Constructing atomic terms (uncoupled representation)

Consider the coupling of two electrons. From eq205, we have

$\hat{J}=\hat{L}\otimes I+I\otimes\hat{S}$

where $\hat{L}=\hat{l}^{(1)}\otimes I+I\otimes\hat{l}^{(2)}$ and $\hat{S}=\hat{s}^{(1)}\otimes I+I\otimes\hat{s}^{(2)}$ .

This implies that the state corresponding to $\hat{L}$ is a coupled representation of $\hat{l}^{(1)}$ and $\hat{l}^{(2)}$ (orbit-orbit coupling), and that the state corresponding to $\hat{S}$ is a coupled representation of $\hat{s}^{(1)}$ and $\hat{s}^{(2)}$ (spin-spin coupling). The overall state can be either a spin-orbit coupled representation $\vert J,M_J,L,S\rangle$ , or a spin-orbit uncoupled representation $\vert L,M_L,S,M_S\rangle$ . As mentioned in a previous article, if we neglect spin-orbit coupling, we can construct atomic terms using the uncoupled representation.

If the two electrons are in the same open subshell, they are called equivalent electrons. An example of such a system is the carbon atom. Even though carbon has six electrons, four of them are in closed shells with zero angular momentum and are therefore ignored when determining atomic terms.

The first step is to tabulate all possible microstates of p²(arrangements of p² electrons)that do not violate the Pauli exclusion principle:

Groups	$\boldsymbol{\mathit{m}}_l$			$\boldsymbol{\mathit{M}}_L$	$\boldsymbol{\mathit{M}}_S$
Groups	+1	0	-1	$\boldsymbol{\mathit{M}}_L$	$\boldsymbol{\mathit{M}}_S$
All up	u	u		1	1
	u		u	0	1
		u	u	-1	1
All down	d	d		1	-1
	d		d	0	-1
		d	d	-1	-1
One up, one down	ud			2	0
		ud		0	0
			ud	-2	0
	u	d		1	0
	u		d	0	0
		u	d	-1	0
	d	u		1	0
	d		u	0	0
		d	u	-1	0

where u and d represent $m_s=+\frac{1}{2}$ and $m_s=-\frac{1}{2}$ respectively.

The above table is re-organised as:

		$\boldsymbol{\mathit{M}}_S$
		+1	0	-1
$\boldsymbol{\mathit{M}}_L$	+2		1
	+1	1	2	1
	0	1	3	1
	-1	1	2	1
	-2		1

where each green number represents the number of microstates corresponding to each $(M_L,M_S)$ combination.

The only microstate with $M_L=+2$ and $M_S=0$ when expressed in the uncoupled representation of $\vert L,M_L,S,M_S\rangle$ is $\vert 2,2,0,0\rangle$ because $L=M_{L,max}$ . It must belong to the term ¹D, since ¹D is when $L=2$ and $S=0$ , with degenerate states of $\vert 2,2,0,0\rangle$ , $\vert 2,1,0,0\rangle$ , $\vert 2,0,0,0\rangle$ , $\vert 2,-1,0,0\rangle$ and $\vert 2,-2,0,0\rangle$ . For accounting purposes, we refresh the above table by removing these 5 states of ¹D, giving:

		$\boldsymbol{\mathit{M}}_S$
		+1	0	-1
$\boldsymbol{\mathit{M}}_L$	+2
	+1	1	1	1
	0	1	2	1
	-1	1	1	1
	-2

Similarly, the only state with $M_L=+1,M_S=+1$ must be one of the 9 degenerate states of ³P. Again, we refresh the above table by removing these 9 states of ³P, leaving behind one state ( $M_L=0,M_S=0$ ), which corresponds to the term ¹S. Therefore, the atomic terms for the p² configuration of carbon are ¹D, ³P and ¹S.

Question

Why are the 9 degenerate states of³P, combinations of $M_L=+1,0,-1$ and $M_S=+1,0,-1$ ?

Answer

Recall that states with same $L$ and same $S$ have the same energy and are grouped into a term. For the term ³P, $L=1$ and $S=1$ with degeneracy 9 (³P₂ has 5, ³P₁ has 3 and ³P₀ has 1). For $S=1$ , we have three spin-coupled basis vectors associated with the quantum numbers $M_S=+1,0,-1$ (formed by the coupling of $s_1$ and $s_2$ ). For $L=1$ , we have another three orbit-coupled basis vectors associated with the quantum numbers $M_L=+1,0,-1$ (formed by the coupling of $l_1$ and $l_2$ ). The total uncoupled microstates $\vert L,M_L,S,M_S\rangle$ are the number of ways to form Kronecker products of basis vectors from the two vector spaces and hence combinations of $M_L=+1,0,-1$ $M_L=+1,0,-1$ and $M_S=+1,0,-1$ .

For the configuration 1s²2s²2p³, e.g. nitrogen, we have

Groups	$\boldsymbol{\mathit{m}}_l$			$\boldsymbol{\mathit{M}}_L$	$\boldsymbol{\mathit{M}}_S$
Groups	+1	0	-1	$\boldsymbol{\mathit{M}}_L$	$\boldsymbol{\mathit{M}}_S$
All up	u	u	u	0	3/2
All down	d	d	d	0	-3/2
One up two down	ud	d		2	-1/2
	ud		d	1	-1/2
	d	ud		1	-1/2
		ud	d	-1	-1/2
	d		ud	-1	-1/2
		d	ud	-2	-1/2
	u	d	d	0	-1/2
	d	u	d	0	-1/2
	d	d	u	0	-1/2
Two up one down	ud	u		2	1/2
	ud		u	1	1/2
	u	ud		1	1/2
		ud	u	-1	1/2
	u		ud	-1	1/2
		u	ud	-2	1/2
	u	u	d	0	1/2
	u	d	u	0	1/2
	d	u	u	0	1/2

We can re-organise the above table as:

		$\boldsymbol{\mathit{M}}_S$
		+3/2	1/2	-1/2	-3/2
$\boldsymbol{\mathit{M}}_L$	+2		1	1
	+1		2	2
	0	1	3	3	1
	-1		2	2
	-2		1	1

The only state with $M_L=0,M_S=+3/2$ must be one of the 4 degenerate states of ⁴S. For accounting purposes, we re-tabulate the above table by removing these 4 states of ⁴S, giving:

		$\boldsymbol{\mathit{M}}_S$
		+3/2	1/2	-1/2	-3/2
$\boldsymbol{\mathit{M}}_L$	+2		1	1
	+1		2	2
	0		2	2
	-1		2	2
	-2		1	1

A good guess for the next term is ²D with a degeneracy of 10, because we are left with states of $M_S=\pm1/2$ , spanning $M_L$ from +2 to -2. The remaining states are:

		$\boldsymbol{\mathit{M}}_S$
		+3/2	1/2	-1/2	-3/2
$\boldsymbol{\mathit{M}}_L$	+2
	+1		1	1
	0		1	1
	-1		1	1
	-2

which obviously belong to the term ²P.

Therefore, the terms for the configuration of nitrogen are ⁴S, ²D, ²P.

For a system with non-equivalent electrons (electrons in different subshells) in open shells, the way to construct atomic terms is similar to a system with equivalent electrons, except that we do not have to apply the Pauli exclusion principle in the construction process. The easiest method of construction is to use the Clebsch-Gordan series by coupling the $l_i$ values to find out the possible $L$ values, and then coupling the $s_i$ values to give the possible $S$ values, and finally combining the possible $L$ and $S$ values to produce the atomic terms. For example, the terms for the 2p¹3p¹ system are ³D, ¹D, ³P, ¹P, ³S and ¹S.

Constructing atomic terms (uncoupled representation)

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