How to sum orbital angular moments to get Russell-Saunders coupling's L?
On this site I have found the following sentence:
Consider, as an example, the excited helium atom in the atomic electron configuration 2p3p. By the triangular conditions the one-electron spins s = ½ can couple to |½−½|, ½+½ = 0, 1 (spin singlet and triplet) and the two orbital angular momenta l = 1 can couple to L = |1−1|, 1, 1+1 = 0, 1, 2.
I sort of understand the spin part but I have trouble with the orbital angular momentum, why can it be equal to 1 if I have two electrons both with l=1. Maybe a better example can be found in Linus Pauling's book titled General Chemistry (and this is what I couldn't understood first but even after trying to find it out on my own I'm stuck.). He shows us an example with a carbon atom. Here's my thought process on getting the Russel-Saunders symbol:
- C's electron config: 1s^2 2s^2 2p^2, we don't have to look at the first two orbitals because both S=0 and L=0
- 2p^2 can be 2px^2, 2px^1 2py^1 (and other basically the same variations), in the first case S=0 in the second S=1 This is what I don't get, both sources says L=0,1,2, but I can't make sense of L=1, (although I can't really make sense about the other values either, but at least I can arithmetically get to those). They are vectors and I understand that px py and pz orbits all point to directions perpendicular to each other, in which case adding them together should result in values 2 (if same direction) or sqrt(2) in every other case. Not even 0 should be a correct answer, so I definitely don't understand something basic here.
Also, if it wouldn't be that big of a trouble, could you please explan what J actually is (the subscript on the right)?
Posted Aug 10, 2017, 2:19 PM UTC · edited
J is the total angular momentum quantum number. J = L + S is the total angular momentum operator. The total angular momentum is always a good quantum number for any rotation ally invariant Hamiltonian. In the special case where all interactions are central, the orbital and spin parts are separately good quantum numbers. But if you include any kind of non-central interaction, the orbital angular momentum is no longer a good quantum number, but the total angular momentum still is.
As for your first question, it's true in general that when you couple two angular momenta j1 and j2, the resulting angular momenta can take on values between |j1 - j2| and j1 + j2 in integer steps. If you apply this result to j1 = j2 = 1, you get 0, 1, and 2.
By the way, since I don't think this deserves a post of its own, how comes that certain pairs of fermions behave like bosons? Like, if the rule is, no two fermions are allowed to occupy the same state, how comes that rule doesn't apply to fermions paired into bosons and occupying the same state a lot?
If you sum the total angular momenta of an arbitrary even number of fermions, you get an integer multiple of
h. So any composite particle consisting of an even number of fermions is a boson.If you force those two composite bosons very close together, you'll find that they have a "hard-core" interaction stemming from the fact that their "innards" are fermions, and fermions obey Pauli exclusion.
An example would be proton-neutron scattering. Protons and neutrons are not bosons, but that's irrelevant to the example. Protons and neutrons are not identical to each other, so there is no Pauli exclusion between a proton and a neutron. However if you smash them together at very high energies (in other words, make them come very "close" together), their interaction becomes repulsive because the quarks which make up the proton start to feel Pauli repulsion from the quarks that make up the neutron.
So, like, a bunch of cooled Helium-4 doesn't collapse into a single super-heavy molecule because it's not "really" a boson, it's a composite particle made out of fermions, but it does behave as a boson otherwise? Is there a rule of thumb that says which boson properties composite bosons have or don't?
When the wavefunctions of the identical fermions overlap significantly in space, Pauli exchange effects will become important. So you have to squeeze composite bosons together very tightly for their constituent fermions to repel.
The way you'd do this is in high energy scattering experiments. Generally speaking, the higher the incident energy, the closer together the particles come, and the more their wavefunctions overlap.