OK, so Feynman has provided a solution, and furthermore one that doesn't look particularly like the one used in the approximation. So the first step is to work the problem using his solution, so that we may understand how it works, and hopefully get some insight as to where it comes from.
Feynman's solution looks like
a[m,n] = exp[ iK(x[m]+x[n]) ] sin k |x[m]-x[n]|
where K = (k2+k1)/2; k = (k2-k1)/2
Now, in the interest of cleaning this up just a bit, we'll use a constant lattice spacing b (as he does in the lecture), so that the above equation becomes
a[m,n] = exp[ iKb(m+n) ] sin kb|m-n|
Now, to begin with, we shuck away the absolute value operation by limiting our examination to those areas in the crystal where m > n (this is sufficient, because |m,n> and |n,m> are the same state; if the solutions work when m>n, we get the others for free simply by switching the indices).
So now we are considering
a[m,n] = exp[ iKb(m+n) ] sin kb(m-n)
And those coefficients get dropped into the Hamiltonian. The drill is to rewrite each coefficient representing a leakage from another state as some factor times the state |m,n>. It's just plug and chug, really. Here's an example:
a[m+1,n]
= exp[ iKb(m+1+n) ]sin kb(m+1-n)
= exp[ iKb(m+n)+iKb ]sin (kb(m-n)+kb)
= exp[ iKb(m+n)+iKb ][cos(kb(m-n))sin(kb) + cos(kb)sin kb(m-n)]
Not quite what we're after. It looks like exp[iKb]cos(kb)a[m,n] plus something weird. But by happy fortune, look at what falls out here
a[m,n+1]
= exp[ iKb(m+n+1) ]sin kb(m-(n+1))
= exp[ iKb(m+n)+iKb ]sin (kb(m-n)-kb)
= exp[ iKb(m+n)+iKb ][cos(kb(m-n))sin(-kb) + cos(kb)sin kb(m-n)]
We get a second weird term that exactly cancels out the weird term from the first equation. Yahtzee!
Therefore, we can express the sum of the leakage terms in the Hamiltonian in just the form we want...
2 * (exp[iKb]+exp[-iKb]) cos(kb) a[m,n]
If we now consider that coefficient in isolation, everything should fall out:
2 * (exp[iKb]+exp[-iKb]) cos(kb)
= (exp[iKb]+exp[-iKb]) 2 cos(kb)
= (exp[iKb]+exp[-iKb])( exp[ikb] + exp[-ikb])
= exp[i(K+k)b]+exp[-i(K-k)b]+exp[i(K-k)b]+exp[-i(K+k)b]
= exp[i(k2)b] +exp[-i(k1)b] +exp[i(k1)b]+exp[-i(k2)b]
= exp[i(k2)b] +exp[-i(k2)b] +exp[i(k1)b]+exp[-i(k1)b]
= 2 cos (k2)b + 2 cos (k1) b
And hey presto, it all works.... BUT this is actually not quite sufficient, as we haven't really covered the case where m = n+1, which means that a pair of terms should be missing from the Hamiltonian. When we work out that equation, we get a[m+1,n] and a[m,n-1], but neither of the other terms. So what happens? Well, if m=n+1, then m-n=1, and we get to play a nice little trick
a[m+1,n] + a[m,n-1]
= exp[ iKb(m+n)+iKb ][cos(kb(m-n))sin(kb) + cos(kb)sin kb(m-n)]
+ exp[ iKb(m+n)-iKb ][cos(kb(m-n))sin(kb) + cos(kb)sin kb(m-n)]
= exp[ iKb(m+n)+iKb ][cos(kb)sin(kb(m-n)) + cos(kb)sin kb(m-n)]
+ exp[ iKb(m+n)-iKb ][cos(kb)sin(kb(m-n)) + cos(kb)sin kb(m-n)]
= exp[ iKb(m+n)+iKb ][2 cos(kb)sin(kb(m-n))]
+ exp[ iKb(m+n)-iKb ][2 cos(kb)sin(kb(m-n))]
= 2 * (exp[iKb]+exp[-iKb]) cos(kb) a[m,n]
And we get exactly the same equation as before for the energy. So everything is working.
August 4, 2005 12:26 AM
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