$O_D(b_0,b_1)+\sqrt{3}O_E(b_0,b_1)+\sqrt{2}O_F(b_0,b_1)$ and $O_G(b_0,b_1)+\sqrt{3}O_H(b_0,b_1)$. Notice that all of the BPS operators from this section and the last can be written as $$O_{\rm BPS}(R,r)=\sum_s \sqrt{d_s} O_{R,(r,s)}(b_0,b_1)$$ where $d_s$ is the dimension of the irreducible representation $s$ of the symmetric group. Finally, notice that if we set $\sqrt{3}O_B(b_0,b_1)-O_C(b_0,b_1)\equiv O_{B-C}(b_0,b_1)$, $\sqrt{2}O_D(b_0,b_1)-O_F(b_0,b_1)\equiv O_{D-F}(b_0,b_1)$, $O_D(b_0,b_1)-\sqrt{3}O_E(b_0,b_1)+\sqrt{2}O_F(b_0,b_1)\equiv O_{DF-E}(b_0,b_1)$ and $\sqrt{3}O_G(b_0,b_1)-O_H(b_0,b_1)\equiv O_{G-H}(b_0,b_1)$, we have [$$\begin{aligned}
\nonumber
DO_{B-C}(b_0,b_1) = &&-2g_{YM}^2(N-b_0)\left[
O_{B-C}(b_0+1,b_1-2)-2O_{B-C}(b_0,b_1)+O_{B-C}(b_0-1,b_1+2)\right]\cr
DO_{D-F}(b_0,b_1) = &&-2g_{YM}^2(N-b_0)\left[
O_{D-F}(b_0+1,b_1-2)-2O_{D-F}(b_0,b_1)+O_{D-F}(b_0-1,b_1+2)\right]\cr
DO_{DF-E}(b_0,b_1) = &&-4g_{YM}^2(N-b_0)\left[
O_{DF-E}(b_0+1,b_1-2)-2O_{DF-E}(b_0,b_1)+O_{DF-E}(b_0-1,b_1+2)\right]\cr
DO_{G-H}(b_0,b_1) = &&-2g_{YM}^2(N-b_0)\left[
O_{G-H}(b_0+1,b_1-2)-2O_{G-H}(b_0,b_1)+O_{G-H}(b_0-1,b_1+2)\right]\end{aligned}$$ ]{} The right hand side is again a discretization of the second derivative.

Numerical Results
=================

[*In this section we describe the result of numerically diagonalizing the dilatation operator.*]{}

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When setting up a numerical computation of the spectrum of the anomalous dimension operator, we need to specify the maximum value for the difference between the number of boxes in the long column and the number of boxes in the short column. Given this value, denoted $a_{\rm max}$, we are able to determine how many operators participate in our problem and we are able to describe the resulting spectrum rather explicitly. We will focus on the case of even $a_{\rm max}$. In this case the difference between the number of boxes in the long column and the 