Understanding Functional Determinants in Mathematics

[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)

 

[Gokul43201]

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

 

[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.

 

[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.

 

[enig_76]

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.

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

 

[Angryphysicist]

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

On the concept of determinant for the differential operators of Quantum Physics http://arxiv.org/abs/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.