Re: Upended quantum physics in the news

anon1@xxxxxxx wrote:
Two electrons that are in _different_ quantum states become bound by
an attractive interaction to form a Cooper pair. ... The attractive
interaction is caused by exchange of phonons in the solid state
lattice, phonons being quantized lattice vibrations.

So you use the lattice structure to compute the energy distribution of
the phonons and from that you compute the depth of the attractive well
caused by them, and then you measure the temperature to learn how high
above the bottom of the well the electrons usually get jostled, and if
that's well below the edge of the well then the capture halflife of any
pair is long enough to treat as a bound pair for long-distance effects?


Let me just answer your easier questions for now. I'll
come back to the others later, or maybe someone else
will have the time to chime in

What you said above is a good summary of part of what
one would like to do in principle: the real calculations are a
hell of a lot harder, because to get the phonons you need
to know about the band (electronic) structure and the two
are coupled to each other not independent. So in general
people don't work from first principles, but use many inputs
to find the parameters in a simplified theory of the bound
states.

But basically you have the right idea, if the temperature goes
too high relative to the strength of the pairing interaction, which
remember needs to exist only for _some_ finite subset of
all the electrons in the solid, then the pairs are jostled apart
and the material is no longer superconducting. This defines
the transition temperature and is the reason why
superconductivity is a low temperature phenomenon
in general. That's the theory in conventional superconductors


The nucleus is bound by the attractive strong interactions between
the nucleons.

Yes, once you explained the phonon-bound system for electron pairs in a
solid, then of course with a nucleus you don't even need the solid, the
nucleus is already bound from the start. So again it's only long range
interations where an integral-spin nucleus can be treated as a single
(pseudo-)boson. And the whole atom if the number of electrons is also
even, such as neutral Helium-4, or single-positive-charged Deuterium
except it's hard to get them to keep their charge when you're packing
them into a Bose-Einstein condensate etc.


Yup. The repulsive forces would be too great for ionized deuterium
at any imaginable density.

And why don't the electrons in an atom form Cooper pairs and collapse
into a single quantum state??
The interactions between electrons in atoms are the purely
repulsive Coulomb interactions so that is not possible, and
in any case the overall wavefunction for the electrons must
satisfy the Pauli principle.

At this point, what I understand is that (1) there is no solid lattice
at the scale *within* a single atom, within the electron shells, it's
just empty space, so there are no phonons to allow bound pairs, and

Yes, there's just empty space in atoms.

The interaction is carried by photons here, and by far the dominant
part is just the basic repulsive force due to electrons all having
negative charge. Single photon exchange can happen
but it's a higher order field theoretic correction to atomic structure
and it's much weaker.

So you just don't get bound states of electrons with each other
separately in atoms. They are all bound to the positively charged
nucleus, and they all infuence each other mostly by their charges,
but of course, also by their spins. Atomic structure itself is
a whole complicated area of theory.


(2)
pairs form effective bosons only for long range interactions, so even
if a Cooper pair of electrons somehow formed inside a single atom
(maybe since neutrinos have rest mass, there could be a crystal of
motionless neutrinos packed closer together than atomic dimensions,
with a single ordinary atom floating in the middle of it, with
neutrino-phonons all over the place), still the Cooper pair wouldn't
have any local effect, so it wouldn't allow collapse of several
virtual-bosons *inside* that one atom.


Neutrinos are so weakly interacting, that it wouldn't even matter
to the structure of the atom if there did happen to be a bunch
of them around!

What happens in an atom is that all of the possible quantum states
for electrons become filled up.

Of course you mean all possible quantum states from the lowest up
partway, however many electrons there are in the atom, get filled,
while the infinite number of higher states remain unfilled except
briefly if an electron absorbs a photon and thereby gets knocked up to
a higher state, right?


Yes, absolutely right.

It's the lining up of electron spins that causes ferromagnetism ...
nuclear magnetic moments are intrinsically much smaller than
electrons spins (magnetic moment of a particle typically goes as
1/mass of particle, and nucleons are far heavier than electrons).

Now *that* (1/mass) is totally counter-intuitive. I thought the "spin"
of an electron was + 1/2 or - 1/2, and likewise the spin of a quark was
+ 1/2 or - 1/2, which in nucleons add up to + 1/2 or - 1/2 spin (and in
mesons with two quarks add up to integral spin). Why should the
magnetic moment not likewise be similar magnitude for all? Is it some
general-relativity effect where EM photons as the exchange particle for
magnetic field are redshifted, or just a gravity-well Keplerian effect
where the deeper the point-mass well is the tighter around it the same
spin is wrapped, so with tighter field in heavier particles the
long-range magnetic field is proportionally less? If not that, then I
return to my "totally counter-intuitive" remark above.


It isn't intuitive, although you can make a kind of classical model
of an electron as a spinning sphere with very special constraints
on its motion.

This is really a result of Special Relativity plus quantum mechanics,
and to really elucidate the connections here requires that we would
go into some deeper physics.

Spin magnetic moments are an idea that had to be invented by human
beings to explain atomic structure by the way!

The 1/m dependence follows from the Dirac equation which describes
spin 1/2 particles in relativity. The equation results from a
combination of quantum mechanics and special relativity. The
magnetic moment points along the direction of the spin, and the
ratio between the spin and the magnetic moment associated
with the spin is then something that depends on the charge of
the particle and its mass.

So you need to learn about the Dirac equation to see how this comes
about in a natural way.

By the way, is it possible to do NMR with deuterons, or is that
impossible because they all have zero spin hence zero nuclear magnetic
moment?


Deuterons actually have spin 1, and yes, it is possible to do NMR
with deuterons.

Here's a reference:

http://prola.aps.org/abstract/PRE/v51/i4/p3332_1

The rest I'll do maybe later tonight, maybe tomorrow.

David

.

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