--- mcintosh@servidor.unam.mx wrote:
Quoting Eugene Salamin <gene_salamin@yahoo.com>:
Now for my question. Suppose I'm only interested in the energy bands, and I don't care about the actual wave function u(x). Then, given some value of E, is there a way to calculate tr(M(E)) without having to solve the differential equation?
As far as I know, which may not be all that much, there is no shortcut.
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.
However, depending on your potential, you may be able to develop a good approximation and relate your answer to the stability chart of the Mathieu functions. That would work if your potential were close to a cosine and splitting the coefficient matrix would give a new equation for the coefficients of the desired solution in terms of Mathieu functions. But if the new coefficient were small, a couple of terms of the power
series for the new solution might suffice.
I've done this for other equations, but never tried it in the Mathieu environment.
- hvm
Could you elaborate a bit more about this "splitting"? 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] [V(x)-E 1]. 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. If V(x) is approximated by a step function, then we can integrate over each step. U(x2) = G0(x2-x1) U(x1), with G0(s) = exp(sK) = cosh(s sqrt(V-E)) + (sinh(s sqrt(V-E))/sqrt(V-E)) K. Then, for 0 < x1 < ... < xn < x, we have the approximation G(x) = G0(x-xn) ... G0(x2-x1) G0(x1-0). If we take as basis in the solution space the functions u1(x), u2(x) such that u1(0) = 1 = u2'(0), u1'(0) = 0 = u2(0), then, in the notation of my previous message, M = G(P). Gene __________________________________ Do you Yahoo!? Yahoo! Mail is new and improved - Check it out! http://promotions.yahoo.com/new_mail