Functional determinant: Difference between revisions

Revision as of 17:03, 6 March 2021

Revision as of 17:09, 6 March 2021

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:<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)} </math>

where ''m'' is a [[complex number|complex]] constant. This expression is a [[meromorphic function]] of ''m'', having zeros when ''m'' equals an eigenvalue of the operator with potential ''V''₁(''x'') and a pole when ''m'' is an eigenvalue of the operator with potential ''V''₂(''x''). We now consider the functions ''ψ^{''{{su|p=''m''}_|b=1}} and ''ψ^{''{{su|p=''m''}_|b=2}} with

:<math> \left(-\frac{d^2}{dx^2} + V_i(x) - m\right) \psi_i^m(x) = 0 </math>

Line 108:

:<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 eigenvalue of the operator number one, then {{nowrap|''ψ^{''{{su|p=''m''}_|b=1}}(''x'')}} will be an eigenfunction thereof, meaning {{nowrap|1=''ψ^{''{{su|p=''m''}_|b=1}}(''L'') = 0}}; and analogously for the denominator. 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>