OK, if we really understand what is going on at this point, we ought to be able to get a rough cut at a system with three down spins.
So following the approach outlined by Feynman in the lecture, we scribble out the Hamiltonian. This is going to have six terms generated by the spin exchange; one for each coordinate l,m,n.
Now we assume, as before, that we can write the solution as the product of a time component, expressing the energy of the system, and a position component. That is, we expect to find solutions of the form
exp[iEt]a(l,m,n)
Following the earlier insight, we consider the three spins to be far enough away that they could be treated independently of each other:
a(l,m,n) = exp[ib lx]exp[ib my]exp[ib nz]
where x,y,z are wave numbers, as before. Operate on this with the Hamiltonian, and we get
E = 6A - 2A cos (bx) - 2A cos(by) - 2A cos(bz)
Now, there's no particular reason that x and l be related; the following expression also works, and gives the same expression for the energy:
a'(l,m,n) = exp[ib ly]exp[ib mx]exp[ib nz]
In fact, there are six expressions that look like this, which are each satisfactory, providing the same energy. So what we really want is some linear combination of these
a(l,m,n)
= exp[ ib(lx+my+nz) ] + B exp[ ib(lx+mz+ny) ]
+ C exp[ ib(ly+mz+nx) ] + D exp[ ib(ly+mx+nz) ]
+ F exp[ ib(lz+mx+ny) ] + G exp[ ib(lz+my+nx) ]
Now you can work these through the Hamiltonian, for each of the cases where l = m+1 etc. Or alternatively you can recognize that what we really need is
a(l,m,n) = 0 whenever l=m, m=n, or l=n.
In other words, we need the solution to be anti-symmetric when we exchange two coordinates. This is pretty easy:
a(l,m,n)
= exp[ ib(lx+my+nz) ] - exp[ ib(lx+mz+ny) ]
+ exp[ ib(ly+mz+nx) ] - exp[ ib(ly+mx+nz) ]
+ exp[ ib(lz+mx+ny) ] - exp[ ib(lz+my+nx) ]
October 6, 2005 12:32 PM
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