Functional determinant: Difference between revisions

Revision as of 17:46, 20 October 2011

Revision as of 10:58, 16 January 2012

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:<math> \Delta(m) = \frac{\psi_1^m(L)}{\psi_2^m(L)}, </math>

which is also a meromorphic function of ''m'', we see that it has exactly the same poles and zeroes as the quotient of determinants we are trying to compute: if ''m'' is an eigenfunctioneigenvalue of the operator number one, then ψ^''m''₁(''x'') will be an eigenfunction thereof, meaning ψ^''m''₁(''L'') = 0; and analogously for the numerator. By [[Liouville's theorem (complex analysis)|Liouville's theorem]], two meromorphic functions with the same zeros and poles must be proportional to one another. In our case, the proportionality constant turns out to be one, and we get

:<math> \frac{\det \left(-\frac{d^2}{dx^2} + V_1(x) - m\right)}{\det \left(-\frac{d^2}{dx^2} + V_2(x) - m\right)} = \frac{\psi_1^m(L)}{\psi_2^m(L)} </math>