Re: orbitals, flowers, quantum puzzlement;


Arnold Neumaier wrote:
> Igor Khavkine wrote:
>
> > Cyberkatru wrote:
> >
> >>>For isolated atoms, rho is spherically symmetric, giving symmetric
> >>>shapes. For molecules, rho is in fact a function of the coordinates of
> >>>all nuclei involved, and there is no longer any reason to have more
> >>>symmetry than the symmetry of the configuration of nuclei has, which is
> >>>very little and often none.
> >>
> >>Hmm. OK. But what determines the relative positions of the nuclei which are
> >>each in and of themselves, basically spherically symmetric?
> >
> > To clarify. An isolated atom/molecule/nucleus/anything will be
>
> An isolated molecule will never be spherically symmetric since the
> nuclei provide preferred directions, and they are already heavy
> enough that the states with different preferred direction do
> not superimpose sufficiently to produce a symmetric state.
>
> Thus your argument (though true for anything) is relevant only for
> single atoms and smaller things.
>
> > perfectly spherically symmetric *only if* it is in a state of definite
> > angular momentum with zero total spin (mathematically, in an eigenstate
> > of the total angular momentum operator with eigenvalue 0). Many nuclei
> > have some definite nonzero total spin, hence cannot be spherically
> > symmetric. In combination with orbiting electrons, a nucleus with
> > nonzero spin may still from an atom of zero total spin (like the
> > singlet state of the hydrogen atom), but it may also not.
>
> This is the idealized situation at absolute zero temperature.
>
> A more realistic situation is a nucleus or atom or electron in a gas
> (or in a beam) at finite temperature. Such a particle is in a mixture
> rather than in a pure state, and the thermal state is radially
> symmetric even for spin > 0 since it averages over all spin directions.
> (The Bloch matrix = spin part of the density matrix is a multiple of
> the identity.) Only after passing a polarizer, the part of the beam
> passing it will have selective spin directions.
>
> > An interesting puzzle can be formulated here for the experts. If a
> > system is rotationally invariant (such as any system in isotropic space
> > and without external fields) will have energy states that are also
> > eigenstates of the total angular momentum operator. But will there
> > always be an energy eigenstate that is an eigenstate totale angular
> > momentum with eigenvalue 0 (i.e. a spherically symmetric state)? It's
> > clear that any fermion (fundamental or composite) can never be in a
> > zero spin state, since its spin has to be half integral. But in what
> > other situations can we put a lower bound on the total spin of the
> > system?
>
> It is easy to see that the lower bound of a composite of n spin 1/2
> particles and arbitrarily many spin 0 particles is 0 if n is even
> and 1/2 if n is odd, and that these can be achieved- simply pair
> up antiparallel spins.
>
> > Physically speaking, given say a long polymer molecule, if the molecule
> > is completely isolated it's Hilbert space of states splits into finite
> > dimensional subspaces with definite total spin. But I would imagine
> > that it'd be very hard to find such a molecule with zero total spin. Is
> > it because 0 is excluded from the spectrum of the total angular
> > momentum operator or is it because such a state would simply have much
> > higher energy than other ones?
>
> Actually, in the center of mass frame, the total spin is typically
> much smaller than the sum of the individual spins, so from a practical
> point of view, the intrinsic spin can be neglected, except for
> extremely small molecules of light atoms (such as H_2), or in
> specially prepared situations (nuclear spin magnetic resonance).
>
> Anyway, since only total spin (including orbital spin) is conserved,
> the intrinsic spin is usually not even sharply determined, so that
> the question of the exact value of the spin does not make physical
> sense.
>

Mr. Arnold Neumaier

In experience, a snow-flake hex type shape results from,

http://en.wikipedia.org/wiki/Molecular_geometry

the 104.5 degree molecule, (H2O), that's seems truthful.

What I can't figure out is why an octogonal shaped snow
flake could not form.

Could you explain why that 104.5 creates hexagons but
not octagons, I'm guessing it's all bonding and less to
do with the angle??
Regards
Ken

.

Relevant Pages

  • Re: orbitals, flowers, quantum puzzlement;
    ... >>>symmetry than the symmetry of the configuration of nuclei has, ... An isolated molecule will never be spherically symmetric since the ... > angular momentum with zero total spin (mathematically, ...
    (sci.physics.research)
  • Re: Nobel Prize for David Thomson?!
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  • Re: Nobel Prize for David Thomson?!
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    (sci.physics)
  • Re: Nobel Prize for David Thomson?!
    ... >>You are not surprised to see the mass of the composite system being ... >>that the spins of two interacting particles must simply add together. ... Two spin 1/2 particles can only add to give a singlet, ... This is true for two non-interacting particles whose total spin ...
    (sci.physics.relativity)
  • Re: orbitals, flowers, quantum puzzlement;
    ... >> also eigenstates of the total angular momentum operator. ... >> never be in a zero spin state, since its spin has to be half ... >> molecule is completely isolated it's Hilbert space of states splits ...
    (sci.physics.research)