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By ill-conditioned I mean it in the mathematical sense that the condition number of the Hamiltonian is large.

One interesting note is that the condition number of a matrix can be written in terms of its singular values as, $$ \kappa(H)=\frac{\sigma_{max}}{\sigma_{min}} $$ But, since the Hamiltonian is Hermitian, the singular values are the same as the absolute values of the eigenvalues, so we have, $$ \kappa(H)=\frac{|\lambda_{max}|}{|\lambda_{min}|} $$ But, in general, there might be infinitely many eigenvalues, so many Hermitian matrices should be ill-conditioned, but they do not seem to be.

So, I have two questions:

  1. How does one determine whether or not a matrix with infinitely many eigenvalues is ill-conditioned?
  2. Are there physical contexts in which it is obvious the Hamiltonian is ill-conditioned because the use of perturbation theory leads to clearly incorrect solutions?

I care a bit more about this second question. One sign of a matrix being ill-conditioned is that a slight perturbation of the matrix leads to a very large change in its eigenvalues. This is obviously bad news for perturbation theory, which operates on this principle, if one is dealing with an ill-conditioned Hamiltonian.

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    $\begingroup$ It seems to me that you're misinterpreting what ill-conditioning means. It entails that small changes to the matrix $A$ can have large changes on $Ax=b$ for some suitable $x$, not that small changes to the matrix change its eigenvalues. Maybe I did lose a beat, though, so: what makes you think the latter is true? $\endgroup$ Feb 28, 2018 at 18:59
  • $\begingroup$ Hmmm... I suppose what I'm thinking is that we are generally dealing with eigenvalue problems so if we slightly perturb $H$, then in perturbation theory $x$ should still be "close" to being an eigenvector of $H$. I am asking if and when this is not the case. In other words, if $b=\lambda x$, and I perturb $H$ to $H'$, then is $b'=\lambda ' x$ close to $b$. That's the connection to eigenvalues I'm imagining. I might be wrong though. $\endgroup$
    – jheindel
    Feb 28, 2018 at 19:06
  • $\begingroup$ The statements "$x$ is close to being an eigenvector of $H$" and "$x$ is close to another eigenvector $x'$ of $H$" are very different. The latter has a very precise meaning ($|x-x'|$ is small) but the former is fuzzy and it ultimately doesn't mean much. If $H$ is ill-conditioned, $Hx$ might change, but that doesn't say anything about the existence of an $x'$ close to $x$ which is now an eigenvalue. Perturbation theory deals with the latter, not the former. $\endgroup$ Feb 28, 2018 at 19:19

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