Re: [Phys-L] zero-point motion at the introductory level

[Moses Fayngold <moshfarlan@yahoo.com>]

  In the wake of illuminating discussion on Feb. 18, I want to discuss the following example illustrating, in my view, the subtleties of the zero-point motion. I would call this example "There is no motion in de Broglie's wave". 
  Consider the celebrated QM state - de Broglie's wave  A exp(kx-(omega)t). It does not describe any motion in the classical sense.  To be specific, suppose that vector k is pointing from Earth to Andromeda Nebulae. Does not it indicate the particle's motion toward Andromeda? As far as we stick with the initial state with definite k, the answer is no. Suppose we perform a position measurement on a pure ensemble of such particles at two different moments of time. The second measurement will not (and cannot) show the average x closer to Andromeda than the first one, for a very simple reason: already the first measurement will show uniform distribution over the whole space with no definite position of the average. The second measurement will only show the same outcome. Formally - it is just an illustration of the well-known statement: "Definite momentum - indefinite position". Once an instant position is indefinite, the notion of velocity loses its
 meaning. The other way to see it is to notice that due to uniformity of Minkowski space, any displacement in x will only cause the phase shift of the initial state, but this state is invariant with respect to phase transformation. In this respect, QM separated momentum from velocity. 
  There arises a very reasonable question: what is then the meaning of "momentum"? My answer is: Direction of k is a potentiality - it shows the potential direction of motion of a wave packet if we add to the same particle at least one new state with the same direction (but different magnitude) of k. That will take the initial stationary state to superposition with different (omega), thus bringing in the motion into the system, with the suitable name - group velocity v=(delta omega)/(delta) t. This is a directly measurable observable.  
  Classically, momentum is a specific characteristic of directed motion. But according to QM, their actual relationship is much more subtle. Specifically, definite non-zero momentum implies no motion, and actual motion implies some indeterminacy of momentum. Our classical concept of particle's motion originates from dealing with wave packets (albeit we had not known this before), and it takes a highly idealized model of de Broglie's wave to see the limits where we cannot extend this concept, or must do it very carefully.

Moses Fayngold,
NJIT
 





On Tuesday, February 18, 2014 4:55 PM, John Denker <jsd@av8n.com> wrote:
 
On 02/18/2014 01:06 PM, Chuck Britton wrote:

> As Far as I Know - SuperFluid flows as well as Superconducting
> currents would continue at Absolute Zero.

They would.

> I am aware of no evidence to the contrary - but the actual experiment
> has not been performed.

Actually, relevant experiments have been performed.  People
have set up persistent currents in a superconductor and
observed no decay over a period of years.  Ditto for the flow
of superfluid helium in a tube, flowing around and around
in a closed loop.  All evidence is that the flow is exceedingly
stable, and only gets more stable as the temperature goes down.

> But perhaps these don’t constitute ‘motion’.

Of course they constitute motion.

OTOH these examples do not demonstrate zero-point motion.
They are not thermal equilibrium states, so AFAICT they are 
not entirely responsive to the original question that was 
asked.  They are good answers to a slightly different question.

I think the original question had more to do with what thermal
/equilibrium/ looks like at low temperatures, where the classical
approximation is no longer valid.

On 02/18/2014 11:25 AM, Bill Nettles wrote:

> I don't know what absolute zero for a neutron star would mean, but
> they are fermions, so that has to provide a clue.

The neutron star is already cold enough to be in the low-temperature
regime, the non-classical regime, the degenerate regime.  Ditto 
for the conduction electrons in a metal, and the electrons within 
a single atom at ordinary temperatures.

The relevant criterion is the actual temperature divided by the
Fermi temperature.  For metals, the Fermi temperature is on the
order of thousands of degrees, so the conduction electrons are
highly degenerate even at room temperature.  Changing the temperature
10% either way does not change the pressure of the electron gas,
because all you are seeing is the zero-point fluctuations.

For a neutron star, the Fermi temperature is something like 10^13
kelvin, so even though the actual temperature is thousands of
degrees, the neutrons are verrrry highly degenerate.  A few
thousand degrees is essentially absolute zero, on the scale of
things.

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