Re: orbitals, flowers, quantum puzzlement;

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.

Arnold Neumaier

.

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