# How do electron configuration microstates map to term symbols?

I am trying to understand energy levels of electron configurations. I visited the NIST web site and discovered that the notation used here are called term symbols.

After reading corresponding wikipedia entry I worked through the carbon example in the "Term symbols for an electron configuration" section. So it appears to me that each of the $\binom{t}{e}$ microstates [(in this case $\binom{6}{2}=15$), where $t$ is number of slots in the outside subshell, and $e$ is the number of electrons] will each have an assigned energy level, or term symbol - since in the end there are 15 1s distributed in three different matrices, but only 5 possible term symbols (1 term symbol for 5 microstates, 3 term symbols for 9 microstates, and 1 term symbol for 1 microstate). However, I don't see how a given microstate maps to a specific term symbol.

For example, $M_L = 0$ and $M_S = 0$ is true for 3 microstates, but how to determine which term symbols correspond to them? Does it even make sense to do this?

UPDATE:

Thanks gigacyan, for the detailed answer. By the wording I am not sure at the end if you mean that $M_L = 0$ and $M_S = 0$ can only have the $^1S$ term. If that is true, then the following must be totally off, but I will take a shot anyway:

So are you saying that ANY given microstate cannot be assigned a term symbol (energy level), or that just certain microstates (such as the $M_L = 0$ and $M_S = 0$ case above) cannot be assigned since there are more than one term symbol possibility?

For example, it seems that when there is only one microstate for a given $M_L$ and $M_S$ combination, it is uniquely determined as long as it falls into a matrix that has only one term symbol - for carbon, say $M_L = 2$ and $M_S = 0$ (row 7 of 15 in the microstate table) then it falls into the 5x1 table, which must be $^1D_2$.

But when there is only one microstate for a given $M_L$ and $M_S$ combination but it falls into a matrix with more than one term symbol - for carbon, say $M_L = 1$ and $M_S = 1$ (row 1 of 15 in the microstate table), it falls into the 3x3 matrix, which means that this microstate must be one of $^3P_2$, $^3P_1$, or $^3P_0$, but it is not known which. Is this correct?

• It is the opposite: any term can have a combination of $M_L$ and $M_S$ equal zero, what I meant is that after assigning different microstates to $^1D$ and $^3P$ you are left with one microstate that corresponds to $^1S$ term. – gigacyan Apr 14 '11 at 16:28

Energy levels are assigned not to microstates but to terms. Terms are electron configurations with well-defined orbital and spin quantum numbers and in case of carbon there will be three terms: $^1D$, $^3P$ and $^1S$. Due to spin-orbit interaction $^3P$ term is split into three closely spaced energy levels $^3P_2$, $^3P_1$ and $^3P_0$ (you cannot say which microstates belong to each of these levels).
It is true – you cannot start with $M_S$ and $M_L$ equal zero and determine which term it belongs to. Instead you start with the largest values of $M_L$ and $M_S$. Note that maximal $M_L$ is +2 which means that there should be a term with $L=2$ (denoted as $D$). All possible $M_L$ values for this term are +2,+1,0,-1,-2 and you should exclude these five microstates from your pool. The next highest $M_L=1$ for which $M_S=1$. This means that there have to be $^3P$ term which is represented by 9 microstates. If you cross them out, the only combination you are left with is $M_L=0$, $M_S=0$ which can only be $^1S$ term.
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 Slater 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. 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 Slater determinant of atomic spin-orbitals, but open shell quantum states with well defined angular momentum typically require a sum of several Slater determinants. There are several simple techniques that show you how to find the right linear combination of Slater determinants. Gigacyan 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 L\"owdin, Adv. Phys. 5, 1 (1956).