John S. Denker
  "Quantum network theory"
  Phys. Rev. A 29, 1419–1437 (1984)
  http://pra.aps.org/abstract/PRA/v29/i3/p1419_1

The result says that the ground state of the atom is in /dynamic/ equilibrium
with the ground state of the EM field.  This is a nontrivial idea, and it
has observable consequences.  For starters, it allows a unified view of the
low-temperature "quantum" fluctuations and the high-temperature "thermal"
fluctuations /and everything in between/ ... which turn out to be all the
same in some fundamental sense.

Let's suppose, hypothetically and temporarily, that the atom were "always"
illuminated by the "same" zero-point fluctuations of the EM field.  To the 
extent that this is always happening, the theory might seem almost trivial.
The atom is always losing energy to radiation and always regaining it, so 
there is no observable net effect, or so the hypothetical story goes.

HOWEVER, you can replace the ground state of the EM field by a /squeezed state/.
In such a state, the RMS fluctuations along one axis are *less* than what
you would see in the ordinary vacuum ground state.  So, in this subspace, 
the radiation is "colder than cold".  The atom, when illuminated by the 
squeezed state, will get smaller in one direction, because the normal S-state 
would lose more energy than it gains.

This may sound crazy, but the experiment has been done!  Once again, QM
gets the right answer.  It is pretty impressive to look at the RMS power
meter as you switch out the squeezed state and switch in a zero-temperature
cold load.  The zero-temperature state has /more/ RMS power, i.e. /more/
fluctuations.

Some discussion of squeezed states, including pictures, can be found at
  http://www.av8n.com/physics/coherent-states.htm