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I am able to accurately calculate the eigenvalues in perturbation theory, as follows:

I perform a similarity transformation on the matrix $M$, given by $M'=S^{-1}MS$ with _{$$S=\frac{1}{2m}\left( \begin{array}{cccc} 0 & 2 m & -\sqrt{m^2+4 n^2}+m-2 n & \sqrt{m^2+4 n^2}+m-2 n \\ 0 & 0 & -\sqrt{m^2+4 n^2}+m-2 n & \sqrt{m^2+4 n^2}+m-2 n \\ 2 m & 0 & 2 m & 2 m \\ 0 & 0 & 2 m & 2 m \\ \end{array} \right).$$} The matrix $M'$ has the same eigenvalues as $M$ and is given by $M'=M_0+M_1$, with _{$$M_0=\left( \begin{array}{cccc} 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & \frac{1}{2} \left(-\sqrt{m^2+4 n^2}+m+2 n\right) & 0 \\ 0 & 0 & 0 & \frac{1}{2} \left(\sqrt{m^2+4 n^2}+m+2 n\right) \\ \end{array} \right),$$ $$M_1=\frac{1}{\sqrt{m^2+4 n^2}}\left( \begin{array}{cccc} -2 \sqrt{m^2+4 n^2} & \sqrt{m^2+4 n^2} & -\frac{\sqrt{m^2+4 n^2} \left(\sqrt{m^2+4 n^2}-m+2 n\right)}{2 m} & \frac{\sqrt{m^2+4 n^2} \left(\sqrt{m^2+4 n^2}+m-2 n\right)}{2 m} \\ \sqrt{m^2+4 n^2} & -\sqrt{m^2+4 n^2} & \sqrt{m^2+4 n^2} & \sqrt{m^2+4 n^2} \\ \sqrt{m^2+4 n^2}-2 n & \frac{1}{2} \left(\sqrt{m^2+4 n^2}-m-2 n\right) & -\frac{3}{2} \sqrt{m^2+4 n^2}-\frac{m}{2}+n & -\frac{1}{2} \sqrt{m^2+4 n^2}-\frac{m}{2}+n \\ \sqrt{m^2+4 n^2}+2 n & \frac{1}{2} \left(\sqrt{m^2+4 n^2}+m+2 n\right) & \frac{1}{2} \left(-\sqrt{m^2+4 n^2}+m-2 n\right) & \frac{1}{2} \left(-3 \sqrt{m^2+4 n^2}+m-2 n\right) \\ \end{array} \right).$$}

Now we note that for large $m,n$ of order $1/\epsilon$, the matrix $M_0$ is of order $1/\epsilon$ while the matrix $M_1$ is of order unity. We can therefore perform perturbation theory in the small parameter $\epsilon$.

Let me first look at the positive eigenvalues. To zeroth order these are given by $\beta_\pm=\frac{1}{2} \left(\pm \sqrt{m^2+4 n^2}+m+2 n\right)$. To first order these are the $(3,3)$ and $(4,4)$ diagonal elements of $M_0+M_1$, which gives the two positive eigenvalues $$\beta_\pm=\frac{1}{2} \left(m+2 n\pm\sqrt{m^2+4 n^2}\right)\pm\frac{m-2 n \mp 3 \sqrt{m^2+4 n^2}}{2 \sqrt{m^2+4 n^2}}.$$ This expression for the positive eigenvalues is quite accurate, see the plot for $m=n$, where the exact expressions (red and green lines) are almost indistinguishable from the perturbative result (blue and orange).

Now for the negative eigenvalues. To first order these are given by the eigenvalues of the $2\times 2$ upper-left block of $M_0+M_1$, so $\alpha_1,\alpha_2$ are the eigenvalues of the matrix ${{-2\;1}\choose{1\;-1}}$, given by $$\alpha_1=\frac{1}{2} \left(-\sqrt{5}-3\right)=-2.618,\;\;\alpha_2=\frac{1}{2} \left(\sqrt{5}-3\right)=-0.382.$$

Here is the plot for $m=n$, again red and green is exact, blue and orange the perturbative result.

So to first order in $\epsilon$ one has $\alpha_1+\alpha_2=-3$. The challenge is to prove that higher order corrections are negative. Incidentally, the restriction to integer $m,n$ does not seem to play any role (in the plot $m=n$ is varied continuously).

I am able to accurately calculate the eigenvalues in perturbation theory, as follows:

I perform a similarity transformation on the matrix $M$, given by $M'=S^{-1}MS$ with _{$$S=\left( \begin{array}{cccc} 1 & 1 & \frac{-\sqrt{m^2+4 n^2}+m-2 n}{2 m} & \frac{\sqrt{m^2+4 n^2}+m-2 n}{2 m} \\ 0 & 0 & \frac{-\sqrt{m^2+4 n^2}+m-2 n}{2 m} & \frac{\sqrt{m^2+4 n^2}+m-2 n}{2 m} \\ \frac{1}{2} \left(-\sqrt{5}-1\right) & \frac{1}{2} \left(\sqrt{5}-1\right) & 1 & 1 \\ 0 & 0 & 1 & 1 \\ \end{array} \right).$$} The matrix $M'$ has the same eigenvalues as $M$ and is given by $M'=M_0+M_1$, with _{$$M_0=\left( \begin{array}{cccc} 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & \frac{1}{2} \left(-\sqrt{m^2+4 n^2}+m+2 n\right) & 0 \\ 0 & 0 & 0 & \frac{1}{2} \left(\sqrt{m^2+4 n^2}+m+2 n\right) \\ \end{array} \right),$$ $$M_1=\left( \begin{array}{cccc} \frac{1}{2} \left(-\sqrt{5}-3\right) & 0 & \frac{\sqrt{m^2+4 n^2}+\left(\sqrt{5}-2\right) m+2 n}{2 \sqrt{5} m} & \frac{-\sqrt{m^2+4 n^2}+\left(\sqrt{5}-2\right) m+2 n}{2 \sqrt{5} m} \\ 0 & \frac{1}{2} \left(\sqrt{5}-3\right) & \frac{-\sqrt{m^2+4 n^2}+\left(\sqrt{5}+2\right) m-2 n}{2 \sqrt{5} m} & \frac{\sqrt{m^2+4 n^2}+\left(\sqrt{5}+2\right) m-2 n}{2 \sqrt{5} m} \\ \frac{-\sqrt{5} \left(\sqrt{m^2+4 n^2}-2 n\right)-m}{2 \sqrt{m^2+4 n^2}} & \frac{\sqrt{5} \left(\sqrt{m^2+4 n^2}-2 n\right)-m}{2 \sqrt{m^2+4 n^2}} & -\frac{3 \sqrt{m^2+4 n^2}+m-2 n}{2 \sqrt{m^2+4 n^2}} & -\frac{\sqrt{m^2+4 n^2}+m-2 n}{2 \sqrt{m^2+4 n^2}} \\ \frac{m-\sqrt{5} \left(\sqrt{m^2+4 n^2}+2 n\right)}{2 \sqrt{m^2+4 n^2}} & \frac{\sqrt{5} \left(\sqrt{m^2+4 n^2}+2 n\right)+m}{2 \sqrt{m^2+4 n^2}} & \frac{-\sqrt{m^2+4 n^2}+m-2 n}{2 \sqrt{m^2+4 n^2}} & \frac{-3 \sqrt{m^2+4 n^2}+m-2 n}{2 \sqrt{m^2+4 n^2}} \\ \end{array} \right).$$}

Now we note that for large $m,n$ of order $1/\epsilon$, the matrix $M_0$ is of order $1/\epsilon$ while the matrix $M_1$ is of order unity. We can therefore perform perturbation theory in the small parameter $\epsilon$.

Let me first look at the positive eigenvalues. To zeroth order these are given by $\beta_\pm=\frac{1}{2} \left(\pm \sqrt{m^2+4 n^2}+m+2 n\right)$. To first order these are the $(3,3)$ and $(4,4)$ diagonal elements of $M_0+M_1$, which gives the two positive eigenvalues $$\beta_\pm=\frac{1}{2} \left(m+2 n\pm\sqrt{m^2+4 n^2}\right)\pm\frac{m-2 n \mp 3 \sqrt{m^2+4 n^2}}{2 \sqrt{m^2+4 n^2}}.$$ This expression for the positive eigenvalues is quite accurate, see the plot for $m=n$, where the exact expressions (red and green lines) are almost indistinguishable from the perturbative result (blue and orange).

Now for the negative eigenvalues. To first order these are given by the $(1,1)$ and $(2,2)$ diagonal elements of $M_0+M_1$, equal to $$\alpha_1=\frac{1}{2} \left(-\sqrt{5}-3\right)=-2.618,\;\;\alpha_2=\frac{1}{2} \left(\sqrt{5}-3\right)=-0.382.$$

Here is the plot for $m=n$, again red and green is exact, blue and orange the perturbative result.

So to first order in $\epsilon$ one has $\alpha_1+\alpha_2=-3$. The challenge is to prove that higher order corrections are negative. Incidentally, the restriction to integer $m,n$ does not seem to play any role (in the plot $m=n$ is varied continuously).

This is not yet an answer but some progress: I am able to accurately calculate the two positive eigenvalues, as follows:

I perform a similarity transformation on the matrix $M$, given by $M'=S^{-1}MS$ with _{$$S=\frac{1}{2m}\left( \begin{array}{cccc} 0 & 2 m & -\sqrt{m^2+4 n^2}+m-2 n & \sqrt{m^2+4 n^2}+m-2 n \\ 0 & 0 & -\sqrt{m^2+4 n^2}+m-2 n & \sqrt{m^2+4 n^2}+m-2 n \\ 2 m & 0 & 2 m & 2 m \\ 0 & 0 & 2 m & 2 m \\ \end{array} \right).$$} The matrix $M'$ has the same eigenvalues as $M$ and is given by $M'=M_0+M_1$, with _{$$M_0=\left( \begin{array}{cccc} 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & \frac{1}{2} \left(-\sqrt{m^2+4 n^2}+m+2 n\right) & 0 \\ 0 & 0 & 0 & \frac{1}{2} \left(\sqrt{m^2+4 n^2}+m+2 n\right) \\ \end{array} \right),$$ $$M_1=\frac{1}{\sqrt{m^2+4 n^2}}\left( \begin{array}{cccc} -2 \sqrt{m^2+4 n^2} & \sqrt{m^2+4 n^2} & -\frac{\sqrt{m^2+4 n^2} \left(\sqrt{m^2+4 n^2}-m+2 n\right)}{2 m} & \frac{\sqrt{m^2+4 n^2} \left(\sqrt{m^2+4 n^2}+m-2 n\right)}{2 m} \\ \sqrt{m^2+4 n^2} & -\sqrt{m^2+4 n^2} & \sqrt{m^2+4 n^2} & \sqrt{m^2+4 n^2} \\ \sqrt{m^2+4 n^2}-2 n & \frac{1}{2} \left(\sqrt{m^2+4 n^2}-m-2 n\right) & -\frac{3}{2} \sqrt{m^2+4 n^2}-\frac{m}{2}+n & -\frac{1}{2} \sqrt{m^2+4 n^2}-\frac{m}{2}+n \\ \sqrt{m^2+4 n^2}+2 n & \frac{1}{2} \left(\sqrt{m^2+4 n^2}+m+2 n\right) & \frac{1}{2} \left(-\sqrt{m^2+4 n^2}+m-2 n\right) & \frac{1}{2} \left(-3 \sqrt{m^2+4 n^2}+m-2 n\right) \\ \end{array} \right).$$} Now we note that for large $m,n$ of order $1/\epsilon$, the matrix $M_0$ is of order $1/\epsilon$ while the matrix $M_1$ is of order unity. We can therefore perform perturbation theory in the small parameter $\epsilon$. To first order the eigenvalues of $M$ are the diagonal elements of $M_0+M_1$, which gives the two negative eigenvalues $\alpha_1=-2$, $\alpha_2=-1$, and the two positive eigenvalues $\beta_+$ and $\beta_-$ given by $$\beta_\pm=\frac{1}{2} \left(m+2 n\pm\sqrt{m^2+4 n^2}\right)\pm\frac{m-2 n \mp 3 \sqrt{m^2+4 n^2}}{2 \sqrt{m^2+4 n^2}}.$$ This expression for the positive eigenvalues is quite accurate, see the plot for $m=n$, where the exact expressions (red and green lines) are almost indistinguishable from the perturbative result (blue and orange).

So to first order in $\epsilon$ one has $\alpha_1+\alpha_2=-3$. The challenge is to prove that higher order corrections are negative. Incidentally, the restriction to integer $m,n$ does not seem to play any role (in the plot $m=n$ is varied continuously).

I am able to accurately calculate the eigenvalues in perturbation theory, as follows:

I perform a similarity transformation on the matrix $M$, given by $M'=S^{-1}MS$ with _{$$S=\frac{1}{2m}\left( \begin{array}{cccc} 0 & 2 m & -\sqrt{m^2+4 n^2}+m-2 n & \sqrt{m^2+4 n^2}+m-2 n \\ 0 & 0 & -\sqrt{m^2+4 n^2}+m-2 n & \sqrt{m^2+4 n^2}+m-2 n \\ 2 m & 0 & 2 m & 2 m \\ 0 & 0 & 2 m & 2 m \\ \end{array} \right).$$} The matrix $M'$ has the same eigenvalues as $M$ and is given by $M'=M_0+M_1$, with _{$$M_0=\left( \begin{array}{cccc} 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & \frac{1}{2} \left(-\sqrt{m^2+4 n^2}+m+2 n\right) & 0 \\ 0 & 0 & 0 & \frac{1}{2} \left(\sqrt{m^2+4 n^2}+m+2 n\right) \\ \end{array} \right),$$ $$M_1=\frac{1}{\sqrt{m^2+4 n^2}}\left( \begin{array}{cccc} -2 \sqrt{m^2+4 n^2} & \sqrt{m^2+4 n^2} & -\frac{\sqrt{m^2+4 n^2} \left(\sqrt{m^2+4 n^2}-m+2 n\right)}{2 m} & \frac{\sqrt{m^2+4 n^2} \left(\sqrt{m^2+4 n^2}+m-2 n\right)}{2 m} \\ \sqrt{m^2+4 n^2} & -\sqrt{m^2+4 n^2} & \sqrt{m^2+4 n^2} & \sqrt{m^2+4 n^2} \\ \sqrt{m^2+4 n^2}-2 n & \frac{1}{2} \left(\sqrt{m^2+4 n^2}-m-2 n\right) & -\frac{3}{2} \sqrt{m^2+4 n^2}-\frac{m}{2}+n & -\frac{1}{2} \sqrt{m^2+4 n^2}-\frac{m}{2}+n \\ \sqrt{m^2+4 n^2}+2 n & \frac{1}{2} \left(\sqrt{m^2+4 n^2}+m+2 n\right) & \frac{1}{2} \left(-\sqrt{m^2+4 n^2}+m-2 n\right) & \frac{1}{2} \left(-3 \sqrt{m^2+4 n^2}+m-2 n\right) \\ \end{array} \right).$$}

Now we note that for large $m,n$ of order $1/\epsilon$, the matrix $M_0$ is of order $1/\epsilon$ while the matrix $M_1$ is of order unity. We can therefore perform perturbation theory in the small parameter $\epsilon$.

Let me first look at the positive eigenvalues. To zeroth order these are given by $\beta_\pm=\frac{1}{2} \left(\pm \sqrt{m^2+4 n^2}+m+2 n\right)$. To first order these are the $(3,3)$ and $(4,4)$ diagonal elements of $M_0+M_1$, which gives the two positive eigenvalues $$\beta_\pm=\frac{1}{2} \left(m+2 n\pm\sqrt{m^2+4 n^2}\right)\pm\frac{m-2 n \mp 3 \sqrt{m^2+4 n^2}}{2 \sqrt{m^2+4 n^2}}.$$ This expression for the positive eigenvalues is quite accurate, see the plot for $m=n$, where the exact expressions (red and green lines) are almost indistinguishable from the perturbative result (blue and orange).

Now for the negative eigenvalues. To first order these are given by the eigenvalues of the $2\times 2$ upper-left block of $M_0+M_1$, so $\alpha_1,\alpha_2$ are the eigenvalues of the matrix ${{-2\;1}\choose{1\;-1}}$, given by $$\alpha_1=\frac{1}{2} \left(-\sqrt{5}-3\right)=-2.618,\;\;\alpha_2=\frac{1}{2} \left(\sqrt{5}-3\right)=-0.382.$$

Here is the plot for $m=n$, again red and green is exact, blue and orange the perturbative result.

So to first order in $\epsilon$ one has $\alpha_1+\alpha_2=-3$. The challenge is to prove that higher order corrections are negative. Incidentally, the restriction to integer $m,n$ does not seem to play any role (in the plot $m=n$ is varied continuously).

This is not yet an answer but some progress: I am able to accurately calculate the two positive eigenvalues, as follows:

I perform a similarity transformation on the matrix $M$, given by $M'=S^{-1}MS$ with _{$$S=\frac{1}{2m}\left( \begin{array}{cccc} 0 & 2 m & -\sqrt{m^2+4 n^2}+m-2 n & \sqrt{m^2+4 n^2}+m-2 n \\ 0 & 0 & -\sqrt{m^2+4 n^2}+m-2 n & \sqrt{m^2+4 n^2}+m-2 n \\ 2 m & 0 & 2 m & 2 m \\ 0 & 0 & 2 m & 2 m \\ \end{array} \right).$$} The matrix $M'$ has the same eigenvalues as $M$ and is given by $M'=M_0+M_1$, with _{$$M_0=\left( \begin{array}{cccc} 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & \frac{1}{2} \left(-\sqrt{m^2+4 n^2}+m+2 n\right) & 0 \\ 0 & 0 & 0 & \frac{1}{2} \left(\sqrt{m^2+4 n^2}+m+2 n\right) \\ \end{array} \right),$$ $$M_1=\frac{1}{\sqrt{m^2+4 n^2}}\left( \begin{array}{cccc} -2 \sqrt{m^2+4 n^2} & \sqrt{m^2+4 n^2} & -\frac{\sqrt{m^2+4 n^2} \left(\sqrt{m^2+4 n^2}-m+2 n\right)}{2 m} & \frac{\sqrt{m^2+4 n^2} \left(\sqrt{m^2+4 n^2}+m-2 n\right)}{2 m} \\ \sqrt{m^2+4 n^2} & -\sqrt{m^2+4 n^2} & \sqrt{m^2+4 n^2} & \sqrt{m^2+4 n^2} \\ \sqrt{m^2+4 n^2}-2 n & \frac{1}{2} \left(\sqrt{m^2+4 n^2}-m-2 n\right) & -\frac{3}{2} \sqrt{m^2+4 n^2}-\frac{m}{2}+n & -\frac{1}{2} \sqrt{m^2+4 n^2}-\frac{m}{2}+n \\ \sqrt{m^2+4 n^2}+2 n & \frac{1}{2} \left(\sqrt{m^2+4 n^2}+m+2 n\right) & \frac{1}{2} \left(-\sqrt{m^2+4 n^2}+m-2 n\right) & \frac{1}{2} \left(-3 \sqrt{m^2+4 n^2}+m-2 n\right) \\ \end{array} \right).$$} Now we note that for large $m,n$ of order $1/\epsilon$, the matrix $M_0$ is of order $1/\epsilon$ while the matrix $M_1$ is of order unity. We can therefore perform perturbation theory in the small parameter $\epsilon$. To first order the eigenvalues of $M$ are the diagonal elements of $M_0+M_1$, which gives the two negative eigenvalues $\alpha_1=-2$, $\alpha_2=-1$, and the two positive eigenvalues $\beta_+$ and $\beta_-$ given by $$\beta_\pm=\frac{1}{2} \left(\sqrt{m^2+4 n^2}+m+2 n\right)\pm\frac{m-2 n-3 \sqrt{m^2+4 n^2}}{2 \sqrt{m^2+4 n^2}}.$$ This expression for the positive eigenvalues is quite accurate, see the plot for $m=n$, where the exact expressions (red and green lines) are almost indistinguishable from the perturbative result (blue and orange).

So to first order in $\epsilon$ one has $\alpha_1+\alpha_2=-3$. The challenge is to prove that higher order corrections are negative. Incidentally, the restriction to integer $m,n$ does not seem to play any role (in the plot $m=n$ is varied continuously).

This is not yet an answer but some progress: I am able to accurately calculate the two positive eigenvalues, as follows:

I perform a similarity transformation on the matrix $M$, given by $M'=S^{-1}MS$ with _{$$S=\frac{1}{2m}\left( \begin{array}{cccc} 0 & 2 m & -\sqrt{m^2+4 n^2}+m-2 n & \sqrt{m^2+4 n^2}+m-2 n \\ 0 & 0 & -\sqrt{m^2+4 n^2}+m-2 n & \sqrt{m^2+4 n^2}+m-2 n \\ 2 m & 0 & 2 m & 2 m \\ 0 & 0 & 2 m & 2 m \\ \end{array} \right).$$} The matrix $M'$ has the same eigenvalues as $M$ and is given by $M'=M_0+M_1$, with _{$$M_0=\left( \begin{array}{cccc} 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & \frac{1}{2} \left(-\sqrt{m^2+4 n^2}+m+2 n\right) & 0 \\ 0 & 0 & 0 & \frac{1}{2} \left(\sqrt{m^2+4 n^2}+m+2 n\right) \\ \end{array} \right),$$ $$M_1=\frac{1}{\sqrt{m^2+4 n^2}}\left( \begin{array}{cccc} -2 \sqrt{m^2+4 n^2} & \sqrt{m^2+4 n^2} & -\frac{\sqrt{m^2+4 n^2} \left(\sqrt{m^2+4 n^2}-m+2 n\right)}{2 m} & \frac{\sqrt{m^2+4 n^2} \left(\sqrt{m^2+4 n^2}+m-2 n\right)}{2 m} \\ \sqrt{m^2+4 n^2} & -\sqrt{m^2+4 n^2} & \sqrt{m^2+4 n^2} & \sqrt{m^2+4 n^2} \\ \sqrt{m^2+4 n^2}-2 n & \frac{1}{2} \left(\sqrt{m^2+4 n^2}-m-2 n\right) & -\frac{3}{2} \sqrt{m^2+4 n^2}-\frac{m}{2}+n & -\frac{1}{2} \sqrt{m^2+4 n^2}-\frac{m}{2}+n \\ \sqrt{m^2+4 n^2}+2 n & \frac{1}{2} \left(\sqrt{m^2+4 n^2}+m+2 n\right) & \frac{1}{2} \left(-\sqrt{m^2+4 n^2}+m-2 n\right) & \frac{1}{2} \left(-3 \sqrt{m^2+4 n^2}+m-2 n\right) \\ \end{array} \right).$$} Now we note that for large $m,n$ of order $1/\epsilon$, the matrix $M_0$ is of order $1/\epsilon$ while the matrix $M_1$ is of order unity. We can therefore perform perturbation theory in the small parameter $\epsilon$. To first order the eigenvalues of $M$ are the diagonal elements of $M_0+M_1$, which gives the two negative eigenvalues $\alpha_1=-2$, $\alpha_2=-1$, and the two positive eigenvalues $\beta_+$ and $\beta_-$ given by $$\beta_\pm=\frac{1}{2} \left(m+2 n\pm\sqrt{m^2+4 n^2}\right)\pm\frac{m-2 n \mp 3 \sqrt{m^2+4 n^2}}{2 \sqrt{m^2+4 n^2}}.$$ This expression for the positive eigenvalues is quite accurate, see the plot for $m=n$, where the exact expressions (red and green lines) are almost indistinguishable from the perturbative result (blue and orange).

So to first order in $\epsilon$ one has $\alpha_1+\alpha_2=-3$. The challenge is to prove that higher order corrections are negative. Incidentally, the restriction to integer $m,n$ does not seem to play any role (in the plot $m=n$ is varied continuously).