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Functional determinant

  1. Aug 3, 2007 #1
    I say what is a functional determinant ??

    for example [tex] Det( \partial ^{2} + m) [/tex]

    is this some kind of Functional determinant?

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

    (the determinant of a Matrix is the product of its eigenvalues)
     
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  3. Aug 3, 2007 #2

    Gokul43201

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    We're talking about an operator with a continuous spectrum, right?
     
  4. Aug 4, 2007 #3

    Chronos

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    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
  5. Aug 4, 2007 #4
    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

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

    if 'A' is an operator Det(A) would be some kind of Functional determinant, which is just the multiplication (after regularization) of its Eigenvalues.
     
  6. Aug 8, 2008 #5
    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
     
  7. Aug 23, 2008 #6
    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. [tex]\zeta(s)=\sum_{n}\lambda_{n}^{-s}[/tex]) 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.
     
    Last edited by a moderator: Apr 23, 2017
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