On 11/22/2012 6:38 AM, Veit Elser wrote:
On Nov 22, 2012, at 2:26 AM, Dan Asimov<dasimov@earthlink.net> wrote:
I have to agree with Brent, since any reasonable measure should respect the symmetry M -> -M (and matrices with at least one 0 eigenvalue should constitute a set of measure 0 among the Hermitians).
Nice!
--Dan That only shows the probability of 5 negative and 9 positive eigenvalues is the same as 9 negative and 5 positive. The claim 2^(-N) assumes much more, that the eigenvalues are independently distributed.
No it doesn't assume that. It only assumes that for every Hermitian matrix there is another one that has the same eigenvalues except for a reversal of all signs and that these two matrices have equal probability. Brent
In fact, they are very far from independent, making the actual probability much smaller: c^(-N^2). This result is asymptotic, for large N (the limit of interest in statistical mechanics). The number c is well known, but it is not 2.
As Andy pointed out, the probability measure should be invariant under arbitrary unitary transformations, i.e. M -> U M U^(-1). But the Hermitian matrices live in an N^2 dimensional space while the space of unitary matrices has only N(N-1) dimensions. The extra N dimensions correspond to the eigenvalues of M.
Wigner had the idea of using the maximum entropy probability distribution, constrained by just two properties: the expectation values of Tr M and Tr M^2. If we want the expectation value of Tr M to be zero, then our probability distribution is simply the Gaussian e^(-Tr M^2) times the unitary-invariant measure.
If you marginalize this distribution on just the eigenvalues (i.e. integrate out the unitary transformations) you get, say in the case of N=3 (unnormalized)
dP = e^(-E1^2-E2^2-E3^2) (E1-E2)^2 (E2-E3)^2 (E3-E1)^2 dE1 dE2 dE3.
It's the product over all eigenvalue pairs -- their differences squared -- that ruins the independence of the eigenvalue distribution. BTW, this very same distribution seems to perfectly model the distribution of Riemann zeta function zeros, but nobody understands why!
-Veit
Brent wrote:
Veit wrote:
Homework: Find the probability a random NxN complex Hermitian matrix has only positive eigenvalues. 2^(-N).
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