the $1^+$ states have $K=1$. We can then use the rotational formula of Bohr and Mottelson (Eq. 4-92) in their book [@bm] ($K_1=0$, $K_2=1$) (We use the notation $\left[\begin{array}{lll}J_1 & J_2 & J\\ M_1 & M_2 &
M\end{array}\right]$ for a Clebsch-Gordan coefficient):

$$\langle K_2I_2 || M(\lambda) || K_1=0 I_1 \rangle
=\sqrt{2(2I_1+1)}\left[\begin{array}{lll}I_1 & \lambda & I_2\\ 0 & K_2 &
K_2\end{array}\right]
\langle K_2 | M(\lambda,\nu=K_2)|K_1=0\rangle$$

From this we can easily deduce

$$r=\frac{B(M1)_{J=1^+, K=1 \rightarrow J=2^+, K=0}}{B(M1)_{J=1^+, K=1
\rightarrow J=0^+, K=0}} =\frac{1}{2}$$

Note, however, that the experimental ratio for $^8Be$ from the $J=1^+$ $T=1$ state at 17.64 $MeV$ (see section 1) is $\frac{0.12}{0.15}=0.8$. Bohr and Mottelson later discuss corrections to the above simple formula.

We can obtain a value for the above ratio by forming an intrinsic state and projecting out states of angular momentum $J=0$ and $J=2$. We assume an extreme prolate shape for $^8Be$ and put the four valence nucleons ($N\uparrow$, $N\downarrow$, $P\uparrow$, $P\downarrow$) in the lowest Nilsson orbit with $\Lambda=0$. The asymptotic wave function is $NrY_{10}$.

The $J=0$ wave function is

$$N'\sum_L \left[\begin{array}{lll}1 & 1 & L\\0 & 0 &
0\end{array}\right]  \left[\begin{array}{lll}1 & 1 & L\\0 & 0 &
0\end{array}\right]  \left[\begin{array}{lll}L & L & 0\\0 & 0 &
0\end{array}\right]  