Tuesday, June 4, 2013

Counting Spin States for Three Electrons

I told you recently that I don't know how to count spin states when I have four or more electrons. I thought at least I could count up to three but it turns out I was wrong. I have a problem counting the spin states with three electrons.

I said when there were four electrons I could specify the total spin state by listing eight numbers, corresponding to the complex amplitudes of each electron being up or down. My problem was that there were actually nine spin states you could make with four electrons...five of spin-2, three of spin-1, and one of spin-0. I posted this problem on stackexchange.com to see if anyone could help me out.

I got a very good answer from one Lagerbeer who unfortunately had a totally different way of counting. He didn't have eight states and he didn't have nine states. He had sixteen states, corresponding to every possible ordered combination of four electrons, listed as being either up or down:


And of course this is right. The way I was counting, if I had n electrons I had to specify 2n parameters. That was wrong. You need to specify 2^n parameters, the amplitudes to be in each of those possible compound states.

In a way that's a little better. I had a real problem specifying nine independent states with only eight numbers. With sixteen numbers at least I have too many parameters, which in a way is better. But there's a new wrinkle. I thought I was OK up to three electrons, because I had six paramters to describe six possible spin states spanning l=(3/2, 1/2) and m=(-3/2...+3/2). It turns out I was wrong. I don't have six parameters, I have eight, corresponding to 2^3. So I can't even count up to three.

But this clarifies the problem tremendously. I know how to work the up/down basis for two electrons...that is, I know how to relate the up/down states to the (l,m) states. I even wrote a couple of blog posts about it. For two electrons it works like this (where the bracketed terms are (l,m):

uu          <---->   (1,1)
ud+du    <---->   (1,0)
ud-du     <---->   (0,0)
dd           <---->   (1,-1)

The l number is of course total spin and the m number is z-axis spin. And I haven't bothered to normalize the mixed states. But you get the idea.

Can I do it for three electrons? Let's list all eight states. For the first and last states, it's obvious what they line up with in terms of (l,m). But for the others it's not so obvious....

1.   uuu    <---->  (3/2, 3/2)
2.   duu
3.   udu     <---->    ?(3/2, 1/2)?
4.   uud                      ?(1/2, 1/2)?
5.   ddu
6.   dud     <---->    ?(3/2, -1/2)?      
7.   ddu                     ?(1/2, -1/2)?
8.   ddd     <---->  (3/2, -3/2)

The problem boils down to this: how do you combine the 2, 3 and 4 states to make the two different m=1/2 states?

Well, by analogy with the two-electron system we can add those three states together: {1} + {2} + {3}. (Again, I'm not bothering to normalize). That definitely gives us a state with z-axis spin of 1/2. But what is the total spin? Is it 1/2 or 3/2? That's not so obvious.

And with two electrons, we got our two different states by taking sums and differences. Well, we've taken the sum, but how do you take the difference of three states? 

I have some ideas on this but I think I'll save that for my next post.

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