Timeline for Regularisation of infinite-dimensional determinants
Current License: CC BY-SA 3.0
12 events
| when toggle format | what | by | license | comment | |
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| Jan 5, 2012 at 19:58 | comment | added | joseph f. johnson | WEll, it sounds like it is worth a try, but, the question was, will this determinant have the property that its zeroes indicate the operator has a kernel? Without that property, it cannot be used to find eigenvalues the way the characteristic polynomial is for finite dimensional determinants.... | |
| Jan 5, 2012 at 2:54 | comment | added | Ron Maimon | @Joseph: If you know the derivative of the log of the determinant function with respect to $\lambda$ then you know the determinant function by integrating and exponentiating. The derivative of log(det(A-\lamdaI)) is the Green's function kernel used in Fredholm theory 1/(A-\lambda I) and you can regulate this using zeta functions. | |
| Jan 4, 2012 at 22:01 | comment | added | joseph f. johnson | This is quite intereting. It still seems that you are using the eigenvalues first and then constructing a function, one which really has nothing to do with the determinant or characteristic polynomial. but what I asked for was a regularisation of the determinant which could then be used to find the eigenvalues. For example, the zeta function regularisation makes sense even if you do not yet know the eigenvalues, and it is a function of lambda. And I asked for something that would work on a unitary operator, so obviously Fredholm determinants are not defined in this context. | |
| Jan 4, 2012 at 15:08 | comment | added | Ron Maimon | I should expand the last comment a little--- suppose the eignevalues are bunched up, so that they accumulate at a finite value. Then it is impossible to have an analytic function which is singularity free in $\lambda$ which gives the limiting characteristic determinant, because the analytic functions have well spaced zeros. But such a bunching up requires the potential to be bounded at infinity, like a H-atom, so that the eigenvalues above the accumulation point become continuous, and you get a continuous line of zeros, like the reciprocal of a function with a cut. | |
| Jan 4, 2012 at 8:39 | comment | added | Ron Maimon | @Joseph: The answer seems to be yes, I didn't work out the limit of the polynomial, but an analytic function is usually specified by its infinite set of zeros and some additional constraints, like a polynomial of infinite degree. I am not sure under what conditions the convergence is guaranteed, and convergence of the determinant is not necessary for any of the physics results, but the mathematical topic is Fredholm theory. | |
| Jan 4, 2012 at 8:27 | comment | added | joseph f. johnson | It seems, then, that the answer to my question is « no », but you are not sure. | |
| Jan 4, 2012 at 7:06 | comment | added | Ron Maimon | @Joseph: yes--- it is not obvious that the polynomial is converging, but for sure the zeros are. But when you get a certain set of zeros, you can write an Euler product formula and make an analytic function which has these zeros, and perhaps this gives a unique correct continuum notion of infinite dimensional determinant, I am not sure. I always think of it regulated. | |
| Jan 3, 2012 at 23:55 | comment | added | joseph f. johnson | you seem to be saying that it is the set of zeroes which has a limit? Not the polynomial? | |
| Jan 3, 2012 at 23:18 | history | edited | Ron Maimon | CC BY-SA 3.0 |
answer question
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| Jan 3, 2012 at 23:16 | comment | added | Ron Maimon | You mean $det(H_L - \lambda I)$. This is a finite degree polynomial in $\lambda$ whose eigenvalues are those of $H_L$. | |
| Jan 3, 2012 at 17:20 | history | edited | Ron Maimon | CC BY-SA 3.0 |
fix error
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| Jan 3, 2012 at 13:34 | history | answered | Ron Maimon | CC BY-SA 3.0 |