Chapter 5
Many-Electron Atoms
5.1Introduction
Add one electron to hydrogen and the problem becomes unsolvable — not merely difficult, but unsolvable in closed form, in the same sense as the three-body problem of classical mechanics. There is no exact expression for the ground-state energy of helium.
The character of the subject therefore changes here. From this point on we work with controlled approximations, and the skill to acquire is judging how good each one is. Fortunately helium is forgiving: three successively better treatments, each a page of algebra, bracket the true answer to within two per cent, and every one of them teaches something that carries forward.
The chapter establishes four things:
Antisymmetry, from which the Pauli principle follows as a theorem rather than a postulate;
Screening — computed, not asserted — which breaks hydrogen's accidental \(l\)-degeneracy and produces the quantum defect;
Exchange, an energy splitting of nearly an electron volt that arises from antisymmetry alone and has no classical analogue;
Term symbols, the notation \(^{2S+1}L_J\) without which no table of atomic data can be read.
At the end we meet a state that can fall apart without absorbing anything — a bound state degenerate with a continuum. That situation is the destination of this book, and helium provides its simplest example.
5.2Identical Particles and Antisymmetry
Electrons are indistinguishable: no measurement can establish which is which. Formally, the exchange operator \(\hat P_{12}\) that swaps particles \(1\) and \(2\) must commute with the Hamiltonian, and since \(\hat P_{12}^{2}=\hat I\) its eigenvalues are \(\pm1\). Nature assigns \(-1\) to all fermions, electrons among them.
The state of a system of identical fermions changes sign under exchange of any two of them:
where \(\xi=(\vec r,m_s)\) collects position and spin.
No two identical fermions can occupy the same one-particle spin-orbital.
Set \(\xi_i=\xi_j\) in Eq. (5.1). The left- and right-hand sides are then the same function, so \(\Psi=-\Psi\), hence \(\Psi=0\). A state with two electrons in the same spin-orbital does not exist.
Exclusion is therefore not an extra rule bolted on to quantum mechanics; it is a two-line consequence of antisymmetry.
For \(N\) electrons occupying spin-orbitals \(\phi_1,\dots,\phi_N\), the normalized antisymmetric state is
The construction works because a determinant changes sign when two rows are interchanged — which is exactly Eq. (5.1) — and vanishes when two columns are equal, which is Theorem 5.2.1.
Write out Eq. (5.2) for \(N=2\) and verify both properties.
Solution.
For two electrons,
Antisymmetry. Swapping \(\xi_1\leftrightarrow\xi_2\) gives
Exclusion. Setting \(\phi_b=\phi_a\),
Normalization. If \(\phi_a\) and \(\phi_b\) are orthonormal, then expanding \(\int|\Psi|^{2}\) gives four terms: two of the form \(\int|\phi_a|^{2}\int|\phi_b|^{2}=1\), each with prefactor \(\tfrac12\), and two cross terms proportional to \(\braket{\phi_a}{\phi_b}=0\). Hence \(\int|\Psi|^{2}=1\), which is why the prefactor is \(1/\sqrt{N!}\).
5.3Helium: The Hamiltonian
Helium is two electrons and a nucleus of charge \(Ze\) with \(Z=2\). Working in atomic units (Chapter 4, Sec. 4.8), the Hamiltonian is
with \(r_{12}=|\vec r_1-\vec r_2|\). In SI units the last term is \(e^{2}/4\pi\epsilon_0r_{12}\).
The operators \(\hat h_1\) and \(\hat h_2\) each involve one electron only, so if \(\hat V_{ee}\) were absent the equation would separate and the solution would be a product of two hydrogenic orbitals with \(Z=2\). The term \(1/r_{12}\) couples the coordinates: it cannot be written as \(f(\vec r_1)+g(\vec r_2)\), no change of variables removes it, and consequently no separable solution exists. That single term is the entire difficulty of many-electron atoms.
The experimental facts to be reproduced are
5.4First Approximation: Independent Electrons
Drop \(\hat V_{ee}\) entirely. The equation separates, and each electron occupies a hydrogenic \(1s\) orbital with \(Z=2\),
which is \(R_{10}Y_0^{0}\) from Chapter 4, Eqs. (4.51) and (4.31), collected into one expression.
Compute the ground-state energy of helium neglecting the electron–electron repulsion.
Solution.
Each electron has the hydrogenic energy of Eq. (4.47) with \(n=1\),
so for two independent electrons
Against the true \(-79.0\) eV this is too low by \(29.8\) eV — an error of \(38\%\). The sign is right and instructive: we omitted a repulsion, and repulsion raises energy, so neglecting it must give a value that is too negative. Any approximation whose error has the wrong sign contains a mistake.
5.5Second Approximation: First-Order Perturbation Theory
Now treat \(\hat V_{ee}=1/r_{12}\) as a perturbation and add its expectation value in the zeroth-order state:
This is a six-dimensional integral, but it can be done exactly, and the method is worth learning because the same trick recurs.
For the normalized hydrogenic \(1s\) orbital of Eq. (5.5),
The charge density of electron \(2\), \(\rho(r)=|\phi_{1s}(\vec r)|^{2}=(Z^{3}/\pi)e^{-2Zr}\), is spherically symmetric. By the classical shell theorem, the potential it produces at radius \(R\) splits into the charge inside, acting as if concentrated at the origin, and the shells outside, each contributing \(1/r\):
Inner integral. With \(\rho\,4\pi r^{2}=4Z^{3}r^{2}e^{-2Zr}\) and the standard result \(\int_0^{R}r^{2}e^{-ar}dr=\frac{2}{a^{3}} -\left(\frac{R^{2}}{a}+\frac{2R}{a^{2}}+\frac{2}{a^{3}}\right)e^{-aR}\) with \(a=2Z\),
Outer integral. With \(\int_R^{\infty}re^{-ar}dr=\left(\frac{R}{a}+\frac{1}{a^{2}}\right)e^{-aR}\),
Combining. Dividing the first by \(R\) and adding,
The \(2Z^{2}R\) terms cancel — a useful check.
Averaging over electron 1. Now
Split into three elementary integrals, using \(\int_0^\infty r^{n}e^{-ar}dr=n!/a^{n+1}\):
Therefore
Use Theorem 5.5.1 to correct Example 5.4.1.
Solution.
For \(Z=2\),
so
We have moved from \(38\%\) too low to \(5.3\%\) too high. Overshooting is expected: the orbitals were not allowed to respond to the repulsion, so each electron is held closer to the nucleus than it really is and feels more repulsion than it really does. Letting the orbitals relax is the next approximation.
5.6Third Approximation: The Variational Method
For any normalized trial state \(\ket{\Psi_t}\) and any Hamiltonian \(\hat H\) with ground-state energy \(E_0\),
Expand \(\ket{\Psi_t}=\sum_nc_n\ket{n}\) in the exact eigenstates, with \(\hat H\ket{n}=E_n\ket{n}\) and \(\sum_n|c_n|^{2}=1\). Then
since every \(E_n\ge E_0\).
The principle turns approximation into optimisation: choose a trial state with adjustable parameters, compute \(\expval{\hat H}\), and minimise. The answer is guaranteed to be an upper bound, which is a valuable property — a calculation returning an energy below the exact one has a bug.
The physical failing of Sec. 5.5 was that each electron still saw the full nuclear charge \(Z=2\) when its partner partly shields it. So let the orbital exponent be a free parameter \(Z'\):
Minimise \(\expval{\hat H}\) over \(Z'\) and evaluate for \(Z=2\).
Solution.
Three expectation values are needed. For a hydrogenic orbital with exponent \(Z'\), Chapter 4's results Eq. (4.58) give \(\expval{1/r}=Z'\), and the virial relation for that orbital gives \(\expval{-\tfrac12\nabla^{2}}=Z'^{2}/2\).
Kinetic energy, two electrons:
Nuclear attraction, two electrons, with the true charge \(Z\) in the operator but the trial exponent \(Z'\) in the state:
Electron repulsion. Theorem 5.5.1 was derived for orbitals of exponent \(Z\); here the exponent is \(Z'\), so
Adding,
Minimise:
Substituting back, and noting that \(2Z-\tfrac58=2Z'\),
For helium, \(Z=2\):
The error is now \(1.9\%\), and — as Theorem 5.6.1 requires — it lies above the exact \(-79.01\) eV.
Read the middle number, because it is the physics. Each electron in helium does not experience charge \(2\); it experiences \(1.69\). Its partner has screened away about \(0.31\) of an elementary charge. That idea — the nuclear charge an electron actually feels is less than \(Z\) — organises the rest of this chapter.
The three treatments, in order: \(-108.8\), \(-74.8\), \(-77.5\) eV against the true \(-79.0\) eV. The pattern is characteristic. A crude model can be badly wrong; first-order perturbation theory usually overcorrects; and allowing the wave function itself to relax recovers most of the remainder. Going further requires trial functions that depend on \(r_{12}\) explicitly, which is where quantum chemistry proper begins and where this course stops.
5.7Exchange: Singlet and Triplet Helium
Now excite one electron, to the configuration \(1s2s\), and call the two spatial orbitals \(\phi_a=\phi_{1s}\) and \(\phi_b=\phi_{2s}\). Antisymmetry of the total state can be achieved in two ways, because the state factorises into a spatial part and a spin part: one of them must be antisymmetric, and either will do.
Both are legitimate. Their energies differ, and the calculation shows exactly why.
\(J\) has a classical reading: it is the electrostatic repulsion between two charge clouds \(|\phi_a|^{2}\) and \(|\phi_b|^{2}\). \(K\) has none — the orbitals are swapped between the two factors, so no classical charge distribution corresponds to it. It exists only because the wave function must be (anti)symmetrised.
For the states of Eq. (5.11),
with \(+\) for the singlet and \(-\) for the triplet. Since \(K>0\) for orbitals with overlapping densities, the triplet lies lower.
The one-electron parts give \(\varepsilon_a+\varepsilon_b\) regardless of symmetrisation. For the repulsion, insert Eq. (5.11) into \(\bra{\Psi_{\pm}}1/r_{12}\ket{\Psi_{\pm}}\) and expand the product of two brackets. Four terms appear. Two are of the form \(\int\!\!\int|\phi_a(1)|^{2}|\phi_b(2)|^{2}/r_{12}\), each with prefactor \(\tfrac12\), summing to \(J\). The other two are the exchange terms, each with prefactor \(\pm\tfrac12\), summing to \(\pm K\).
The Hamiltonian (5.3) contains no spin operator whatsoever, yet the two states differ in energy by \(2K\) and are distinguished only by their spin arrangement. The spin is not causing the splitting; antisymmetry is. Requiring the total state to change sign under exchange ties the spin arrangement to the spatial arrangement, and it is the spatial arrangement that the Coulomb repulsion sees.
The mechanism is visible in Eq. (5.11): setting \(\vec r_1=\vec r_2\) makes the antisymmetric spatial function vanish identically. The triplet electrons therefore avoid each other, are further apart on average, and repel less. This is the exchange interaction, and it has no classical counterpart of any kind.
The \(1s2s\) levels of helium are observed at \(19.820\) eV (\(^{3}S\)) and \(20.616\) eV (\(^{1}S\)) above the ground state. Extract \(K\) and comment on its size.
Solution.
From Eq. (5.14) the separation is \(2K\):
The triplet is lower, confirming \(K>0\).
Compare this with a magnetic interaction between two electron spins, which at a separation of a few \(a_0\) is of order \(10^{-4}\) eV — four orders of magnitude smaller. Whatever splits these levels, it is not magnetism. It is electrostatics, routed through antisymmetry.
The consequence is chemical as well as spectroscopic: the rule that electrons in different orbitals prefer parallel spins — Hund's first rule — is this same \(-K\), and it is why oxygen's ground state is a triplet.
Historically the two families were thought to be different substances and named parahelium (singlet) and orthohelium (triplet); the names survive. Note also that \(1s2s\ ^{3}S\) cannot reach the \(1s^{2}\ ^{1}S\) ground state by one photon — the transition would have to change \(S\), which the dipole operator of Chapter 3 cannot do — so it is metastable, with a lifetime of about two hours. Compare hydrogen's \(2s\) in Example 3.6.1: a second instance of a forbidden route creating a long-lived state.
5.8Heavier Atoms: Screening and the Quantum Defect
For more than two electrons the practical approach keeps the one-electron picture and repairs the potential. Each electron is taken to move in a central average field produced by the nucleus and all the others, so Chapter 3's machinery applies unchanged: states are labelled \(nlm\) and Eq. (3.43) is solved with a modified \(V(r)\).
The essential feature is the behaviour at the two extremes:
An electron far away sees net charge \(+1\) and is hydrogenic. One that penetrates sees much more and is bound much more tightly. And Chapter 3 already told us who penetrates: the centrifugal barrier \(l(l+1)/2r^{2}\) in Eq. (3.43) excludes high \(l\) from the core entirely, while \(l=0\) has no barrier at all. Hence
The accidental \(l\)-degeneracy of Chapter 4 therefore breaks, and in a definite order. It is parameterised by one number per \(l\).
The energies of a series of states of the same \(l\) converging to an ionization threshold \(E_{\mathrm{th}}\) are written
with \(\mathcal{E}_{\mathrm{Ryd}}=13.606\) eV. The number \(\delta_l\) is the quantum defect and \(n^{*}\equiv n-\delta_l\) the effective quantum number.
Two properties make this useful. First, \(\delta_l\) is nearly independent of \(n\) within a series: it is a property of the core, sampled only during the small fraction of the orbit spent inside, so measuring it once predicts every member. Second, inverting Eq. (5.17) converts a measured binding energy directly into \(n^{*}\):
Sodium has one electron outside a closed-shell neon-like core, so it is hydrogen-like in structure and is the standard demonstration. Its ionization energy is \(E_{\mathrm{th}}=5.139\) eV above the ground state. The \(3s\) ground state is bound by \(5.139\) eV, the \(3p\) state by \(3.035\) eV, and the \(3d\) state by \(1.522\) eV. Find \(n^{*}\) and \(\delta_l\) for each.
Solution.
Apply Eq. (5.18) three times.
The ordering is Eq. (5.16), and the magnitudes are striking. The \(3d\) electron is hydrogenic to a third of a per cent: its centrifugal barrier holds it so far outside the core that it cannot tell sodium from hydrogen. The \(3s\) electron has \(n^{*}=1.63\) — it behaves like a hydrogen state with \(n\) between \(1\) and \(2\), because on every orbit it dives through the neon core and briefly feels a charge far larger than \(1\).
One number, \(\delta_l\), and one mechanism, penetration, account for the entire structure of the alkali spectrum. Keep hold of this: every Rydberg series in this book, including the molecular ones of Chapter 9, is described the same way.
5.9Term Symbols
A configuration lists which orbitals are occupied — carbon is \(1s^{2}2s^{2}2p^{2}\). A configuration is not a state: the electrons' angular momenta can couple several ways, giving several energies. The resulting states are labelled by a term symbol.
where \(S\) is the total spin, \(L\) the total orbital angular momentum written as a capital letter (\(L=0,1,2,3\to S,P,D,F\)), \(J\) the total angular momentum obtained by coupling \(L\) and \(S\), and \(2S+1\) the multiplicity (\(1\) singlet, \(2\) doublet, \(3\) triplet).
A filled subshell has \(S=0\) and \(L=0\), and therefore contributes \(^{1}S\) to a term symbol. A subshell containing one hole has the same \(L\) and \(S\) as one containing one electron.
In a filled subshell every \(m_l\) from \(-l\) to \(+l\) is occupied twice, once with each spin. Hence \(M_S=\sum m_s=0\) and \(M_L=\sum m_l=0\), and since this is the only state of the configuration, \(S=L=0\). For the second claim, removing one electron leaves \(M_L=-m_l\) and \(M_S=-m_s\) of the removed electron, so the available \((M_L,M_S)\) values are those of a single electron with reversed signs — the same set, hence the same \(L\) and \(S\).
These two shortcuts remove almost all the labour. In particular, ionizing a closed-shell atom or molecule always leaves one hole and therefore always gives a doublet, \(2S+1=2\).
Assign term symbols to helium's ground state, sodium's ground state, and sodium's first excited state.
Solution.
Helium, \(1s^{2}\). A closed shell, so by Theorem 5.9.1 \(S=0\), \(L=0\), hence \(J=0\) and the term is \(^{1}S_0\) — a singlet, consistent with Sec. 5.7, where the ground configuration admitted only the spin-antisymmetric combination.
Sodium, \(1s^{2}2s^{2}2p^{6}3s^{1}\). All shells closed except one \(3s\) electron, so \(S=\tfrac12\), \(L=0\), \(J=\tfrac12\): the term is \(^{2}S_{1/2}\).
Sodium, \(\dots3p^{1}\). Now \(L=1\) and \(S=\tfrac12\), so \(J\) can be \(L+S=\tfrac32\) or \(L-S=\tfrac12\), giving two terms \(^{2}P_{3/2}\) and \(^{2}P_{1/2}\). These are split by the spin–orbit interaction, and the two transitions down to \(^{2}S_{1/2}\) are the sodium D lines at \(589.0\) and \(589.6\) nm — the yellow of a street lamp, resolved into a doublet by an effect that exists only because the electron has spin.
5.10States Inside a Continuum
One more feature of many-electron atoms must be introduced, because the last chapters of this book depend on it entirely. It cannot occur in hydrogen.
Helium's ionization threshold is \(24.59\) eV: above that energy the system can exist as He\(^{+}\) plus a free electron of any kinetic energy — a continuum. Now excite both electrons, to the configuration \(2s2p\). That state lies at \(60.15\) eV, well above the threshold. It is a perfectly good bound state of the two-electron system — both electrons are attached, and it has a definite energy — and it is simultaneously degenerate with the continuum.
Such a state cannot survive. The repulsion \(\hat V_{ee}\) of Eq. (5.3) connects it to the continuum at the same energy: one electron drops to \(1s\) while the other takes up the released energy and leaves. No photon is absorbed or emitted.
Autoionization is the decay of a bound state that is degenerate with a continuum, driven by an interaction internal to the system rather than by absorption of radiation.
Because the state decays, it is not sharp: it has a width \(\Gamma\) related to its lifetime by \(\Gamma\tau=\hbar\). For helium's \(2s2p\) resonance \(\Gamma\approx37\) meV, so
This configuration — a discrete state embedded in and coupled to a continuum — produces the asymmetric spectral profiles that Fano explained in 1961, using precisely this helium resonance as his example. We lack the tools to treat it yet: it needs the continuum normalization mentioned at the end of Chapter 3 and the time-dependent methods of Chapter 6. Chapter 10 does it properly.
For now, retain the picture and the number. A doubly excited atom is a bound state inside a continuum; it lives femtoseconds rather than nanoseconds; and the cause is the same \(\hat V_{ee}\) that made helium's ground state impossible to compute exactly. The term we could not handle in Remark 5.3.1 turns out to drive the physics this course is aimed at.
5.11Summary
Fermion states are antisymmetric under exchange (Def. 5.2.1); Pauli exclusion (Thm. 5.2.1) and the Slater determinant (Def. 5.2.2) follow.
Helium's Hamiltonian, Eq. (5.3), is not separable because of the single term \(1/r_{12}\).
Three treatments: independent electrons give \(-108.8\) eV; first-order perturbation theory, using \(\expval{1/r_{12}}=5Z/8\) (Thm. 5.5.1), gives \(-74.8\) eV; the variational method with a screened exponent gives \(Z'=Z-\tfrac{5}{16}=1.6875\) and \(E=-Z'^{2}=-77.5\) eV, against the exact \(-79.0\) eV.
The variational principle (Thm. 5.6.1) guarantees an upper bound, so a result below the exact energy indicates an error.
Antisymmetry ties spin to spatial symmetry, giving \(E_{\pm}=\varepsilon_a+\varepsilon_b+J\pm K\) (Thm. 5.7.1). The triplet lies lower by \(2K\); for helium's \(1s2s\), \(2K=0.796\) eV — four orders of magnitude larger than any magnetic spin interaction. This is exchange, and it has no classical analogue.
In heavier atoms each electron moves in a screened central field running from \(-Z/r\) to \(-1/r\). Penetration, controlled by the centrifugal barrier, breaks hydrogen's \(l\)-degeneracy in the order \(s>p>d>f\).
This is parameterised by the quantum defect, \(E_n=E_{\mathrm{th}}-\mathcal{E}_{\mathrm{Ryd}}/(n-\delta_l)^{2}\). For sodium: \(\delta_s=1.373\), \(\delta_p=0.883\), \(\delta_d=0.010\).
Term symbols are \(^{2S+1}L_J\); closed shells contribute \(^{1}S\) and a single hole behaves like a single electron (Thm. 5.9.1), so ionizing a closed shell always gives a doublet.
A doubly excited state can be degenerate with a continuum and decay into it without radiating — autoionization. Helium's \(2s2p\) at \(60.15\) eV has \(\Gamma\approx37\) meV, hence \(\tau\approx18\) fs.
5.12Exercises
The determinant. Write out the Slater determinant for three electrons in orbitals \(\phi_a,\phi_b,\phi_c\). Verify that interchanging \(\xi_1\) and \(\xi_3\) changes the sign, and that setting \(\phi_c=\phi_a\) gives zero.
The Coulomb integral. Reproduce the proof of Theorem 5.5.1, filling in the two standard integrals used. Then evaluate \(5Z/8\) in eV for \(Z=1\), \(2\) and \(3\), and explain physically why the repulsion grows with \(Z\) even though the electrons are being pulled closer together.
The variational calculation.
Derive Eq. (5.10), stating where each of the three terms comes from.
Minimise it to obtain \(Z'=Z-\tfrac{5}{16}\) and \(E_{\min}=-Z'^{2}\).
Apply the result to H\(^{-}\) (\(Z=1\), two electrons). What \(Z'\) and what energy do you get? Compare with the energy of a hydrogen atom plus a free electron (\(-0.5\) a.u.) and comment on whether this trial function predicts H\(^{-}\) to be bound.
Exchange.
Carry out the expansion in the proof of Theorem 5.7.1, showing all four terms explicitly.
Explain in three or four sentences why the spatially antisymmetric state has lower energy, referring to what happens at \(\vec r_1=\vec r_2\).
The measured splitting of helium's \(1s2p\) levels is \(0.25\) eV, smaller than the \(0.796\) eV of the \(1s2s\) levels. Using Eq. (5.13), explain why a \(2p\) orbital gives a smaller exchange integral with \(1s\) than a \(2s\) orbital does.
Quantum defects from data. Using \(E_{\mathrm{th}}=5.139\) eV for sodium and Eq. (5.18), compute \(n^{*}\) and \(\delta\) for the \(4s\) state (bound by \(1.947\) eV) and the \(4d\) state (bound by \(0.851\) eV). Compare with the \(3s\) and \(3d\) values of Example 5.8.1 and state whether \(\delta_l\) is behaving as Def. 5.8.1 claims.
Term symbols. Give the ground-state term symbol for He (\(1s^{2}\)), Li (\(\dots2s^{1}\)), Be (\(\dots2s^{2}\)) and F (\(\dots2p^{5}\)). For fluorine, use Theorem 5.9.1 rather than coupling five electrons.
5.13Project: Screening and the Quantum Defect
The problem. Take the numerical radial solver built in Chapter 4 and put a screened potential into it. Show that screening breaks the \(l\)-degeneracy, produce quantum defects from your own eigenvalues, and compare with sodium.
A convenient model potential — crude but with the right limits — is
in atomic units, with \(Z=11\) for sodium and \(d\) a parameter setting the core size.
On paper. Verify that Eq. (5.21) satisfies both limits of Eq. (5.15). Sketch it against \(-1/r\) and \(-Z/r\), and say where the crossover sits in terms of \(d\).
On paper. Before computing, predict the sign of the effect: will \(3s\) move up or down relative to hydrogen's \(n=3\), and will \(3d\) move more or less? Justify with the centrifugal barrier in two sentences, and write the prediction down.
On the computer. Solve Eq. (3.43) with Eq. (5.21) for \(l=0,1,2\), starting from \(d=0.6\,a_0\). Extract the eigenvalues near \(n=3\) and convert each to a quantum defect using Eq. (5.18) with \(E_{\mathrm{th}}=0\), so the binding energy is simply \(|E|\).
On the computer. Compare with sodium's measured \(\delta_s=1.373\), \(\delta_p=0.883\), \(\delta_d=0.010\). Tune \(d\) to fit \(\delta_s\), then report what the other two do. Does fitting one spoil the others? State which features of the real atom this one-parameter model captures and which it cannot.
On the computer. Plot \(|U_{3l}(r)|^{2}\) for \(l=0,1,2\) in your fitted potential, with the core region \(r<d\) shaded. The figure should make the mechanism visible at a glance: the \(s\) state has weight inside the shading and the \(d\) state almost none. Quantify it by computing \(\int_0^{d}|U_{3l}|^{2}dr\) for each \(l\).
What has to move. One animation: increase the screening smoothly from none (\(Z=1\), pure hydrogen) to \(Z=11\), showing the three \(n=3\) levels starting degenerate and fanning apart with the \(s\) level diving. That fanning is the breaking of the accidental degeneracy.
The check. At \(Z=1\) your three levels must be degenerate at \(-1/18\) a.u. and all three defects must vanish. If not, the error is in the solver, not the physics — fix it there before switching the screening on.
Be ready to answer. Your model has one adjustable parameter and three defects to reproduce. Suppose you can fit all three by hand. What would that tell you — and what would it not tell you — about whether Eq. (5.21) is a good description of the sodium core?