Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
The OP is apparently already aware of the Gelfand-Yaglom method judging from page 3 in his own paper – 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

I was at a talk a while back by Gerald Dunne where he talked about the Gelfand-Yaglom theorem. He used it for calculating some Euler-Heisenberg type effective actions. A paper of his with Hyunsoo Min on the subject is A comment on the Gelfand–Yaglom theorem, zeta functions and heat kernels for PT-symmetric Hamiltonians and he's got some nice lecture notes: Functional Determinants in Quantum Field Theory (also see a wider spanning set of lectures of the same name).

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.

This basic result can be generalised to more general boundary conditions, coupled systems of ODEs and higher order linear ODEs.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.