Total spin

If the Hamiltonian is independent of spin, then it is clear that the total spin of an $N$-particle system

$${\bf S}= \sum_{i=1}^N {\bf S}_i$$

will be a constant of the motion, but so will the individual spins, ${\bf S}_i$ of the individual particles.

What happens, however, when the Hamiltonian is spin dependent. Consider the case of the hydrogen atom with relativistic corrections. It can be shown (see problem set # 2) from the Dirac equation that when relativistic corrections are accounted for, a term in the Hamiltonian appears that is explicitly spin dependent and takes the form

$$H_{\rm so} = f(r){\bf L}\cdot{\bf S}$$

which is known as the spin-orbit coupling.

Let us look at the commutator of this Hamiltonian with the $z$ components of ${\bf L}$ and ${\bf S}$. First note that

$$H_{\rm so} = f(r)\left(L_x S_x + L_y S_y + L_z S_z\right)$$

Therefore,

$$\left[L_z,H_{\rm so}\right] = f(r)\left(\left[L_z,L_x\right]S_x + \left[L_z,L_y\right]S_y + \left[L_z,L_z\right]S_z\right)$$

$$= f(r)\left(i\hbar L_y S_x - i\hbar L_x S_y\right) \neq 0$$

Also,

$$\left[S_z,H_{\rm so}\right] = f(r)\left(L_x\left[S_z,S_x\right] + L_y\left[S_z,S_y\right] + L_z\left[S_z,S_z\right]\right)$$

$$= f(r)\left(i\hbar L_x S_y - i\hbar L_y S_x\right) \neq 0$$

However, if we add these together

$$\left[L_z,H_{\rm so}\right] + \left[S_z,H_{\rm so}\right] = \left[(L_z+S_z),H_{\rm so}\right]$$

$$= i\hbar f(r)\left(L_y S_x - L_x S_y + L_x S_y - L_y S_x\right) =0$$

Thus, $L_z+S_z$ is a constant of the motion. The same can be shown to be true for the $x$ and $y$ components, thus it follows that

$${\bf J}= {\bf L}+ {\bf S}$$

is a constant of the motion. Recall that ${\bf J}$ is the total angular momentum. It is often true for spin-dependent Hamiltonians that the total angular momentum is still conserved.

Thus, we see that total angular orbital angular momentum, total spin, and total angular momentum are all important quantities in quantum mechanics. When ${\bf J}$ is conserved, then $J^2$ and $J_z$ are good quantum numbers. In general, it remains to discuss how to derive a set of basis vectors appropriate for total angular momenta of any type. We expect that they can be composed of tensor products of the basis vectors of the corresponding individual angular momenta but will not be equal to them. We will show that they are equal to in the next lecture.