# Functional determinant

1. Aug 3, 2007

### Klaus_Hoffmann

I say what is a functional determinant ??

for example $$Det( \partial ^{2} + m)$$

is this some kind of Functional determinant?

then i also believe (althouhg it diverges ) that $$Det( \partial ^{2} + m)= \lambda_{1}\lambda_{2}\lambda_{3}\lambda_{4}.....$$

(the determinant of a Matrix is the product of its eigenvalues)

2. Aug 3, 2007

### Gokul43201

Staff Emeritus
We're talking about an operator with a continuous spectrum, right?

3. Aug 4, 2007

### Chronos

My gut reaction is in agreement with Gokul's. Your assumption is risky, but interesting. Can you expand upon it? I disagree with your closing remark, but may have taken in out of context.

Last edited: Aug 4, 2007
4. Aug 4, 2007

### Klaus_Hoffmann

Wouldn't call it 'risky' is just a direct generalization of what a Determinant should be on oo-dimensional spaces.

In fact an operator (integral or differential) is nothing but a oo-dimensional matrix with Trace and determinant, i believe these 'FUnctional determinants' appear when evaluating Gaussian integrals

$$\int \mathcal D[x]e^{ixAx^{T}}=B/ Det(A)$$

if 'A' is an operator Det(A) would be some kind of Functional determinant, which is just the multiplication (after regularization) of its Eigenvalues.

5. Aug 8, 2008

### enig_76

I noticed your comment that from the point of view of operator theory, an integral or derivation is nothing by an oo-dimensional matrix. I can intuitively come close to that conclusion, but I would love to see a more rigorous mathematical treatment, can you post one or recommend a good introductory book giving such a treatment?

Thanks

6. Aug 23, 2008

### Angryphysicist

Here's the paper I learned something like this from:

On the concept of determinant for the differential operators of Quantum Physics arXiv:hep-th/9906229

It requires using the zeta trace of the function (i.e. $$\zeta(s)=\sum_{n}\lambda_{n}^{-s}$$) when you take it's derivative and then set s=0, you get the sum of the logarithms of the eigenvalues of the operator. You exponentiate this, and you get your determinant.