Lecture 14 - The spin in atoms with many electrons

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In the last lecture \cite{electrons} we have seen that we can typically treat complex atomic systems within the central field approximation. \begin{align} \label{eq:lsvsjjhamiltonian} \hat{H} &&=& \underbrace{\sum_i^N \left( \frac{1}{2} \vec{\nabla}^2_{\vec{r}_i} + V_\textrm{cf} (r_i) \right)}_{\hat{H}_0} + \underbrace{\sum^N_{j>i} \left( \frac{1}{r_{ij}} - S(r_i) \right)}_{\hat{H}_1} \end{align}

So we we can can treat atoms through the shell structure known from the atom, but the screening lifts the *l* degeneracy. For a single outer electron, we have even seen how this screening can be described by the quantum defect.

We would now like to go beyond this simple picture and discuss the following questions:

How should the residual term $\hat{H}_1$ be taken into account?

How do we properly take into account the Pauli principle ?

How can we treat the fine-splitting ?

# On the residual coupling

If we ignore the residual coupling, we obtain a spherically symmetric problem, which implies that the angular momentum $\vec{l}_i$ of each electron is conserved. This conservation will be broken by $\hat{H}_{1}$. However, these forces are internal, which implies that the total angular momentum $\vec{L} = \sum_i \vec{l}_i$ is conserved. So we should label the states in the complex Hamiltonian by $\vec{L}$.

The total angular momentum will then set the symmetry of the spatial wavefunction. As already discussed in some detail for the He atom, this has wide-reaching consquence on the spin degree of freedom through exchange interaction.

# The Pauli principle and spin

According to the Pauli principle, each single-particle state can be occupied only by one electron. After distributing all electrons over different single-particle eigenstates (“orbitals”), the resulting state needs to be fully antisymmetrized (Slater determinant).

There is a simplification for atoms with many electrons: The angular momenta and spins of a complete subshell with

*n*,*l*, {*m*_{−l}, ⋯,*m*_{l}} add to zero and can be ignored in the further considerations (“shell structure”). Note that this is often broken in molecular binding!Alkali atoms are the simplest atoms with shell structure: All but one

*valence*electron add to*L*= 0,*S*= 0. The ground state thus has*L*= 0,*S*= 1/2.For more complex atoms, the valence electrons couple to a total orbital angular momentum

*L*with a given symmetry according to particle exchange.

Let us have a look at two examples for light atoms, starting with :

1

*s*^{2}→*L*= 0,*S*= 0. The corresponding term is^{1}*S*1

*s*2*s*→*L*= 0, {*S*= 0,*S*= 1}. The corresponding terms are^{1}*S*and^{3}*S*. .

The electronic configuration of is: \begin{align}
\underbrace{1s^2 2s^2 2p^6 3s^2}_{L=0,\,S=0} 3p^2
\end{align} Per valence electron we have *l* = 1 and *s* = 1/2. So we get *L* = 0, 1, 2 and *S* = 0, 1. Here *S* = 1 means symmetry and *S* = 0 antisymmetry with respect to particle exchange. In principle we can form the following terms: \begin{align}
^1S,\,^3S,\,^1P,\,^3P,\,^1D,\,^3D
\end{align} Which of these terms can be fully antisymmetrized? Here, only the terms ^{1}*S*, ^{3}*P* and ^{1}*D* fulfill Pauli’s principle. In general the exchange interaction (seen in the discussion of He), will then lower the energy of the states with high spins.

## Optional: Symmetry of the *L* states

We can construct the following *L*-states for them: \begin{align}
\ket{L=2,M_L=2} &= \ket{\overbrace{1}^{l_1},\overbrace{1}^{m_{l_1}};\overbrace{1}^{l_2},\overbrace{1}^{m_{l_2}}} \label{eq:lstate1}
\end{align} \begin{align}
\ket{L=1,M_L=0} &= \frac{1}{\sqrt{2}} ( \ket{\overbrace{1}^{m_{l_1}},\overbrace{-1}^{m_{l_2}}} - \ket{-1,1} ) \label{eq:lstate2}
\end{align} \begin{align}
\ket{L=0,M_L=0} &= \frac{1}{\sqrt{3}}(\ket{\overbrace{1}^{m_{l_1}},\overbrace{-1}^{m_{l_2}}} - \ket{0,0} + \ket{-1,1} ) \label{eq:lstate3}
\end{align}

The states and are symmetric and the state is antisymmetric with respect to particle exchange.

Lecture 15 - Diatomic molecules

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Lecture 13 - Atoms with many electrons

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