# 9.9: Multiplicity and Degeneracy of Excited States

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Ignoring electron-electron interaction, all 1s2s and 1s2p states have the same energy. The perturbation $$(e^2/4\pi \epsilon_0 r_{12})$$ lifts that degeneracy, and we can treat it with degenerate perturbation theory. Rather than evaluating the integral in the 4x4 matrix exactly, we can use a physical argument: $$(e^2/4\pi \epsilon_0 r_{12})$$ is not an external potential, and so applies no net torque or force on the electrons. The perturbation cannot change the angular momentum, so it cannot mix states with different $$l$$ or $$m$$. The theta integral will be $$\delta_{ll'}$$, and the phi integral $$\delta_{mm'}$$, total angular momentum remains a good quantum number: L=0(1s2s) or L=1(1s2p). Since the 2s state has finite probability of being at the nucleus, and the 2p has zero probability of being there, the 2s state is less well screened from the nuclear charge by the 1s and will have lower energy.

For a given spatial excited state the possible normalised spin wavefunction combinations, consistent with the antisymmetry requirement are a spin triplet and a spin singlet.

$\Phi_3 = (\phi_{nlm,n' l'm'} − \phi_{n' l'm' ,nlm})(\uparrow \uparrow )/ \sqrt{2} \\ (\phi_{nlm,n' l'm'} − \phi_{n' l'm' ,nlm})(\downarrow \downarrow )/ \sqrt{2} \\ (\phi_{nlm,n' l'm'} − \phi_{n' l'm' ,nlm})(\uparrow \downarrow + \downarrow \uparrow )/2 \nonumber$

$\Phi_1 = (\phi_{nlm,n' l'm'} + \phi_{n' l'm' ,nlm})(\uparrow \downarrow − \downarrow \uparrow )/2 \nonumber$

Where $$|\phi_{nlm,n' l 'm'} \rangle$$ represents electron 1 in a hydrogenic state with quantum numbers $$n$$, $$l$$ and $$m$$ and electron 2 with $$n'$$, $$l'$$, and $$m'$$. The subscripts on the $$\Phi$$ label spin multiplicity (2S+1) Figure $$\PageIndex{1}$$

Again whole effect of the potential is contained in the spatial part, the spin integral will be $$\delta_{\sigma \sigma'}$$. so off-diagonal matrix elements are all zero. We need to evaluate

$J_{nl} = \langle \phi_{nlm,n' l'm'} |(e^2/4\pi \epsilon_0 r_{12})|\phi_{nlm,n' l'm'} \rangle \text{ - the direct integral.} \nonumber$

$K_{nl} = \langle \phi_{nlm,n' l'm'} |(e^2/4\pi \epsilon_0 r_{12})|\phi_{nlm,n' l'm'} \rangle \text{ - the exchange integral.} \nonumber$

with which perturbation theory gives an energy shift in the $$1s^12s^1$$ state of:

$\frac{1}{2} \frac{e^2}{4\pi \epsilon_0} ( \langle \phi_{100,200} | 1/r_{12}|\phi_{100,200} \rangle + \langle \phi_{200,100} |1/r_{12}|\phi_{200,100} \rangle \pm \langle \phi_{100,200} |1/r_{12}|\phi_{200,100} \rangle \pm \langle \phi_{200,100}|1/r_{12}|\phi_{100,200} \rangle) \nonumber$

where the + applies to the singlet state and the − to the triplet. The direct integral, electronelectron repulsion, increases the energy, but the exchange integral can either increase of decrease energy.

Thus the energy levels are split by different direct interactions into L=0 and L=1 and again through exchange interaction into singlet and triplet. The final degeneracies of states with one electron excited to n=2 are 3,1,9 and 3. The spectroscopic notation in the figure gives the quantum numbers as: $$(nl)(n' l')^{2S+1}L_J$$

Again, the most useful quantum number labels are the total spin and angular momentum: we could write the perturbation energy as $$\Delta E = J_{nl} − (2S − 1)K_{nl}$$, even though the perturbing potential does not act on the spin. The ‘exchange force’ selects preferred spin state via the requirement of overall antisymmetry.

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