Functional determinant approach to perturbation

[Sangoku]

given the functional integral with 'g' small coupling constant

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

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

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

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

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

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:

$$\lambda _{n} =\lambda _{0}+g\lambda _{n}^{1} +g^{2}\lambda_{n}^{2}+...$$

so $$det= \prod_{n} \lambda_{n}$$

 

[Emmanuel114]

Um,

shouldn't your infinite product have some additional conditions to ensure convergence?

 

[olgranpappy]

given the functional integral with 'g' small coupling constant

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

I don't see any 'g' in the above...

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

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

It appears that the functional integral you wrote can not be equal to the above determinant because the above determinant appears to depend on 'phi' whereas the value of the functional integral does not depend on 'phi'

 

[Sangoku]

To 'Emmanuelle 14' = product DIVERGES however using zeta regularization you can attach a finite value to it equal to $$exp(-\zeta (0))$$

My question 'Olgranppapy' is to know if there is a generalization of the Functional determinant to the case of Non-linear operators as the produt of the eigenvalues for n=01,2,3,4,... given by

$$\Delta \phi +\phi^{k-1}k=\lambda_{n}\phi = \partial _{t} \phi$$

 

[Emmanuel114]

To 'Emmanuelle 14' = product DIVERGES however using zeta regularization you can attach a finite value to it equal to $$exp(-\zeta (0))$$

Are you saying $$det = \prod_{n} \lambda_{n} = exp(-\zeta (0))$$ ?

How did you get $$\prod_{n} \lambda_{n} = exp(-\zeta (0))$$ to work? The article at http://en.wikipedia.org/wiki/Zeta_regularization doesn't say anything about it. Do you have another reference to zeta regularization?

 

[olgranpappy]

My question 'Olgranppapy' is to know if there is a generalization of the Functional determinant to the case of Non-linear operators...

Perhaps formally... I don't know--But dealing with non-linear terms (interactions) is always the whole problem in every problem in physics, isn't it?

 

[Sangoku]

You can see at http://mathworld.wolfram.com/Zeta-RegularizedProduct.html

Or ... http://fr.wikipedia.org/wiki/Noyau_de_la_chaleur

In French (although it can be understood easily) :)

when you digest the articles try searching in google 'Heat kernel regularization' or 'Zeta egularization and path integrals' ..there is a good and comprehensible article by S. Hawking... good luck 'Emmanuelle'

 

[Emmanuel114]

Thanks, sangoku, looks interesting!