Addition of spin angular momenta

[epsilonjon]

Hi

I am working my way through Griffith's Introduction To Quantum Mechanics and I have got to the section on addition of spin angular momenta. I'll copy and paste the bit I'm struggling with, as it's easier than paraphrasing:

http://img714.imageshack.us/img714/2783/80183858.png

http://img442.imageshack.us/img442/1294/20886932.png

I'm assuming \chi_1 is the spin state of the electron and \chi_2 is the spin state of the proton. So why is the total state of the atom written as \chi_1\chi_2? Why not \chi_1 + \chi_2 or something different?

Secondly, he says that S^{(1)} acts only on \chi_1, and S^{(2)} acts only on \chi_2. Why is this, and how come S \equiv S^{(1)}+S^{(2)}? If the two particles are both spin 1/2 then aren't the matrices S^{(1)} and S^{(2)} both the same (i.e. just given by the 2x2 Pauli spin matrices)?

I think if I was clearer on those two points maybe the rest will make sense, so I will leave it there.

Thanks for any help!
Jon.

 

[Vanadium 50]

The Chis are wavefunctions. The wavefunction of two particles is the product of the individual wavefunctions.

 

[epsilonjon]

Oh right. It's strange that he's just thrown that in there with no explanation. Is it something to do with the fact that P(A \cap B)=P(A)P(B) when the probabilities are not connected?

Up to that point (for spin 1/2) the chis were represented by two-element vectors and the spin operators were 2x2 matrices, but this seems to have gone completely out the window?

Thanks for your help.

 

[Chopin]

This point confused me for a really long time when first learning it as well. The reason it doesn't quite seem to follow logically is that they're taking some shortcuts to avoid a big digression into linear algebra. The shortcuts end up working, but if you don't understand what's going on underneath, it can seem a bit arbitrary.

What actually happens when you do angular momentum addition is that you're taking the tensor product of two Hilbert spaces. What that means is that if the first particle lives in the Hilbert space \mathcal{H_1}, and the second particle lives in the Hilbert space \mathcal{H_2}, then the combined system lives in the Hilbert space \mathcal{H_1}\times\mathcal{H_2}. An element of this space looks like (h_1, h_2), where h_1\in\mathcal{H_1}, h_2\in\mathcal{H_2}. As part of the definition of the tensor product, we also define (ah_1, h_2) = a(h_1, h_2), and (h_1, bh_2) = b(h_1, h_2).

You can also construct operators on this new combined space, out of the operators from the original spaces. Specifically, if S_1 is an operator on \mathcal{H_1}, then you can construct an operator on the combined Hilbert space by making (S_1, 1), where we define (S_1,1)(h_1, h_2) = (S_1h_1, h_2) = (s_1h_1,h_2) = s_1(h_1, h_2). A similar process can be done for \mathcal{H_2}. Therefore, in this notation, the total spin operator is really S = (S_1, 1) + (1, S_2). By substituting in these definitions, you should be able to see how this operator leads to eigenvalues which are the sums of the eigenvalues of the individual states.

The operators I constructed here are special cases of the general concept of an operator on the combined space, which is called a bilinear form. The general case is (S_1, S_2), which, if S_1h_1 = s_1h_1 and S_2h_2=s_2h_2, leads to (S_1, S_2)(h_1, h_2) = (s_1h_1, s_2h_2) = s_1s_2(h_1, h_2). Defined this way, the combined operator is linear with respect to variations of either h_1 or h_2, hence the name. In the sections immediately following the one you're currently reading, you'll learn how to change the basis states of the combined Hilbert space into one where the bilinear form for spin separates the states out into two sub-spaces (the spin singlet and the spin triplet), using a process called a Clebsch-Gordan decomposition.