The Hamiltonian of an atom commutes with the square of the total angular momentum $\hat{J} = \hat{L} + \hat{S}$ and one component, disregarding very small hyperfine splittings. Knowing the energy $E$, $J$ and $J_z$ amounts to solving the fine structure of the atom. Since $\hat{J}$ is an angular momentum there are 2J+1 degenerate components in the absence of an external magnetic field. For non heavy atoms the Hamiltonian also approximately commutes with $\hat{L}^2$, $\hat{L}_z$, $\hat{S}^2$ and $\hat{S}_z$, the orbital and spin angular momenta, so you can have approximately well defined energy and $L$, $M_L$, $S$ and $M_S$ quantum numbers. These states have the right (or approximately right) symmetry, so you want to have your eigenstates with all those labels, say $|E,L,M_L,S,M_S,J,M_J\rangle$.
In typical Quantum calculations, however, you write the n-electron wave function as a <PERSON> determinant of {\it atomic orbitals}, which have well defined one-electron orbital and spin angular momenta, $l_j$, $m_{l,j}$, $s_j$ and $m_{s,j}$. The number of micro-states that you can find specifying the lower case quantum numbers $l_j$, $m_{l,j}$, etc. (you only need to consider electrons that don't fill a "shell") is the same as the number of "real" states (or microstates) that you find specifying the upper case quantum numbers $L$, $M_L$, etc. {\bf but} the wave functions are not the same. However you can transform from one representation to the other by a unitary operator. 976
In the case of two electrons (or holes) in the outer open shell, the coefficients of this transformation are the well known Clebsch-Gordan coefficients.
In particular the micro-states specification is related to a single <PERSON> determinant of atomic spin-orbitals, but open shell quantum states with well defined angular momentum typically require a sum of several <PERSON> determinants. There are several simple techniques that show you how to find the right linear combination of <PERSON> determinants. <PERSON> pointed to the most usual one. Since for the highest possible $M_L$ there is only one possible single determinant, you can generate the linear combination for the other $M_L$ states with the same $L$ by using ladder angular momentum operators. The same procedure can be used to generate the different $M_S$ states sharing equal $S$. Somewhat more generally, you can also use projection operators. See for instance <PERSON>, Adv. Phys. 5, 1 (1956). 469