5
$\begingroup$

In high energy physics we oftentimes encounter the following problem. For a given parametrized matrix $\{H_{ij}(\Lambda)\}$, we know that in the limit $\Lambda\to\infty$ some of its entries become large:

$$ \text{For some $(i,j)$: }H_{ij}(\Lambda)\to\infty\text{ as }\Lambda\to\infty. $$

We also know that in the limit $\Lambda\to\infty$ the lowest eigenvalues in the spectrum of $\{H_{ij}(\Lambda)\}$ — which we are most interested in — tend to converge to particular values (while the larger eigenvalues may potentially diverge).

Have we had access to an "infinite-precision" computer, we could just diagonalize the matrix for increasing values of $\Lambda$ extrapolate to $\Lambda\to\infty$. In reality, this is impractical.

I am wondering if there exist any well-known ways to deal with such a situation, at least for some specific cases (such as divergent elements staying on the diagonal and matrix being sparse).

$\endgroup$
7
  • $\begingroup$ Isn't a basis where the entries are not divergent, even though the eigenvalues are not bounded above. $\endgroup$
    – nicoguaro
    Commented Jun 9, 2022 at 22:35
  • $\begingroup$ I'm afraid that finding such a basis may be as hard as finding these eigenvalues. $\endgroup$
    – mavzolej
    Commented Jun 9, 2022 at 22:51
  • $\begingroup$ Write the eigenvalues as a function of $\Lambda$, and then do extrapolation techniques or Taylor expansion in $1/\Lambda$. $\endgroup$ Commented Jun 10, 2022 at 23:08
  • $\begingroup$ How can we do this analytically for a matrix of size larger than $4\times 4$? In reality, the matrix is VERY large, even though its entries as analytic functions of $\Lambda$ are known. Doing this numerically this also is not an option as choosing the $\Lambda$-divergent entries to be much larger than others leads to computational issues. $\endgroup$
    – mavzolej
    Commented Jun 11, 2022 at 5:48
  • $\begingroup$ Is the matrix symmetric? What do you mean by "lowest" otherwise? $\endgroup$ Commented Jun 12, 2022 at 15:32

0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Browse other questions tagged or ask your own question.