Find parameter values for which a 3x3 matrix has a triple eigenvalue

Question

An Exceptional point generally occurs in eigenvalue problems in which the matrix is dependent on some parameter(s). The particular point in which the eigenvalues become degenerate for the parameter(s) gives the Exceptional point. An alternate way of thinking for such points is that the real part and the imaginary part of the eigenvalues coincides at some specific parameter(s), and such parameter(s) are called Exceptional points.

I have the following 3x3 Matrix $M3$ with the following eigenvalues $e1$, $e2$, and $e3$

$$ M3=\begin{bmatrix} -i\omega+\frac{\Gamma}{2} & -ig_1 & 0\\ -ig_1 & -i\omega+\frac{\kappa_1}{2} & -ig_2\\ 0 & -ig_2 &-i\omega+\frac{\kappa_2}{2} \end{bmatrix} $$

$$ e1 = \frac{1}{3}\bigg[p+\Sigma-\frac{u_2}{2^{\frac{8}{3}}\Sigma}-3i\omega\bigg] \\ e2 =\frac{1}{3}\Bigg[p-\frac{1}{2}\bigg[\Sigma-\frac{u_2}{2^{\frac{8}{3}}\Sigma}\bigg]+\frac{i\sqrt{3}}{2}\bigg[\Sigma+\frac{u_2}{2^{\frac{8}{3}}\Sigma}\bigg]-3i\omega\Bigg]\\ e3 = \frac{1}{3}\Bigg[p-\frac{1}{2}\bigg[\Sigma-\frac{u_2}{2^{\frac{8}{3}}\Sigma}\bigg]-\frac{i\sqrt{3}}{2}\bigg[\Sigma+\frac{u_2}{2^{\frac{8}{3}}\Sigma}\bigg]-3i\omega\Bigg]\\ $$

Whereby

$$ p =\frac{\Gamma+\kappa_1+\kappa_2}{2} \\ \kappa_\pm = \frac{\kappa_1\pm\kappa_2}{4} \\ u_1 = 36g_1^{2}(3\kappa_2-2p)+\big(36g_2^{2}+(2p-3\kappa_1)(2p-3\kappa_2)\big)\big(4p-12\kappa_+)\big)\\ u_2=2^{\frac{2}{3}} \left(12 g_1^2+12g_2^2-4(p-3\kappa_+)^{2}-12\kappa_-^{2}\right)\\ \Sigma=\Bigg(\frac{u_1+\sqrt{u_1^{2}+u_2^{3}}}{16}\Bigg)^{\frac{1}{3}} $$

As one can see, the eigenvalues are quite nasty and long and it took some time obtaining them through Mathematica (it can be checked that the eigenvalues are correct since their corresponding eigenvectors diagonalizes $M3$).

My goal is to find the exceptional point(s) among the eigenvalues $e1$, $e2$, and $e3$ in terms of the parameters ($\Gamma$, $\kappa_1$, $\kappa_2$, $g_1$, $g_2$) with $\omega\rightarrow0$ for $g_1$ and/or $g_2$ (the only tunable parameters) for which $e1=e2=e3$. A straightforward attempt for solving for such a $g_1$ using Mathematica suggests that

$$ g1\rightarrow\pm\frac{\sqrt{-12g_2^{2}+4(p-3\kappa_+)^{2}+12\kappa_-^{2}}}{2\sqrt{3}} $$

which seems to imply that $u_2=0$ upon substituting the $g_1$ above into $u_2$ (for all intents and purposes, we are taking the $g_1>0$ solution). But for the same $g_1$ value, $u_1$ reduces to $$ u_1 = -(2p-3\kappa_2)^{3}+108g_2^{2}(2p-\kappa1-2\kappa2) $$

which can be zero or non-zero (depending on $g_2$).

Here's the problem: From a purely mathematical perspective (without relying on Mathematica), the eigenvalue equations do not appear to be degenerate when $u_2=0$. Referring to $\Sigma$, setting $u_2=0$ gives

$$ \Sigma=\frac{1}{2}(u_1)^{\frac{1}{3}} $$

So the eigenvalues $e1$, $e2$, and $e3$ gives (when $\omega\rightarrow0$):

$$ e1=\frac{1}{3}\bigg[p+\frac{1}{2}(u_1)^{\frac{1}{3}}\bigg]\\ e2=\frac{1}{3}\bigg[p+e^{-i\frac{\pi}{3}}\frac{(u_1)^{\frac{1}{3}}}{2}\bigg]\\ e3=\frac{1}{3}\bigg[p+e^{i\frac{\pi}{3}}\frac{(u_1)^{\frac{1}{3}}}{2}\bigg] $$

which shows that the eigenvalues are not degenerate. Well, unless, $u_1=0$ in which case $e1=e2=e3=\frac{p}{3}$ and we would have found our exceptional point. However, setting $u_1=u_2=0$ leads to indeterminate issues since when $u_1=u_2=0$, $\Sigma\rightarrow0$ and the term $\frac{u_2}{2^{\frac{8}{3}}\Sigma}\rightarrow\frac{0}{0}$ in $e1$, $e2$, and $e3$. What is the problem here? How should I go about finding such parameters of $g_1$ and/or $g_2$ for which the eigenvalues are degenerate?

Edit: I tried to plot the real part and imaginary part of the eigenvalues as a function of $g_1$ to use it to guide me in finding the exceptional point in the expressions. It's clear that there must exist at least one exceptional point (at approximately $g_1=0.64$): Real part of eigenvalues vs $g_1$, Imaginary part of eigenvalues vs $g_1$ but I am unable to recover the expression for the exceptional point for $g_1$ in terms of the other parameters. How should I approach this overall issue? Any thoughts?

Answer 1

Expand $(M3 - \frac13 \operatorname{Tr}(M3))^3 = 0$. This gives you simpler equations than the ones you are using.

Note that this equation must hold in the triple point, because $M3 - \frac13 \operatorname{Tr}(M3)$ has only the zero eigenvalue (with a Jordan block of size at most 2).

Setting $a=\frac{\Gamma}{2}-\frac{\Gamma + \kappa_1 + \kappa_2}{6}$, $b=\frac{\kappa_2}{2}-\frac{\Gamma + \kappa_1 + \kappa_2}{6}$ (so that the diagonal of $M3 - \frac13 \operatorname{Tr}(M3)$ is $a,-a-b,b$), the (1,1) and (3,3) entries of $$(M3 - \frac13 \operatorname{Tr}(M3))^3 = 0$$ give two single-variate equations $$a^3 - ag_1^2 + bg_1^2 = b^3 + ag_2^2 - bg_2^2 = 0,$$ which one can solve to get $$ g_1 = \pm a\sqrt{\frac{a}{a-b}}, \quad g_2= \pm b \sqrt{\frac{-b}{a-b}}. $$ One can check (I did it using a CAS) that after these substitutions all the other matrix entries are zero, too. So there is indeed a triple eigenvalue.

With the data in your plot of $\Gamma=0.01,\kappa_1=2.0,\kappa_2=3.25$, one gets $$ g_1 = \pm 0.639393402070183,\quad g_2 = \pm 0.508610273905719, $$ which is not too far from your guessed value of $g_2\approx 0.5$.

Answer 2

Equating $(\tfrac{1}{3}{\rm tr}\,M)^3={\rm det}\,M$ for $\omega=0$ gives $$g_1=\pm\tfrac{1}{6}(3\kappa_2)^{-1/2}\sqrt{\Gamma^3+3 \Gamma^2 (\kappa_1+\kappa_2)+3 \Gamma \left(\kappa_1^2+\kappa_2^2-7 \kappa_1 \kappa_2-36 g_2^2\right)+(\kappa_1+\kappa_2)^3}.$$ As a test, I took the values from your plot, $\Gamma=0.01,\kappa_1=2,\kappa_2=3.25,g_2=0.5$, and find $g_1=0.639414$, in agreement with your estimate that $g_1\approx 0.64$.

Note that this formula for $g_1$ is different from the one in the OP, and no longer implies that $u_2=0$, hopefully resolving the contradiction.