Re: [Phys-L] foundations of QM : fluctuations

[John Denker <jsd@av8n.com>]

On 02/18/2014 10:30 AM, David Bowman wrote:
> The physics only says the average over a large ensemble of squared 
> momentum measurements for an ensemble of identically prepared atomic
>  systems is positive.

If /any/ measurement of the velocity /ever/ comes up 
nonzero, for any member of the ensemble, then it proves 
that the velocity is not always zero.

The experiment has been done.  The velocity is not 
always zero.  (In fact it is never zero, but that's
harder to prove.)

The thing has velocity.  It has momentum.  It has kinetic
energy.  Calling this 'not motion' is ludicrous.

If you average over a million nonzero observations and 
get zero, that's just willful blindness.  That's throwing
away a million valid proofs to obtain one irrelevant 
number that proves nothing.

> But *nothing* about that distribution nor the distribution of the 
> multiple possible values of outcomes of position (or any other 
> observable property) measurements depends on time.

A single measurement *can* tell me something about the
time-dependence of the velocity, in at least two ways:
  a) Since we know even before we do the measurement that
   the /average/ velocity must be zero, even one non-zero
   measurement of the actual velocity proves that the 
   velocity must be changing.
  b) We can measure the acceleration operator directly.
   Even one nonzero measurement on /any/ element of the
   ensemble proves that the acceleration is not always
   zero.

The experiment has been done.  There is a nonzero acceleration.
The velocity is changing.

> The length of time is not relevant in this situation.

Actually it is relevant.  Given that the electron has a
nonzero velocity, if it maintained that velocity for a long
time, it would fly out of the atom ... which is not what
happens.  This is one of the ways that we know that the
velocity must be fluctuating, even if we haven't directly
measured the acceleration.

> I don't see how you can measure inbound or outbound radiation without
> measuring at least one quantum of it.

Just because you don't see how to do it doesn't mean it
can't be done.

In fact, the experiment /has/ been done.  Reference below.

Hints:
  a) Use a voltmeter, not a photon counter.
  b) Use an AC voltmeter at a nice high frequency, microwave
   or above, so that quantum fluctuations are relatively large.
  c) Use a nice low temperature so that thermal fluctuations
   are relatively less of a problem.
   Remember: Planck's constant h equals 21 gigahertz per kelvin.
  d) Use a reasonably large bandwidth.

In more detail:  In the language of second-quantized creation
and annihilation operators, a voltmeter measures 
    V  =  a† + a
To a first rough approximation, the photon counter measures
the long-time average of V^2/2 ... which means that for small 
V, the photon counter is vastly less sensitive.  If the voltage
is small, the photon number is small squared.  This begins to 
explain why it doesn't see the small fluctuations.

But wait, it's even worse than that.  The long-time average
of V^2/2 is (a†a + aa†)/2 which is equal to (a†a + a†a + 1)/2
which is (a†a + 1/2).  I warned you that this is only a first
rough approximation to what the photon counter measures.  In
fact, it measures only a†a ... not (a†a + 1/2).  So if you
want to know how much energy there is, you would either need
to use a voltmeter directly ... or count photons and then 
add a half.  In particular, you add a half ℏω per mode, and
if you make a broadband measurement, summing over many modes, 
you see a /lot/ of fluctuations.

An early reference:
  Yurke et al.
  "Observation of 4.2-K equilibrium-noise squeezing via a Josephson-parametric amplifier"
  Phys. Rev. Lett. 60, 764–767 (1988) 
  http://prl.aps.org/abstract/PRL/v60/i9/p764_1