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added heuristic argument for the square-root-N law, thanks to Tim Kokkeler, a student in my RMT class at the Les Houches school.
Carlo Beenakker
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I interpret the question in a bit more general terms, as a request to "shed light" on the phenomenon that "oftentimes many eigenvalues lie exactly on the real axis".

The key thing to notice is that the spectrum of the Lindbladian ${\cal L}$ is invariant under complex conjugation. So if we consider a parameter dependence ${\cal L}(t)$, an eigenvalue that is on the real axis for some $t=t_0$ cannot move into the complex plane when $t$ is increased slightly. It must first collide with another real eigenvalue, to become a complex conjugated pair.

This mechanism generically produces a number $N_{\rm real}$ of real eigenvalues that scales with the square root of the total number $N_{\rm tot}$ of eigenvalues. Precise calculations can be done in the Ginibre ensemble of random matrices, where $$\mathbb{E}[N_{\rm real}]=\sqrt{2N_{\rm tot}/\pi}+{\cal O}(N_{\rm tot}^{-1}),$$ but the same square root law is observed in other ensembles (see, for example, appendix A.4 of arXiv: 1305.2924 ).

A heuristic argument for the square root law goes as follows:$^\ast$ The rate of change $dN_{\rm real}/dt=G-L$ contains loss terms and gain terms: A loss term $L$ appears when, upon increasing $t$, two real eigenvalues collide and move into the complex plane as a complex conjugate pair. A gain term $G$ appears when a complex conjugate pair of eigenvalues merges and splits on the real axis. It is natural to assume that $L\propto N_{\rm real}^2$, because it involves two independent real eigenvalues, while $G\propto N_{\rm tot}-N_{\rm real}$, since it involves only a single independent complex eigenvalue. A stationary state is reached when $L=G$, hence when $$N_{\rm real}\simeq\sqrt{N_{\rm tot}},\;\;\text{for}\;\;N_{\rm tot}\gg 1.$$


$^\ast$ I am indebted to Tim Kokkeler for this argument.

Carlo Beenakker
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