# Functional determinant approach to perturbation

1. Aug 29, 2007

### 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 Feynmann 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 $$det= \prod_{n} \lambda_{n}$$

2. Aug 29, 2007

### Emmanuel114

Um,

3. Aug 30, 2007

### olgranpappy

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

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'

4. Aug 30, 2007

### 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

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

5. Aug 30, 2007

### Emmanuel114

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?

6. Aug 30, 2007

### olgranpappy

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?

7. Aug 30, 2007

### Sangoku

8. Aug 30, 2007

### Emmanuel114

Thanks, sangoku, looks interesting!