given the functional integral with 'g' small coupling constant(adsbygoogle = window.adsbygoogle || []).push({});

[tex] \int \mathcal D [\phi]exp(iS_{0}[\phi]+\int d^{4}x \phi ^{k}) [/tex]

so k >2 then could we use a similar 'Functional determinant approach' to this Feynmann integral ?? in the sense that the integral above will be equal to

[tex] Cx(Det[\partial _{\mu}+k\phi^{k-1})^{-1/b} [/tex]

where C and 'b' are constant and the determinant is defined as an infinite product of eigenvalues

[tex] \zeta (s) \Gamma(s)= \int_{0}^{\infty}dt t^{s-1}Tr[e^{-sH_{0}-sgV_{int}] [/tex]

where the index '0' means the quadratic part of our Hamiltonian /action and so on..

since 'g' is small then we can express for every eigenvalue:

[tex] \lambda _{n} =\lambda _{0}+g\lambda _{n}^{1} +g^{2}\lambda_{n}^{2}+........ [/tex]

so [tex] det= \prod_{n} \lambda_{n} [/tex]

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# Functional determinant approach to perturbation

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