= Solution
The <nuclear quadrupole vibration> has five one-phonon magnetic substates, $m=-2,-1,0,1,2$. <Bose-Einstein statistics> restrict three identical <phonons> to the <symmetric power> $S^3(V_2)$, of dimension $\binom{5+3-1}{3}=35$. Count occupation states $(n_{-2},\ldots,n_2)$ with $\sum n_m=3$ and $\sum mn_m=M$. For nonnegative $M$ the counts are
$$
\begin{array}{c|rrrrrrrr}M&0&1&2&3&4&5&6&7\\\hline d_M&5&4&4&3&2&1&1&0\end{array}.
$$
Each spin-$J$ <irreducible representation> contributes one state at every projection from $-J$ to $J$. <magnetic-substate subtraction> therefore gives multiplicity $d_J-d_{J+1}$: one each at $J=0,2,3,4,6$, and none at $1,5$. Thus the <three-quadrupole-phonon multiplet> is
$$
\boxed{S^3(V_2)=V_0\oplus V_2\oplus V_3\oplus V_4\oplus V_6,\qquad J^P=0^+,2^+,3^+,4^+,6^+}.
$$
The dimension check is $1+5+7+9+13=35$. The <parity> is positive because each constituent <phonon> has positive <parity>. Pairwise angular-momentum triangle rules alone would include states incompatible with complete boson symmetrization.
Back to article page
