Gelfand-Yaglom theorem for functional determinants

What is the 'Gelfand-Yaglom' Theorem? I have heard that it is used to calculate Functional determinants by solivng an initial value problem of the form

$Hy(x)-zy(x)=0$ with $y(0)=0$ and $y'(0)=1$. Here $H$ is the Hamiltonian and $z$ is a real parameter.

Is it that simple? If $H$ is a Hamiltonian, could I use the WKB approximation to solve the initial value problem and to be valid for $z$ big?

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The OP is apparently already aware of the Gelfand-Yaglom method judging from page 3 in his own paper vixra.org/abs/1007.0005 –  Qmechanic Jul 16 '11 at 18:24
Jose, would you care to explain Qmechanic's comment? –  David Z Jul 17 '11 at 4:23
i used it to compute a proposed Hamiltonian operator ... but i did not invent it :) i had some doubts about it , if you want erase the question :S i did not want to importunate –  Jose Javier Garcia Jul 20 '11 at 14:59
Basically, it's a way of calculating the determinant of a 1-dimensional operator $\det(H)=\prod_i \lambda_i$ with out calculating, let alone multiplying, any of its eigenvalues $H \psi_i = \lambda_i \psi_i$.
To state the original theorem: assume that you have a Schrodinger operator (or Hamiltonian) $H = -\frac{d^2}{d x^2} + V(x)$ on the interval $x\in[0,L]$ with Dirichlet boundary conditions: $$H \psi_i(x) = \lambda_i \psi_i(x) \,, \quad \psi(0)=\psi(L)=0 \ .$$ Then we can compute its determinant by solving the related initial value problem $$H \phi(x) = 0\,, \quad \phi(0)=0\,,\quad \phi'(0) = 1 \ ,$$ so that $$\det H \approx \phi(L) \,,$$ where the final result is only $\approx$ as we can only really calculate the ratio of two determinants.