The Hamiltonian matrix will have smaller dimensions (a factor of 10 or more) in the $J$-scheme than in the $M$-scheme. On the other hand, as we'll show in the next slide, the $M$-scheme is very easy to construct with Slater determinants, while the $J$-scheme basis states, and thus the matrix elements, are more complicated, almost always being linear combinations of $M$-scheme states. $J$-scheme bases are important and useful, but we'll focus on the simpler $M$-scheme.

The quantum number $m$ is additive (because the underlying group is Abelian): if a Slater determinant $\hat{a}_i^\dagger \hat{a}^\dagger_j \hat{a}^\dagger_k \ldots | 0 \rangle$ is built from single-particle states all with good $m$, then the total $$M = m_i + m_j + m_k + \ldots$$ This is not true of $J$, because the angular momentum group SU(2) is not Abelian.