Nail-Tinted Glasses

Picture worth a thousand words

via Daily Kos, but the picture deserves more play...

from http://icasualties.org/

Interlude

So we've got a couple demonstrations that Feynman's solution is correct; what does this tell us about going forward?

My feeling is that the important observation along the way was expressing the coefficient a[m,n] as

a[m,n] = exp[ ib( (k2)m+(k1)n ) ] - exp[ ib( (k1)m+(k2)n ) ]

or, putting it in a form more akin to 15.18

a[m,n] = exp[ib(k2)m]exp[ib(k1)n] - exp[ib(k1)m]exp[ib(k2)n]

The first term is just two completely independent particles coasting along; the second is... well, it's really just the same thing, isn't it, with the relationship between the indicies flip flopped. There's no inherent reason why k2 and m should go together (after all, spin down looks the same everywhere).

In other words, understanding what's going on you could almost guess the solution - you are looking for a superposition of two possible solutions, so you stick some variable coefficients in front, and let the boundary value calculations tell you what the coefficients must be.

Or alternatively you eyeball it, based on symmetry arguments, and double check at the end that you've gotten the sign right (which seems much more in the spirit of Feynman.

Checking the solution, part 2

In the first pass of verifying Feynman's solution, we took a moment at the very end to convert part of the expression from cosines to exponentials. Does solving the problem get any easier if we do that right away?

Again, 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

and when we clean it up a bit, it becomes:
a[m,n] = exp[ iKb(m+n) ] sin kb|m-n|

converting this to an exponential form, we get
a[m,n] = exp[ iKb(m+n) ] ( exp[ ikb|m-n| ] - exp[ -ikb|m-n| ] ) / 2i

Now, the 1/2i here is just a multiplicative factor; since we are solving a linear equation, we can just drop it out of consideration.

a[m,n] = exp[ iKb(m+n) ] ( exp[ ikb|m-n| ] - exp[ -ikb|m-n| ] )

Once again, we restrict our inspection to the states where m>n, and go to work

a[m,n]
= exp[ iKb(m+n) ] ( exp[ ikb(m-n) ] - exp[ -ikb(m-n) ] )
= exp[ ib(Km+Kn+km-kn) ] - exp[ ib(Km+Kn-km+kn) ]
= exp[ ib(K+k)m ]exp[ ib(K-k)n] - exp[ ib(K-k)m ]exp[ ib(K+k)n ]
= exp[ ib(k2)m ]exp[ib(k1)n] - exp[ib(k1)m]exp[ib(k2)n]
= exp[ ib( (k2)m+(k1)n ) ] - exp[ ib( (k1)m+(k2)n ) ]

to keep things uncluttered, we define
A = exp[ ib( (k2)m+(k1)n ) ]
B = exp[ ib( (k1)m+(k2)n ) ]

and so it follows that
a[m,n] = A - B
a[m+1,n] = exp[ib(k2)]A - exp[ib(k1)]B
a[m,n+1] = exp[ib(k1)]A - exp[ib(k2)]B
a[m-1,n] = exp[-ib(k2)]A - exp[-ib(k1)]B
a[m,n-1] = exp[-ib(k1)]A - exp[-ib(k2)]B

and again, rearranging produces 2 cos b(k2) + 2 cos b(k1) when m and n are apart. When they are adjascent, the middle terms go away. Then what?

a[m+1,n] = exp[ib(k2)]A - exp[ib(k1)]B
a[m,n-1] = exp[-ib(k1)]A - exp[-ib(k2)]B

a[m+1,n] + a[m,n-1]
= exp[ib(k2)]A - exp[ib(k1)]B + exp[-ib(k1)]A - exp[-ib(k2)]B

= exp[ib(k2)]A - exp[ib(k2)]B + exp[ib(k2)]B
- exp[ib(k1)]B + exp[ib(k1)]A - exp[ib(k1)]A
+ exp[-ib(k1)]A - exp[-ib(k1)]B + exp[-ib(k1)]B
- exp[-ib(k2)]B + exp[-ib(k2)]A - exp[-ib(k2)]A

= exp[ib(k2)]A - exp[ib(k2)]B + exp[ib(k2)]B
+ exp[ib(k1)]A - exp[ib(k1)]B - exp[ib(k1)]A
+ exp[-ib(k1)]A - exp[-ib(k1)]B + exp[-ib(k1)]B
+ exp[-ib(k2)]A - exp[-ib(k2)]B - exp[-ib(k2)]A

And here's the joke - when m=n+1:
exp[ib(k2)]B = exp[ib(k1)]A
so the last term in each row is cancelled away, leaving

= exp[ib(k2)]A - exp[ib(k2)]B
+ exp[ib(k1)]A - exp[ib(k1)]B
+ exp[-ib(k1)]A - exp[-ib(k1)]B
+ exp[-ib(k2)]A - exp[-ib(k2)]B

= 2 cos (k1)b + 2 cos (k2)b

Checking the solution, part 1

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.

Overview of Spin Waves

In Feynman's lecture on the independent particle approximation (vol III, ch 15 - available in volume 4 of the audio lectures), he describes spin waves. Spin waves are one kind of boojum that can propagate through a crystal lattice.

The fun problem that he glosses over is the interaction of two spin waves in a single lattice. In the lecture he remarks that he was able to solve it, but didn't bring the solution with him. The text contains the exact solution in a parenthetical remark, with no further discussion of where the solution comes from.

So the next few posts will be describing my adventures in understanding what's going on.

Half Angles

Working on yet another physics exercise (more on that later, I hope), I found myself needing the trigonometric formulas for half angles.

Demonstration below the fold....

Aside: long ago, I demonstrated some of the consequences of the Law of Cosines.

One of those consequences is the double angle formula for cosines
cos( 2x )
= cos ( x - (-x) )
= cos(x)cos(-x) + sin(x)sin(-x)
= cos(x)^2 - sin(x)^2
= cos(x)^2 - ( 1 - cos(x)^2)
= 2cos(x)^2 - 1
= 2(1-sin(x)^2) - 1
= 1 - 2sin(x)^2

Now, let x = y/2. We then have the following equalities
cos(y)
= 2cos(y/2)^2 - 1
= 1 - 2sin(y/2)

rearranging each of these gives a half angle formula.
cos(y/2) = ( [ 1 + cos(y) ] / 2 ) ^1/2
sin(y/2) = ( [ 1 - cos(y) ] / 2 ) ^1/2

These can be verified using the obvious tests ( y=0, y=π, 1=cos(y/2)^2 + sin(y/2)^2). We didn't use sin double angle formula, so that's another fair test
2cos(y/2)sin(y/2)
= 2 ( [ 1 + cos(y) ] / 2 ) ^1/2 ( [ 1 - cos(y) ] / 2 ) ^1/2
= 2 ( [ 1 + cos(y) ] [ 1 - cos(y) ] / [2*2] ) ^1/2
= ( 1 - cos(y)^2 ) ^1/2
= ( sin(y)^2 ) ^ 1/2
= sin(y)

Keep in mind that you have to exhibit some care over the sign of the result; the angle 2π = the angle 0, but 2π/2 != 0/2.