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Selasa, 04 Juni 2013

Using Three Electrons to Make L = 1/2, M = 1/2

I told you in my last post about how I was trying to line up the spin states of three electrons with the (l,m) descriptions of spin. The real problem was when you had a z-axis spin of 1/2, how to tell whether your total spin was 3/2 or 1/2.

I figured out that you can basically create either of these states only by combining states with two electrons up and one down, which I am going to label A, B, and C:

A = duu
B = udu
C = uud

The one obvious way to create a z-axis spin of 1/2 was to add the three states: A+B+C. But what is the total spin of this system (the l-number)? Is it 3/2 or 1/2?

I'm pretty sure it's 3/2. Here is my reasoning. To get a spin of 1/2, you would want to start by combining two electrons into a single state, and then adding a third electron. Something like this:

{ud -du}*u = udu - duu  = B - C    (Equation 1)

To get a total spin of 3/2, on the other hand, you would want to take a spin-1 combination and add an electron pointing down the other way:

{uu}*d =  uud   =  A                   (Equation 2)

But in fact, neither of these systems is right. The problem is that electrons don't really have separate identities. You just can't say the first electron is doing this and the second electron is doing that. You have to come up with a description whereby it doesn't matter which electron is which. That's not always so easy.

But in the case of Equation 2, we can actually do it. We just have to consider all permutation of whether the first, second or third electron is the one  pointing down. It looks like this:

uud + udu + duu  

But this is just the combination A + B + C (actually in reverse order this time) which I said was my likely candidate for the (l,m) state of (3/2, 1/2).

Can we do the same thing then for the (1/2, 1/2) state? We wrote an expression in Equation 1 for combining a spin-up electron with a pair in the singlet state. (It's a fascinating story as to why the difference ud-du gives you the singlet state! but that's one for another day.) By analogy, all we need to do is symmetrize it by combining all possible permutations of A, B and C. Generalizing  Equation 1, we get:

{B-C} + {C-A} + {A-B} =  .......0  ????

Horrifyingly, just when we've got a nice symmetric expression, we find it adds up to....zero!

But there's one possible way out. I've taken all permuations of {B-C} and added them up (actually I've taken only the even permutations, as defined in Group Theory, but don't worry about that). Who said we had to add them? Why not subtract them? Well, there's a problem subtracting three things from each other, but I'm going to borrow some ideas from the Theory of Equations that I wrote about last summer. We're going to use the three cube roots of unity, which I will call {1, w, and w^2}. Here is the combination I will try:

{B-C} + w{C-A} + w^2{A-B}

It's like I've taken three vectors in the complex plane separated by 120 degrees and applied them to my spin states. It turns out you can reduce this expression with algebra and it comes to:


{B-C} + w{C-A} + w^2{A-B} = -i{A + wB + w^2C}

Interesting results. The phase of a quantum state is of course arbitrary, so this is the same as:

{A} + w{B} + w^2{C}    =  (?)  (1/2, 1/2) ???

That's the question. Have we written a description of the (1/2, 1/2) state in terms of three electrons? I think we probably have, but we'll consider the implications later. Because there are some very important parity conditions that have to be observed, and it's not at all clear that we've done so.


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