# Stability of eigenvalues

1. Jul 16, 2013

### hilbert2

Suppose we have a matrix A that has eigenvalues λ1, λ2, λ3,... Matrix B is a matrix that has "very small" matrix elements. Then we could expect that the eigenvalues of sum matrix A + B would be very close to the eigenvalues λi. But this is not the case. The eigenvalues of a matrix are not necessarily stable with respect to small changes in the matrix elements, and a relatively small change in them could displace the eigenvalues considerably in the complex plane. This is because the eigenvalues are the zeroes of characteristic polynomial, and the roots of a polynomial are not always stable with respect to small changes in the coefficients of its terms.

For hermitian matrices one can prove (AFAIK) a theorem that states that a small perturbation that keeps the matrix hermitian causes a correspondingly small change in the eigenvalues.

Does this property also always hold in the infinite-dimensional case, i.e. finding the eigenvalues of a hermitian operator acting in Hilbert space? Suppose we have a hamiltonian operator for a macroscopic system of very many (like NA) particles interacting by the Lennard-Jones potential for example. Does a small change in the LJ parameters necessarily result in a small change in the predicted thermodynamic properties of the system?

I was reading about the anthropic principle and an article listed the lower density of ice as compared to the density of liquid water as an "anthropic coincidence". Is it possible that a very small change in values of fundamental constants (Planck constant, elementary charge,...) could remove this special property of water?

2. Jul 16, 2013

### tom.stoer

I think the problem is how to define "small perturbation" for an operator. This cannot be done by looking at a small parameter only. In the finite dimensional example one could write A+B = A+εb with small ε and study the solutions λ(ε) of |A+εb - λ1| = 0 as functions of ε for small ε≈0.

In the infinite dimensional example I don't see a general approach. Look the Hamiltonians H = p2 and H(ε) = p2 + εx2. They do not even exist in the same Hilbert space, and there is no smooth limit ε → 0 for the spectrum.

3. Jul 16, 2013

### hilbert2

Yeah, when ε=0 the spectrum is continuum and otherwise its discrete...

4. Jul 16, 2013

### tom.stoer

So for an Hamiltonian H(ε) = H0 + ε H' we have to find necessary and sufficient conditions for a smooth limit ε → 0 for spectrum and eigenstates.