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Verify the quoted eigenvalues by calculation using the operatorSz=S(1)z+S(2)z.
We expect to be able to formeigenstates ofS^2 from linear combinations of these four states. From
pure counting of the number of states for eachSzeigenvalue, we can guess that we can make one
s= 1multipletplus ones= 0 multiplet. Thes= 1 multiplet has three component states, two
of which are obvious from the list above. We can use the lowering operator toderive(see section
21.8.2)the other eigenstates ofS^2.
χs=1,m=1 = χ(1)+χ(2)+χs=1,m=0 =1
√
2
(
χ(1)+χ(2)− +χ(1)−χ(2)+)
χs=1,m=− 1 = χ(1)−χ(2)−χs=0,m=0 =1
√
2
(
χ(1)+χ(2)− −χ(1)−χ(2)+)
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As a necessary check, we operate on these states withS^2 andverify(see section 21.8.3)that they
are indeed the correct eigenstates.
Note that by deciding to add the spins together, we could not change the nature of the electrons.
They are still spin^12 and hence, these are all still eigenstates ofS(1)^2 andS(2)^2 , however, (some of) the
above states are not eigenstates ofS(1)zandS(2)z. This will prove to be a general feature of adding
angular momenta. Our states of definite total angular momentum and z component of total angular
momentum will still also be eigenstates of the individual angular momenta squared.
21.2 Total Angular Momentum and The Spin Orbit Interaction
Thespin-orbit interaction(between magnetic dipoles) will play a role in the fine structure of
Hydrogen as well as in other problems. It is a good example of the need for states of total angular
momentum. The additional term in the Hamiltonian is
HSO=
Ze^2
2 m^2 c^2L~·S~
r^3If we define thetotal angular momentumJ~in the obvious way we can write~L·S~in terms of
quantum numbers.
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J~ = ~L+S~
J^2 = L^2 + 2L~·S~+S^2
~L·S~ =^1
2(J^2 −L^2 −S^2 )→
̄h^2
2(j(j+ 1)−ℓ(ℓ+ 1)−s(s+ 1))