$1p$ shell states. This is because the number of different $M$ states availbale for particles (3) coincides with the number of possible directions for oscillator quanta $(a^{\dagger}_x,~a^{\dagger}_y,~a^{\dagger}_z)$ and a single creation operator correponds to each particle.

In more detail, the Casimir operator is $\tilde{C_2}=Q\cdot Q~-3\vec{L} \cdot
\vec{L}$. Hence,

$$\begin{aligned}
\langle-\chi Q\cdot Q\rangle_{\lambda~\mu~L} & = &\bar{\chi}\left[-\langle
\tilde{C_2}\rangle_{\lambda~\mu}~+~3L(L+1)\right]\\
& = &
\bar{\chi}\left[-4(\lambda^2+\mu^2+\lambda\mu+3(\lambda+\mu)+3L(L+1)\right]\end{aligned}$$

where $\bar{\chi}=\chi \frac{5b^4}{32\pi}$ with $b$ the harmonic oscillator length parameter ($b^2=\frac{\hbar}{m\omega}$). The magnetic dipole modes in the $L~S~T$ representation are:

$L=1$ $S=0$ $T=0$ = $L=0$ $S=1$ $T=1$ (isovector spin mode)\
$L=1$ $S=0$ $T=1$ (scissors mode) $L=1$ $S=1$ $T=0$\
$L=0$ $S=1$ $T=0$ $L=1$ $S=1$ $T=1$\

With the $Q\cdot Q$ interaction that we have chosen, transitions from the $L=0~S=0$ ground state in $^8Be$ to all of these modes except one will vanish. The only surviving mode is the $L=1~S=0~T=1$ scissors mode.

Let us give briefly the energies and some properties of the states in $^8Be$ ($\bar{\chi}=0.1865$):

\(a) \[f\]=\[4,0\] ($\lambda,\mu$)=(4,0) Ground State Band

$L$ = $S$ = $T$ = $\frac{E}{\bar{\chi}}$ = $E^*(MeV)$\
0 0 0 $-112$ 0\
2 0 0 $-94$ 3.36\
4 0 0 $-52$ 11.19\

\(b) \[f\]=\[3,1\] ($\lambda,\mu$)=(2,1) -contains the scissors mode ($L=1$, $S=0$, $T=1$).

Note that the ($S,T$) combinations (0,1), 