Re: Angular Momentum & Energy Levels
- From: whopkins@xxxxxxxxxxx
- Date: 2 May 2005 01:48:43 -0700
Sidney wrote:
> Now my inquiry is.
>
> What the heck has the angular momentum of the electron got
> to do with the creation of different orbitals??
Look up the recent thread "The Return of Bohr: Orbits as Orbitals".
The energy of the orbital is NOT related to the angular momentum. It
is directly related to the mean radius, a, of the orbit (which is also
a well-defined eigenvalue in the energy eigenstate).
The value of a is given by the usual Kepler formula.
The angular momentum is NOT related to the energy level of the orbital,
so your question is false. It's related to the SHAPE of the orbit --
the eccentricity, e, of the orbit ... which is also a well-defined
eigenvalue in the energy/angular momentum eigenstates. The values of e
for the lowest lying orbitals were listed in the thread above.
> Also the angular momentum of the electron is how fast it
> rotates.
Nobody ever said that. You're freshman-paraphrase-filtering what you
heard.
> But then electron doesn't really rotate. So what
> does the angular momentum represent.
That which is left over:
S = L - (r x p)
after subtracting out from the angular momentum L the kinetic
contribution arising from the vector product of the position (r) and
momentum (p) is the internal angular momentum, which THEREFORE has no
connection to any rotation of any sort, when the system in question is
fundamental.
Nobody ever said that a fundamental system had to have 0 internal
angular momentum (= spin), just because it has no components; and
nobody ever said angular momentum had to be connected to any kind of
rotation.
The only thing you can say is that S must be internal, since (unlike
the vector product r x p or the angular momentum vector L) it doesn't
depend on the location of the origin r = 0 to the coordinate system and
is therefore an invariant of the particle, itself. So, S is the
coordinate-independent residue left over after the coordinate-dependent
part of L is removed.
It also bears explaining that
(a) spin has absolutely nothing specific to do with
Relativity. The Galilean analogue of the Wigner
classification also involves spin. Therefore
spin is also pertinent in *non-relativistic*
quantum theory.
The relevant equation for spin 1/2 is the Pauli-Schroedinger equation
which, for a Lorentz force with potentials (A, phi) is:
(sigma.(p - eA))^2/2m psi = (H - e phi) psi
where ().() is the vector product and
sigma=(sigma_1,sigma_2,sigma_3)
the vector of Pauli matrices.
.... and ...
(b) spin has nothing specific to do with Quantum Physics
EITHER! The same Wigner-style classification also
yields (in the general case) CLASSICAL particles with
spin.
For the classical system the relevant Poisson bracket for the spin
vector S = (S1, S2, S3) is simply
{f(S),g(S)}
= @(f,g)/@(S1,S2) + @(f,g)/@(S2,S3) + @(f,g)/@(S3,S1)
where
@ is my rendition of the curly d partial derivative symbol
@(A,B)/@(C,D) = @A/@C @B/@D - @A/@D @B/@C
In spherical coordinates, with
S1 = s cos(phi) sin(theta)
S2 = s sin(phi) sin(theta)
S3 = s cos(theta)
this becomes (for s > 0):
{f,g} = @(f,g)/@(theta,phi) * (1/s sin(theta))
= @(f,g)/@(phi, s cos(theta))
yielding
S3 = s cos(theta)
is the conjugate momentum to the anglular coordinate phi.
For each s > 0, this yields a separate dynamics. So, classical systems
are classified by a "spin", just as well as quantum systems.
.
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