Re: [Phys-L] collisions

[Moses Fayngold <moshfarlan@yahoo.com>]

   As already mentioned by J. D., in case of identical atoms fired toward each other, they, while being structurally identical, must differ in their internal states, e.g., an H-atom launched from the left must be in an excited state whereas H-atom from the right - in its ground state; or they must differ in their spin state etc. The particle's inner state is a personal ID number serving as a black dot for a classical object when the latter's path is hidden behind an opaque screen. But even with an ID attached to the quantum particle, there are many other factors complicating the game. First of all, it is very hard if not impossible to prevent the atoms from just passing by each other with only very small deflection (large impact parameter). One of the reasons, apart from the small atomic size, is the lateral spread of atomic wave packet. For a collision starting with initial separation 2 m, at a low temperature corresponding to velocity only 1 m/s,
 when the atoms meet, their lateral spread will exceed their initial (Gaussian) packet width by about 13 orders of magnitude. (See, e.g., Sec. 11.4 of our recent book "Quantum Mechanics and Quantum Information", - it already got a good review by  Nick Herbert, known by his "Flash proposal" and as the author of "Quantum Reality" ). So the atoms, no matter how accurately we try to line up their initial velocities, will indeed pass one through the other almost without deflection. We could model this as motion of one atom with reduced mass in the field of the other (approximated by Lennard-Jones potential), by applying the quasi-stationary scattering theory. The advantage is the possibility to calculate the probability amplitude, and it will tell you exactly the probability of deflection of either atom for any angle. But this requires low collision energy to avoid spin or energy state exchanges. And even so, the math will be very heavy. I would suggest
 qualitative prediction that at small scattering angles one could safely gamble expecting the marked atom to appear on the forward side (forward scattering). Of course, in the reference frame attached to their center of mass, the output directions will always be opposite to each other. At 90 degrees there will be 50 - 50 chance to see either atom moving along the given direction in the plane perpendicular to their initial velocities. . 
  A detailed description of the relevant questions can be also found in the Feynman Lectures on Physics, Vol. 3.

Moses Fayngold,
NJIT




On Friday, January 24, 2014 2:37 PM, John Denker <jsd@av8n.com> wrote:
 
On 01/24/2014 09:50 AM, Carl Mungan wrote:
> 2. Let's expand a bit on the oven ideas. I assume one starts with a 
> little oven (or fridge) full of a large number of hydrogen atoms. So 
> the temperature of the gas is well defined. Then I assume the 
> launchers consist of drilling small holes in the oven (opening into
> a vacuum chamber).

OK.  That sort of oven-with-hole has been used as a source in 
atomic-beam experiments for a verrry long time.

Just to clarify my understanding of the question:  Suppose
we replace
 the atoms with photons.  Then there are two
answers:

  Two flashlights   is equivalent to    one flashlight + beam splitter

  Two lasers      is NOT equivalent to   one laser + beam splitter

because the laser might have a significant amount of coherence,
whereas the flashlight does not.

Returning to the original question:  an oven is like a flashlight,
not like a laser.  If you wanted to, you could design a launch
system to produce correlated atoms, but it wouldn't look like an
oven.  It would be vastly more complicated than that.

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

> I can compute lambda1 = h/p where p is a momentum of a
 single atom. I
> can also compute lambda2 = h/sqrt(2*pi*m*k*T) while an atom is in the
> oven, where T is well defined.

Lambda1 is the plain old wavelength.

The thermal de Broglie length (lambda2) does not behave like
a wavelength;  it has more to do with the envelope-size of
the wavepacket.  This is spelled out, with diagrams, at:
  http://www.av8n.com/physics/thermal-wave-packet.htm

See also
  http://www.av8n.com/physics/exchange.htm

There is an unfortunate tradition of calling lambda2 the thermal
de Broglie
 «wavelength» but AFAICT that is a complete misnomer,
the kind of misnomer that leads to deep-seated misconceptions.

Suggestion:  Train yourself -- and your students -- to call it
the thermal de Broglie /length/.  This won't solve all the world's
problems, but it will at least not create new ones.

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