Jumat, 17 Mei 2013

Having trouble counting spin states

There is something funny about the fact that spin is supposedly "quantized" along the z axis. The thing is that it only really "works" because of a funny geometrical property of spheres. Let's say you have a spherical apple and you cut it into 1/4-inch slices. Everyone knows that you get a different amount of apple in each slice, but the funny thing get the same amount of peel. What this says in quantum mechanics is that if you have a spin-11 system whose z-axis spin is randomly distributed among all values between plus and minus eleven, then those random spins are equally distributed among all spherical directions. It's a property of spheres that is intimately related to the fact that the surface area of a sphere is exactly four times the cross-sectional area.

I started trying to write about this...oh, maybe two months ago. I finally cobbled something together and posted it but I swept something under the rug that is very embarassing. It turns out I don't know how to count spin states. I thought I did, but I didn't. I'm going to try and explain the problem.

If you have one electron, there are two spin states. People say the spin must be either up or down, but that's nonsense. The spin can be in any direction, but those directions are expressible  as the superposition of an "up" state and a "down" state. The total spin is one-half.

If you have two electrons, there are now four spin states. Both up, both down, and one-up/one-down. Wait...that's only three spin states. Not turns out that depending on the relative phase of the up-down combination, they can form a spin-zero state. They don't have to...they can form a spin-one state in a superposition of left- and right-spin. That's not spin-zero. There is a unique spin-zero state that has no spin anywhere, called the "singlet state". I've written about it here.

So for two electrons, you can describe the total spin by giving the complex amplitudes for the up- and down- spings of both electrons...that's four "basis" states. Or you can be more "elegant", as they say, and describe the total spin as the three amplitudes for the triplet states...that is, (+1, 0, or -1) along the z-axis, plus the amplitude for the singlet states. Either way, it's four basis states.

With three electrons, it's the same. You have now six amplitudes to describe the individual electrons. Or, you can group them in states...they can form either a spin-3/2 state with four z-axis levels, or a spin-1/2 state with two z-axis levels. Six states.

With four bear with's the same.; You now have eight amplitudes to describe the individual electrons. Or, you can group them into spin-2 states (five z-axis levels), spin-1 states (3 z-axis levels), and a spin-zero state. Eight states.

No...wait. There are nine states if you group them into total-spin states, and only eight if you just list the individual electron states.

And it gets worse. For five electrons, you have ten numbers to describe the individual spins. Or you can group them by total spin: spin-5/2 (six states), spin-3/2 (four states), and spin-1/2 (two states).

I don't think you can have more spin states than you get by listing the states of each individual electron. But at the same time, I don't see why there is any redundancy in describing a box of 21 electrons as having such-and-such amplitude to be l=11, m=6, such-and-such amplitude to be l=9, m=7, plus whatever. Because it works perfectly for a two-electron system. You have such-and-such amplitude to be in each of the triplet states, plus an amplitude to be in the singlet state. Why does this system break down when we add the fourth electron?

So there it is. I don't know how to count spin states. But at least I know that I don't know how to count them. That counts for something, doesn't it?

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