so if the we look at an electron in an atom with the angular quantum number 1, it could have L_z = hbar, 0, -hbar, then if we want total angular momentum quantum numbers from both L and S of the electron we could have 3/2 or 1/2. So a measurement of total angular momentum squared should return 15/4 hbar^2 or 3/4 hbar^2. But are the probabilities for each of these equal? I believe there is some small degree of energy splitting between the two states but I am not sure this causes a probability split.
If we write T as total can say
T^2 = (S + L)•(S + L) = S^2 + L^2 + 2L•S. So I guess you could treat a single electron as a tensor product of the spinless part and the spin part, so something like |1 0>|1/2 1/2> for one of the states. Then for example on the state |1 1>|1/2 1/2>
T^2 |1 1>|1/2 1/2> = |1 1>S^2 |1/2 1/2> + L^2 |1 1>|1/2 1/2> + 2L•S |1 1>|1/2 1/2>, and
L•S |1 1>|1/2 1/2> = (L_x |1 1>)(S_x |1/2 1/2>) + (L_y |1 1>)(S_y |1/2 1/2>) + (L_z |1 1>)(S_z |1/2 1/2>), using L_x = 1/2(L+ + L-) and L_y = i/2(L- - L+) this becomes:
1/2 hbar^2 |1 1>|1/2 1/2>
so sanity check that indeed does give
T^2 |1 1>|1/2 1/2> = (3/4 + 2 + 1) hbar^2 |1 1>|1/2 1/2> = 15/4 hbar^2 |1 1>|1/2 1/2>
= t(t + 1) hbar^2 |1 1>|1/2 1/2>, where t is the total angular momentum quantum number. But when I try this on other states they are not eigenvalues except |1 -1>|1/2 -1/2> with same eigenvalue as the other.
Then I know the other eigen values should be linear combinations of |1 0>|1/2 1/2>, |1 1>|1/2 -1/2> and |1 0>|1/2 -1/2>, |1 -1>|1/2 1/2> but the algebra for determining the constants gets super fucked.
Sanity check? Why is this marked easy in my textbook? Are you just supposed to assume that the Clebsh-Gordon coeffs also hold even tho L isn't spin? afaik the textbook didn't explicitly explain this kind of thing.