Total $142.8$ $e^2fm^4$\
4. Right hand Side ($\frac{9\chi}{16\pi}=0.0647$): $\mu_N^2MeV$\

For $^{10}\mbox{Be}$ we are also curious to see what happens if we use only the lowest $2^+$ state in the right hand side of the sum rule. But we have to be careful! It turns out that there is substantial $B(E2)$ strength to the two lowest $J=2^+$ states. This can be understood from the fact that with a $Q \cdot Q$ interaction in a small space calculation ($(0s)^4(0p)^6$) the two lowest $2^+$ states are exactly degenerate. The states belong to the $[f]=[42]$ representation. The $Q \cdot Q$ interaction fails to remove the degeneracy of these states. Another way of stating this is that the ($\lambda \mu$) values for both states are (22), and the allowed values of the $K$ quantum number in the Nilsson scheme are $K=\mu$, $\mu-2$, etc. Thus the $2^+$ states have $K=0$ and $K=2$.

When we go to the large space calculation with a $Q \cdot Q$ interaction, limiting the excitations to $2\hbar\omega$, the degeneracy is removed but the states are still fairly close together. The calculated values are:

= = $E2(1,1)$ = $E2(1,-1)$\
$2^+_1$ $E=2.08$ $MeV$ 64.94 12.32\
$2^+_2$ $E=2.92$ $MeV$ 93.38 10.11\

Thus, using the calculated 