Re: Selection rules

Dear Joel,

let me start with a simple description of the photon. Photons are quanta
of the electromagnetic field that can be parameterized by the
electrostatic potential "phi" and the vector potential "A" (the magnetic
field "B" is "curl A", for example).

In relativity, we unify them into a 4-dimensional vector. It has 4
components, but because of the gauge invariance - and the corresponding
Gauss' law (essentially div E = rho), two of the polarizations are
unphysical. Because it is a 4-vector, is has "1" index, which is the
origin of the "spin 1" that photons have.

Physical photons with a well-defined momentum (frequency and direction)
can only have two different states. Electromagnetic waves are transverse
waves, i.e. a wave that moves in the "z" direction can only oscillate (its
electric field) in the "x" or "y" directions (and the magnetic field is
transverse to the electric field). There is a two-dimensional space of
possible photon polarizations for each momentum. You can choose the basis
"x" and "y", but you may also choose their combinations

        R = x + iy,     L = x - iy

which are the right-handed and the left-handed photon, respectively. If
the photon moves in the "z" direction, these two possible modes have "m =
-1" and "m = +1". There is no "m=0" polarization of a photon with respect
to the axis of motion. The polarized photon always carries an angular
momentum, and typical light is just a mixture of left- and right-handed
photons, and therefore it carries no *macroscopic* angular momentum.

(In a similar fashion, general covariance kills most of the polarizations
of the gravitational field, and gravitons also have 2 polarizations in 4
dimensions only: m=-2 and m=+2, respectively. The spin "2" of gravitons
essentially comes from two indices of the metric.)

The most typical electronic transition of atoms is the "dipole radiation".
The electric dipole is nonzero if the negative charge is separated from
the positive charge. If the separation oscillates in the "z" direction,
electromagnetic waves are emitted almost everywhere (except for the z
direction itself) according to the sin^4(theta) rule.

This also holds in quantum mechanics. The amplitude for a transition
A -> B is proportionnal to the matrix element of the electric dipole "p"
between the states A,B:   <A|p|B>.  This amplitude often vanishes if the
conservation/addition rules of angular momentum OR parity conservation
does not allow this transition.

By the rules of adding momenta, we see that "p" carries spin 1 (and m=+1
or m=-1 with respect to the axis of motion), and therefore you are
guaranteed that "m" of A,B *can* differ by +1 or -1. Can they also differ
by "0"?

The answer is "yes, they can". The rule is easy, the z-component of the
dipole carries m=0, while the x+-iy components carry m=-1,+1.

You don't need to study complicated atoms.  Just think of the Hydrogen in
the "2p" state. For electric dipole transitions, the spin plays no role -
it only interacts with the magnetic field. So you only consider the
orbital angular momentum L. Of course, "2p" has L=1, and it means three
polarizations m=-1,0,+1. All of them are equally good - they are related
by the SO(3) symmetry - and it is therefore clear that if some of them are
unstable, all of them must be unstable. And this is the case: all three
"2p" states of the Hydrogen are unstable, and they can decay to "1s", I
hope. ;-)

The photon that moves in the "z" direction has m=-1 or m=+1 with respect
to the "z" axis. But if it moves in another direction - for example the
"x" axis - it also has some probability amplitude to carry m=0 with
respect to the axis "z". Therefore "m" does not have to change. The
invariant statement is that "m" changes by -1,0, or +1. You can
equivalently say that the total "scalar" orbital momentum "L" changes by
-1,0, or +1 - all possibilities are allowed by the rules of adding the
angular momenta. The photons carry spin 1, in a sense, and tensor
multiplication of spin "1" with spin "L" decomposes into states with
L+1, L, or L-1 (the latter is of course missing for L=0).

The dipole radiation always wants to change "L" of a single electron -
which also means changing the total "L" - but because the electrons are
not independent, they can rearrange their total "L" after the emission.
Yes, once the spin-orbital interaction becomes important, you can no
longer classify the energy eigenstates by L,S, and you must use the total
"J" instead. Of course, J changes by -1, 0, or +1, and all possibilities
are allowed again.

Parity however does not really allow delta L= 0, which is another
selection rule that I will discuss below.

Which type of spin exactly changes depends on which interactions you can
neglect. As long as you neglect the magnetic interaction between the
nucleus and the electrons, you can say that the angular momentum of the
nucleus is conserved separately, and the angular momentum of the electrons
changes by +1,0,-1. If you can neglect the spin-orbit interactions, it is
only the orbital momentum of the electrons that is changed. If you can
neglect the interactions between all electrons (which really does not
quite happen), it is really one electron whose "L" changes, and so on.

If you make a more precise calculation, you must consider the fact that
you have also magnetic radiation, electric quadrupole radiation, and so
forth, but they are becoming increasingly weaker.

> http://chsfpc5.chem.ncsu.edu/CH331/help/sel_rule/sel_rule.html

This web page implicitly only considers the z-component of the dipole
moment. The z-component carries m=0, and therefore m is conserved here.
However the dipole can oscillate in other directions, too.

> http://tesla.phys.unm.edu/phy537/3/node8.html

This page correctly considers both "z" as well as "x+iy" and "x-iy". In
the latter case, they correctly say that "m" changes by +1 or -1.

> http://farside.ph.utexas.edu/teaching/qm/perturbation.pdf

I have not found the electric dipole radiation in this file, sorry.

Note that if you allow L to be changed by +-1, its projection "m" with
respect to a general axis can also change, simply because "m" is always
allowed to go between -L and L, and claiming that "m" is conserved while
"L" is not conserved is not a SO(3) invariant statement! It is only if you
consider the special z-component of the dipole radiation, you can obtain a
new rule that "m" does not change.

> However, I'm quite confused about some of that site.  In particular, on page
> http://math.ucr.edu/home/baez/spin/node10.html he says that Delta J (the
> total angular momentum) can be +1,-1 OR 0.  At the very top he says that
> little l changes by only +1,-1

Such a restriction on delta L occurs, but it comes about not because of
angular momentum conservation, but because of parity conservation. You
know, the parity of the spherical harmonics is (-1)^l, and if the emitted
photon has a negative parity (and a vector such as the dipole moment
simply HAS negative parity - vectors reflect under x -> -x), "l" must
change by an odd number so that the parity changes its sign. Different
types of transitions have different parities.

> - what's the difference between this and big
> L (single electron vs multiple?

Yes, I have not looked at this page, but the standard notation is what you
say, while J also includes the spin.

> Another thing that's bugging me is that he and quite a few other
> (particularly not-so-physics-y) sites explain these selection rules by the
> vector addition of angular momentum, by considering a photon having spin 1
> (e.g. http://www.umsl.edu/~jjl/P471/10eselection.pdf page 4).

Photons definitely have spin 1.

> The general tactic is that since the photon has angular momentum 1,
> the total angular momentum must go up or down by one - handwaving
> only.

By the rules of the addition of angular momenta, the angular momentum can
change by +-1 or 0. But then you have parity that constrains some types
of transition.

> V_{J} * V_{1} = V_{J-1} + V_{J} + V_{J+1}
> However, this allows for the total angular momentum of the system to be J=0
> as well as J=+1/-1.

That's correct. But once again, parity conservation allows l=+1,-1 only
for these electric dipole transitions.

> The only thing I could think of is if there's a subtlety here with the
> photon actually being restricted to only the m=+1/-1 states (i.e. can't have
> m=0)? (I _think_ this is right? This is also something I don't understand!)

Yes, photons moving in the "z" direction can only have m=-1,+1, not zero.
However they can also have an amplitude to have m=0 with respect to
another axis than the axis of their motion.

> If I do that I get the right number of states on each side if I cut out
> V_{J} and the photon m=0 state, but I'm not sure I can get everything to
> close under the action of the ladder operators.  (Heck, what does V_1 do if
> I just chop out that middle state?)

Not sure what you exactly want to close under the action of ladder
operators. Ladder operators are usually meant to give you states with m=+L
(or m=-L) from one another, not all states. The other states, |m| < L, are
obtained by SO(3) rotations from the highest weight states m=+L.

> * Definitely, what are the selection rules for electronic transitions (and
> in what situations are they different, such as the discrepency over whether
> Delta m=0 only)

"m" can be -1,0,+1, if you consider all components of the dipole moments
(and, which is related, all directions of the photon).

> * Can these all be derived by taking the dipole moment operator (which I
> assume is a linear combination of the position operators for each electron
> (& nuclei?), or something close)

Yes, the selection rules are derived from the transformations of this
operator under SO(3) and under parity - both of them are conserved.

> * Can these all be derived by applying conservation of angular momentum and
> angular momentum addition rules, and if so, how?

Yes, I hope it is above. Don't forget the parity.

Cheers,
Lubos
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