# Atomic terms

### LS scheme

The LS scheme is appropriate for light atoms $\left(Z<40\right)$ , when spin-orbit hamiltonian can be treated as a perturbation to the hartree hamiltonian with non spherical corrections. The good quantum number are fixed by the necessity of diagonalizing non-spherical corrections hamiltonian and are ${n_{i},l_{i}},L,S,m_{L},m_{S}$ . Spin orbit effects will split these states according to their $J$ value. The problem of finding the atomic terms of a particular electronic configuration simply reduces to the determination of the total angular momentum $\mathbf {L}$ eigenvalue and the total spin $\mathbf {S}$ . The term is then of the form:

{\begin{aligned}^{2S+1}L\quad L&=S\quad {\mbox{if }}L=0\\&=P\quad {\mbox{if }}L=1\\&=D\quad {\mbox{if }}L=2\\&=F\quad {\mbox{if }}L=3\\&=\dots \end{aligned}} Let us discuss a concrete example in order to point out the general procedure to follow. Consider the atom $^{7}N$ . An electronic configuration is an expression of the form $1s^{2}2s^{2}2p^{3}$ , which specifies the number of electrons in each atomic orbital. The previous expression is the electronic configuration of nitrogen ground state. Another possible configuration for our atom is the excited one $1s^{2}2s^{1}2p^{3}3d^{1}$ . Now, how can we find the terms corresponding to these configurations?

• First of all remove the closed shells: they do not participate in the determination of the atomic terms
• Find the number of possible states for each atomic orbital. The reason for doing this will be clear in the following discussion. The number is easy to find with the little help of combinatorial analysis. Let us call $g=2\cdot \left(2L+1\right)$ and n the number of electrons in the L-orbital. Then
${\mbox{number of states }}={\binom {g}{n}}={\frac {g!}{n!\left(g-n\right)!}}$ For the ground-state configuration $1s^{2}2s^{2}2p^{3}$ the number of states for the $2p$ orbital with 3 electrons in it is ${\frac {6!}{3!3!}}=20$ • Put the electrons in the orbitals and find the possible values of $m_{S},m_{L}$ . According to the values found you can deduce that of L and S, since $-L\leq m_{L}\leq L$ and $-S\leq m_{S}\leq S$ . In our example we find for the nitrogen ground state: • Check that the number of states given by the terms is the same you previously calculated.
• The exercise may ask you to find the term which correspond to the lowest energy. According to Hund's rules :
• The term with the maximum spin has the lowest energy
• If there still are more terms, choose the one with the highest angular momentum
• For the ground-state terms the lowest energy term is thus $^{4}S$ • In doing exercises remember that terms are symmetric with respect to the half-full level: $p^{4}$ terms are the same of $p^{2}$ and so on.
• Let us consider the excited configuration of nitrogen $1s^{2}2s^{1}2p^{3}3d^{1}$ . We now have three not completely full orbitals. If you are asked for this configuration's terms we can proceed as follows:
• Find the terms for each orbital as previously done for the ground state configuration. We get:
Orbital Terms Number of states
$2s^{1}$ $^{2}S$ 2
$2p^{3}$ $^{4}S\quad ^{2}P\quad ^{2}D$ $4+6+10=20$ $3d^{1}$ $^{2}D$ 10
• Sum the terms according to angular momenta addition rules:
Configuration First terms Second terms Resulting terms
$2s^{1}2p^{3}$ $^{2}S$ $^{4}S,^{2}P,^{2}D$ {\begin{aligned}^{2}S+^{4}S\rightarrow ^{3}S,^{5}S\\^{2}S+^{2}P\rightarrow ^{3}P,^{1}P\\^{2}S+^{2}D\rightarrow ^{3}D,^{1}D\end{aligned}} $2s^{1}2p^{3}3d^{1}$ $^{2}D$ $^{3}S,^{5}S,^{3}P,^{1}P,^{3}D,^{1}D$ {\begin{aligned}^{2}D+^{3}S\rightarrow ^{2}D\quad ^{4}D\\^{2}D+^{5}S\rightarrow ^{4}D,^{6}D\\^{2}D+^{3}P\rightarrow ^{4}P,^{2}P,^{4}D,^{2}D,^{4}F,^{2}F\\^{2}D+^{1}P\rightarrow ^{2}P,^{2}D,^{2}F\\^{2}D+^{3}D\rightarrow ^{2}S,^{4}S,^{2}P,^{4}P,^{2}D,^{4}D,^{2}F,^{4}F,^{2}G,^{4}G\\^{2}D+^{1}D\rightarrow ^{2}S,^{2}P,^{2}D,^{2}F,^{2}G\end{aligned}} • If you are asked for the term with lowest energy use Hund's rules to select the term. In this example it's $^{6}D$ . However if the exercise do not ask to find every term you can simplify the problem selecting in each orbital the lowest energy term and then adding just these terms
• If we consider spin-orbit effects as a perturbation to the hamiltonian we have to introduce an additional index representing the total angular momentum eigenvalue,$J$ . For the ground state term $^{4}S\quad {\frac {3}{2}}-0\leq J\leq {\frac {3}{2}}+0={\frac {3}{2}}$ and we can write the term as:
$^{4}S_{\frac {3}{2}}$ For the excited term $^{6}D\quad L=2,S={\frac {5}{2}}$ so that ${\frac {1}{2}}\leq J\leq {\frac {9}{2}}$ and the possible terms are:
$^{6}D_{\frac {1}{2}}\quad ^{6}D_{\frac {3}{2}}\quad ^{6}D_{\frac {5}{2}}\quad ^{6}D_{\frac {7}{2}}\quad ^{6}D_{\frac {9}{2}}$ Which of these terms has the lowest energy? The third Hund's rule saves us from complex calculations: if the shell is less than half-full the lowest energy term is the one with the lowest value of J, otherwise the one with the highest value of J. In our example the shell less than half-full so the lowest energy term is $^{6}D_{\frac {1}{2}}$ .

### JJ scheme

For heavy atoms $\left(Z>40\right)$ the spin-orbit effects can be stronger than non- spherical corrections, so we must include $H_{SO}$ in the hartree hamiltonian and then treat non-spherical corrections as a perturbation.The good quantum numbers describing the single-particle unperturbed states are $n,l,j,m_{j}$ . In order to take into account non-spherical corrections we then have to find $J,M_{J}$ . Let us discuss the general procedure through a concrete example. Consider the lead atom $^{82}Pb$ in the ground state $[Xe]4f^{14}5d^{10}6s^{2}6p^{2}$ . In the following discussion we'll not take into account 4f orbital.

• Find the j's for each orbital corresponding to a single electron placed in it:
Hartree orbital l s j jj orbital max number of electrons
5d 2 ${\frac {1}{2}}$ ${\frac {3}{2}},{\frac {5}{2}}$ $d_{\frac {3}{2}}\quad d_{\frac {5}{2}}$ $4\quad 6$ 6s 0 ${\frac {1}{2}}$ ${\frac {1}{2}}$ $s_{\frac {1}{2}}$ $2$ 6p 1 ${\frac {1}{2}}$ ${\frac {1}{2}},{\frac {3}{2}}$ $p_{\frac {1}{2}}\quad p_{\frac {3}{2}}$ $2\quad 4$ • Place the electrons in the jj-orbitals so found:10 electrons in 5d, 2 in 6s, 2 in 6p
$[]5d_{\frac {3}{2}}^{4}5d_{\frac {5}{2}}^{6}6s_{\frac {1}{2}}^{2}6p_{\frac {1}{2}}^{2}6p_{\frac {3}{2}}^{0}$ • Find J-values for the shell which is not completely full. The others would give J=0 and do not contribute.In this case we have to find J for the shell $6p^{2}\rightarrow 6p_{\frac {1}{2}}^{2},6p_{\frac {3}{2}}$ . We can have just $m_{J}=0$ so that $J=0$ • If you're asked for the lowest energy configuration choose the J-value according to Hund's third rule.Here we have just one possible value of J so there is no need for doing this.However we write the lowest energy state for lead in the jj-scheme in order to highlight the notation:
$\left([]5d_{\frac {3}{2}}^{4}5d_{\frac {5}{2}}^{6}6s_{\frac {1}{2}}^{2}6p_{\frac {1}{2}}^{2}6p_{\frac {3}{2}}^{0}\right)_{0}$ Let us discuss a more complex situation with 3 electrons in the $d_{\frac {5}{2}}$ jj-orbital: $d_{\frac {3}{2}}^{4}d_{\frac {5}{2}}^{3}$ . There are ${\binom {6}{3}}=20$ states corresponding to this configuration. Keeping in mind that for pauli's exclusion principle we can have just one electron with fixed $l,j,m_{j}$ we have these possibilities: We have selected time by time the configurations with the highest $m_{J}$ . Indeed,rigorously speaking if we find $m_{J}={\frac {5}{2}}$ we can just say that $J\geq m_{J}$ . However since we have already counted the states with $J={\frac {9}{2}}$ and no other value of J between ${\frac {9}{2}}$ and ${\frac {5}{2}}$ is allowed because of Pauli's exclusion principle we can say that $m_{J}={\frac {5}{2}}\Rightarrow J={\frac {5}{2}}$ . Counting the states allows us to verify that these configurations exhaust the multiplicity of $d_{\frac {5}{2}}^{3}$ We then have these possible terms:

$\left(d_{\frac {5}{2}}^{3}\right)_{{\frac {9}{2}},{\frac {5}{2}},{\frac {3}{2}}}$ If we are now looking for the lowest energy term we take into account that a $d$ orbital can have a maximum of 10 electrons so that with 7 electrons we are in a more than half-full shell and we have to pick the highest value of J. The lowest energy term is thus
$\left(d_{\frac {5}{2}}^{3}\right)_{\frac {9}{2}}$ 