Re: Does a rotating electric dipole radiate?
- From: Jim Black <fmlast3@xxxxxxxxxxxxxxxx>
- Date: Mon, 24 Aug 2009 13:55:25 -0700
On Mon, 24 Aug 2009 06:52:57 -0700 (PDT), Darwin123 wrote:
On Aug 23, 11:58 pm, Jim Black <fmla...@xxxxxxxxxxxxxxxx> wrote:
On Sun, 23 Aug 2009 09:10:45 -0700 (PDT), Darwin123 wrote:Thank you for the correction. Your answer is based on quantum
On Aug 22, 10:11 pm, eric gisse <jowr.pi.nos...@xxxxxxxxx> wrote:
Edward Green wrote:
Not about its symmetry axis. What implications does this have for the
spectrum of HCl?
...yes?
Applying classical E&M to a quantum system is a setup for failure, by the
way.
He is asking about a symmetry property. Symmetry properties are
usually independent of the specific dynamics. That is, if the system
has a certain symmetry you can say things about it that are valid both
in classical physics and quantum physics.
For instance, one could say based on this symmetry of his that
the rotational energy states are highly degenerate. For every given
energy, there are an infinite (very large?) number of energy states
with the same energy.
You just proved his point. There are 2L+1 energy states with the same
energy, where sqrt(L(L+1)) hbar is the angular momentum of the molecule.
mechanics. My infinite states was based on classical mechanics.
However, consider:
A spinning electric dipole is also a magnetic dipole. The magnetic
dipole is in the direction of spin. Therefore, states that you are
referring to can be split using a magnetic field. If you apply the
magnetic field in any direction orthogonal the the dipole, then you
break the degeneracy. Those L(L+1) states differ in energy.
Also, I think I made a mistake concerning the electric field. I
am not sure the electric field will cause a change in energy of the
rotational states. It think it will cause a change in energy of the
vibrational states.
Take the electric field in the z-direction. For the L=1 states, it's
pretty easy to see that the first-order perturbation in the energy
(<psi|V|psi> with V proporitional to z) is zero. Less mathematically, in
any of these states, without an electric field, the hydrogen has an equal
probability of being on either side of the molecule's center of mass. But
turning on the electric field pulls the hydrogen atom to one side more
often, so you'd expect a second-order change in its energy.
The two L=1 states with angular momentum m=+/-1 about the z-axis stay
exactly degenerate due to symmetry. I don't see any reason why the m=0 and
m=+/-1 states would remain degenerate at second order in perturbation
theory, but you'd have to work it out to check.
For the states with higher L, you should get a splitting at first order,
but I haven't worked out which L that first happens at.
But a magnetic field ought to give you a splitting at first order in
perturbation theory for all of the degenerate levels.
--
Jim E. Black (domain in headers)
How to filter out stupid arguments in 40tude Dialog:
!markread,ignore From "Name" +"<email address>"
[X] Watch/Ignore works on subthreads
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