The upshot is that

• It is easy to construct a Slater determinant with good total \( M \);
• It is trivial to calculate \( M \) for each Slater determinant;
• So it is easy to construct an \( M \)-scheme basis with fixed total \( M \).

Note that the individual \( M \)-scheme basis states will not, in general, have good total \( J \). Because the Hamiltonian is rotationally invariant, however, the eigenstates will have good \( J \). (The situation is muddied when one has states of different \( J \) that are nonetheless degenerate.)