Atomic and Molecular Physics
Level scheme of two-electron atoms
Taking into account the electric dipole approximation, it is possible to show that radiative
transitions between singlet and triplet spin states are not possible as long as we consider
the spin-orbit interactions are negligible. That is particularly true for atoms or ions whose
atomic number is low enough (under 40), which is the case of the Helium atom (the one
concerning us). For these kind of atoms the energy spectrum consist of two nearly independent systems of levels, one formed of para states, where S = 0, and the other of ortho states,
where S = 1. This fact is shown in Figure 1, which is a representation of the first energy
levels of the atom we are studying. That is the discrete part of the spectrum.
Figure 1 – Experimental values of the lowest energy levels of helium. The ionisation threshold correspond to the zero energy level. All the levels represented here are of the form 1s
nl.
Let us denote the sum of the orbital angular momentum of the electrons by L = L1 + L2 .
The eigenvalues of L2 will be represented as L(L + 1) (in atomic units), and those of Lz will
be denoted as ML = −L, −L + 1..., L − 1, L. A common way of expressing the energy levels
is the next one:
2S+1
L
where for a given value of L is associated a letter following the rule:
L = 0 1 2 3 4 5
↕ ↕ ↕ ↕ ↕ ↕
SP DF GH
1
Atomic and Molecular Physics
It is easily deduced that the quantity 2S + 1, situated on the left, gives the multiplicity of
the state, and takes the value 1 when S = 0 (singlet states), and 3 when S = 1 (triplet states).
There are two interactions, which are the spin-orbit effect (concerning the interaction
between the spin and the angular momentum) and the spin-spin effect (coming from the
interaction of the spins of the electrons), whose direct consequence is the fine structure splitting of the levels shown before. Nevertheless, this fact cannot be appreciated in the previous
figure. If we wanted to eliminate partially the degeneracy of the triplet states, we could
define the total angular momentum as J = L + S (let us remind the eigenvalues of the J2
operator are J(J + 1) and those of Jz are Mz ), so that each energy level of the triplet states
(unless the 3S state) splits into three different ones very close to each other, corresponding
to the three different values of J = L − 1, L, L + 1.
A full spectrum for the He++ nucleus and the two electrons is shown in the following
figure. We choose the origin of the energy so that E=0 is the minimum energy necessary to
ionize two electrons.
Figure 2 – Complete energy level spectrum of helium. In contrast with the previous figure,
the threshold for the ionisation of the two electrons is the new zero energy.
In the diagram can be appreciated the discrete levels of helium are between −79.0eV
(the energy of the ground state of He) and −54.4eV (the ground state of the He+ ). Consequently, we can establish that the ionisation potential for the helium ground state (the
energy necessary for ionizing the atom) is, obviously, the difference of the energies of these
two levels:
Ip = E0 (He+ ) − E0 (He) = −54.4 + 79 = 24.6eV
2
(1)
Atomic and Molecular Physics
This spectrum is comparable for any other two-electron ions with an atomic number
higher than two. Also we can consider the example of the hydrogen anion H − with only one
bound state and a ionisation potential around 0.75eV , which tell us that this ion is not very
stable and can easily dissociate into a neutral hydrogen atom and a free electron.
Figure 3 – Ground state of H − .
d⟨Nij ⟩
= αij
dt
!
X
k̸=i
⟨Nik ⟩ +
X
⟨Njk ⟩
k̸=j
3
− [(N − k − 1)αmin + kαmax ]⟨Nij ⟩
(2)