--- mcintosh@servidor.unam.mx wrote:
Quoting Eugene Salamin <gene_salamin@yahoo.com>:
I had come to the same conclusion myself. If a simple cosine potential leads to complicated Mathieu function solutions, then the appearance of Mathieu functions in tr(M) seems unavoidable, and the full machinery of solving the differential equation should be required just to bring these Mathieu functions into existence.
Are Mathieu functions such bad little critters? Mainly they just don't get used much unless you specialize in the things we are talking about. They´re just lumpy sines and cosines.
I'm just speaking complexity theoretically here. The machinery needed to go from rational functions to logarithms is integration.
Assuming I can't avoid solving the differential equation, rewrite it as a first order matrix differential equation. Let U(x) be the column vector [u(x) u'(x)]. Then U'(x) = K(x) U(x) with
K(x) = [0 0] [ <--- 1,1 element is 1] [V(x)-E 1].
That was indeed a typo. I should have said K(x) = [0 1] [V(x)-E 0].
The wave function can be written symbolically as U(x) = G(x) U(0), but because the commutator [K(x1),K(x2)] is nonzero, the expression of G(x) in terms of V(t), 0<=t<=x, is nontrivial. Indeed, we know that if V(x) is a cosine, G(x) has Mathieu functions.
This show up in my treatment where G is written as a sum, the nice half is solved, but then it must be used to transform the second half. Sometimes the process can be repeated, usually not. That is, it always can be, but the result may not be pleasant. With the Dirac Harmonic Oscillator you get some nice spirals.
If V(x) is approximated by a step function, then we can integrate over each step. U(x2) = G0(x2-x1) U(x1), with [...]
Authors have suffested this, I have done it myself. Even though the half intervals are readily soluble, you may need more intervals, and then the hyperbolic trigonometry becomes oppressive, and the convergence may not be as good as for other methods. But I don't know than anyone
has made a really systematic study. It works well for the Kronig-Penny combs, where your lattice is made up of delta functions, and that is good enough to deduce bands.
- hvm
For Kronig-Penny, or more generally, when V(x) is stepwise constant plus delta functions (finitely many of each), my method exactly integrates the Schroedinger equation. An alternative to increasing the number of intervals is to approximate V(x) by piecewise linear. Then the hyperbolic/trig functions in G0 are replaced by Airy functions. Better yet, use a professional numerical differential equation solver. Gene __________________________________ Do you Yahoo!? New and Improved Yahoo! Mail - 100MB free storage! http://promotions.yahoo.com/new_mail