For the $2_1^+$ state in $^8\mbox{Be}$ we obtain (in our large space calculation) $B(E2;1,1)=196.76$ and $B(E2;1,-1)=0$ (because the $2_1^+$ state has $T=0$). The right hand side becomes $20.30$ $\mu_N^2MeV$. We get a answer using the lowest $2^+$ state than we do if we use all $2^+$ states in the $0\hbar\omega$ and $2\hbar\omega$ space. The reason for this is that when the lowest $2^+$ state is excluded, the isovector $B(E2)$ is larger than the isoscalar $B(E2)$.

Can we also get agreement for the sum rule in the small space $(0s)^4(0p)^4$ for $^8\mbox{Be}$? Now the numbers are:

= 1. Energy weighted isovector orbital $M1$ strength: = = $\mu_N^2MeV$\
2. The isoscalar summed strength $B(E2;1,1)$: $72.54$ $e^2fm^4$\
3. The isovector summed strength $B(E2;1,-1)$: $10.37$ $e^2fm^4$\
4. The right hand side ($\frac{9\chi}{16\pi}=0.1032$): $\mu_N^2MeV$.\

We get perfect agreement.

We next consider $^{10}\mbox{Be}$. The relevant numbers for the large space calculation are:

= 1. Energy weighted $M1$ strength: = $J=0^+,~T=1$ $\rightarrow$ $J=1^+,~T=1$ = $ 3.811 $ = $\mu_N^2MeV$\
$J=0^+,~T=1$ $\rightarrow$ $J=1^+,~T=2$ $ 2.602 $ $\mu_N^2MeV$\
Left Hand Side $\mu_N^2MeV$\
2. $\sum B(E2;1,1)$ $J=0^+,~T=1$ $\rightarrow$ $J=2^+,~T=1$ $251.4$ $e^2fm^4$\
3.(a) $\sum B(E2;1,-1)$ $J=0^+,~T=1$ $\rightarrow$ $J=2^+,~T=1$ $94.02$ $e^2fm^4$\
  (b) $\sum B(E2;1,-1)$ $J=0^+,~T=1$ $\rightarrow$ $J=2^+,~T=2$ $48.78$ $e^2fm^4$\
  (c) $\sum B(E2;1,-1)$ 