Re: How is spin in quantum systems a direct result of special relativity?

The short answer to your question is that electron spin is relativistic because it automatically pops out of the relativistic Dirac equation.

This, of course, is very unsatisfying, so I'll go into some more detail to try to answer "why" this would be the case.

In non-relativistic quantum mechanics, spin is a simple addition to the equations; any amount of spin can be easily added or removed as desired. But when Schrodinger proposed to make the equations relativistic, he found he was unable to "add" the electron spin the way it had been done in the non- relativistic case. The problem boils down to the relationship between energy and momentum. In the non-relativistic world, E = p²/2m. But with relativity, E²=p²c² + m² c⁴; Energy and momentum are now linear (at relativistic speeds), which changes the nature of the equations. All Schrodinger was able to write down was the relativistic quantum equation for a zero-spin particle; the equations wouldn't allow the addition of spin.

Then Dirac came along and started from a very simple premise: that a relativistic Hamiltonian would be linear in both momentum and mass. Other people had tried to do this but had found there was no possible solution. But Dirac was "thinking outside the box" and decided the constants in the equations didn't have to be numbers, but could be matrixes instead. The simplest matrixes that solved the equation were 4-by-4 matrixes, which therefore led to 4 different solutions of his equation: two electron solutions and two positron solutions (therefore, Dirac predicted the existence of antimatter!) The difference between the two electron solutions turned out to be a matrix that was already known to be the spin operator in quantum mechanics -- spin had naturally popped out of his equation.

Still, this doesn't answer WHY it popped out, and I'm not sure I can answer this without going into the math. (Which means, as Feynman once said, that I don't really understand it at all.) I do recognize it seems strange that spin SHOULD pop out of the math, so I'll try to make it seem less strange, and that will be about all I can do.

First of all, Orbital angular momentum L is equal to the cross product r x p. This is not spin, but writing it in this form makes it obvious that relativistic corrections can affect angular momentum, as r is the position vector and p is the momentum vector.

As it turns out, Dirac's relativistic equation does effect L -- so much that L isn't even a constant of the motion anymore. You have to add some other matrix S to make the quantity (L + S) a constant of the motion. But what is S? S, amazingly enough, is the mysterious matrix that happens to be the difference between the two electron solutions which popped out of the Dirac equation. Because it is natural to think that total angular momentum must be conserved (as in the classical world), the natural assumption is that S stands for some other angular momentum that isn't included in L. That's why S is called the "spin angular momentum operator".

However, it's important to keep in mind that the electron isn't literally spinning around like an infinitesimal top. "Spin" is just what we call the observed values of S. It's a truly quantum parameter that doesn't necessarily have a classical analogy. In this case, the analogy to a spinning top is probably a pretty good one, but that doesn't mean the two concepts are exactly the same.

If you're interested in looking into this in more detail, try this web page for the details of the math and perhaps some additional insight.