Every electron is replaced by one that feels an average. That single substitution turns an unsolvable problem into the radial equation of chapter 9, wearing a different potential.

Feynman’s question opens the chapter: why do chemists count funny? Instead of 1, 2, 3, 4 they say hydrogen, helium, lithium, beryllium.

The answer is that they are counting something real. Every atom with $Z$ electrons is identical to every other — same size, same ionization energy, same willingness to react. And that sameness is resilient: knock an atom about with a photon or another atom and it returns to precisely the state it left. There are only about a hundred kinds of thing, and each kind is perfectly reproducible.

No classical model gives you that. A solar system with six planets can have them anywhere. This chapter shows where the sameness comes from, using nothing new: ch09’s central-potential machinery and ch10’s exclusion principle, combined.

The problem, and why it cannot be solved

An atom is $Z$ electrons attracted to a nucleus and repelling each other. For helium that reads

V(\mathbf r_p,\mathbf r_q) = -\frac{2e^2}{4\pi\epsilon_0 r_p} - \frac{2e^2}{4\pi\epsilon_0 r_q} + \frac{e^2}{4\pi\epsilon_0|\mathbf r_p - \mathbf r_q|}\tag{11.1}

and the eigenvalue equation is

\left[-\frac{\hbar^2}{2m_e}\left(\nabla_p^2 + \nabla_q^2\right) + V(\mathbf r_p,\mathbf r_q)\right]\psi(\mathbf r_p,\mathbf r_q) = E\,\psi(\mathbf r_p,\mathbf r_q)\tag{11.2}

Look at the last term of Eq. (11.1). It couples $\mathbf r_p$ to $\mathbf r_q$ , so the equation does not separate — and everything this book has done so far has depended on separating. Two electrons already put six coordinates into one partial differential equation; uranium puts 276.

Accurate numerical solutions exist for small atoms. For anything larger, the approximation below is not a convenience but the only way in.

The central field approximation

The idea is to stop tracking what the other electrons are doing individually and replace them by their average effect: each electron moves independently in one spherically symmetric potential that already includes everybody else.

What should that potential look like? Both ends are known exactly.

Far out, an electron sees the nucleus $Ze$ with $Z-1$ other electrons between it and the nucleus, cancelling all but one unit of charge:

V_B(r) = -\frac{e^2}{4\pi\epsilon_0 r}\tag{11.3}

Close in, it is inside the other electrons entirely and sees the bare nucleus:

V_C(r) = -\frac{Ze^2}{4\pi\epsilon_0 r}\tag{11.4}

The truth interpolates. Eq. (11.5) does it with one exponential:

V_A(r) = -\frac{z(r)e^2}{4\pi\epsilon_0 r}, \qquad\text{where}\qquad z(r) = (Z-1)e^{-r/a} + 1\tag{11.5}

Check the limits: $z(0) = Z$ and $z(\infty) = 1$ , exactly as required. The screening radius $a$ says how quickly one becomes the other, and Fig. 11.1 uses $a = a_0/2$ with $Z = 6$ .

This is mean-field approximation, and you have probably written one

Replacing pairwise interactions by each element’s average effect on the others is a standard move well outside physics. It is how PageRank treats the link graph, how mean-field variational inference treats a joint distribution, and how every $n$ -body simulation with a Barnes–Hut tree treats distant masses.

The structure is always the same: an $O(N^2)$ coupled problem becomes $N$ independent $O(1)$ problems, at the cost of a self-consistency requirement — the field depends on the solution, and the solution depends on the field.

The resolution is also always the same, and the book states it plainly: guess a potential, solve for the electrons, recompute the potential from where they turned out to be, and repeat until input matches output. That is the self-consistent field method, and it is a fixed-point iteration on a function. Hartree published it in 1928; you would recognise the loop immediately.

Here are the three potentials. Turn on the comparison to see $V_A$ bracketed by its two limits, exactly as Fig. 11.1 draws it.

Fig. 11.1 — the central potential for carbon, bracketed by its two limits

compare with bare / fully screened

atomic number ZZ = 6screening radius aa = 0.50 a₀

orbital quantum number l — moves the centrifugal barrierl = 0

Chapter 11's central field approximation, Eq. (11.5). The nucleus is bare at short range and screened to a single unit of charge at long range, and everything chemistry cares about lives in between. With l = 0 there is no centrifugal barrier at all — the potential is purely attractive, and a classical particle would fall straight in. Solved by the same grid → hamiltonian → eigh pipeline as every other potential on this site — because Eq. (9.9) is one-dimensional.

The payoff: this is an equation we have already solved

Given any central potential, each electron obeys

\left[-\frac{\hbar^2}{2m_e}\nabla^2 + V(r)\right]\psi(\mathbf r) = E\,\psi(\mathbf r)\tag{11.6}

Eq. (11.6) is Eq. (9.5). Not similar to it — the same equation, with a different $V(r)$ . Everything ch09 built therefore transfers intact: the separation into radial and angular parts, the spherical harmonics, the radial equation with its centrifugal barrier, and the quantum numbers $n$ , $l$ , $m_l$ , $m_s$ . Single-particle states get the atomic-physics name orbitals and the spectroscopic labels 1s, 2s, 2p.

One thing does change, and it is the whole of chemistry

For hydrogen, energy depended on $n$ alone. Levels with the same $n$ and different $l$ were degenerate — §9.2 called this an accidental degeneracy, special to the $1/r$ potential.

$V_A(r)$ is not $1/r$ . So that accident does not happen, and

E \longrightarrow E_{nl}

The energy now depends on $l$ as well as $n$ , and it increases with $l$ .

The mechanism is §9.1’s centrifugal barrier $l(l+1)\hbar^2/2m_er^2$ . A higher $l$ means a stronger barrier, so the electron is held further out — where the nucleus is well screened and the attraction is weaker. Higher $l$ , less penetration, weaker binding, higher energy.

If hydrogen’s degeneracy had survived, every shell would fill in one step and there would be no periodic table.

Solve it yourself. This is the same grid → hamiltonian → eigh pipeline from §0.3 that has generated every result on this site — now with Eq. (11.5) as the potential.

Eq. (11.6) solved in the screened potential — raise a to remove screening

compare with bare / fully screened

atomic number ZZ = 6screening radius aa = 0.50 a₀

orbital quantum number l — moves the centrifugal barrierl = 0

E = -8.609573 = -17.219 E_R, so Z* = 4.150

Reproduce it in NumPy

The control matters as much as the result: with screening switched off the same solver must return -Z^2/n^2 exactly, and it does to three decimals. Only then is it worth trusting on the screened potential, where no closed form exists to check against. Sturm bisection counts eigenvalues below a trial energy in O(N), which is what makes a 60,000-point grid affordable here.

import numpy as np

# Atomic units: hbar = m_e = e^2/4pi eps0 = 1. Energies in Hartree; 1 Hartree = 2 E_R.
Z, a = 6, 0.5                                    # carbon, screening radius a_0/2
zeff = lambda r: (Z-1)*np.exp(-r/a) + 1          # Eq. (11.5)
VA   = lambda r: -zeff(r)/r                      # screened
VC   = lambda r: -Z/r                            # Eq. (11.4), bare nucleus

def sturm(d, e2, x):
    """How many eigenvalues lie below x. O(N) for a tridiagonal matrix -- the
       same trick ch00 used, and what makes a 60,000-point grid affordable."""
    q = d[0] - x; n = 1 if q < 0 else 0
    for i in range(1, len(d)):
        q = d[i] - x - (e2/q if q != 0 else 1e300)
        if q < 0: n += 1
    return n

def level(V, l, j, rmax=80.0, N=60000):
    """j-th eigenvalue of -1/2 u'' + [l(l+1)/2r^2 + V] u = E u, in Rydbergs.
       Starting the grid at dr rather than 0 imposes u(0) = 0 exactly."""
    dr = rmax/N; r = np.linspace(dr, rmax, N)
    d  = 1/dr**2 + V(r) + l*(l+1)/(2*r*r)
    e2 = (0.5/dr**2)**2
    lo, hi = -4000.0, 5.0
    for _ in range(200):
        m = (lo+hi)/2
        if sturm(d, e2, m) > j: hi = m
        else: lo = m
    return (lo+hi)/2 * 2

print("Eq. (11.6) is Eq. (9.5). Same solver, new potential.")
print("=" * 76)
print("  First, the control: switch screening OFF and we must recover hydrogen-like")
print("  levels -Z^2/n^2 exactly, because a bare 1/r potential is what ch09 solved.\n")
print(f"    {'orbital':<10}{'computed':>14}{'-Z^2/n^2':>12}")
for lab, l, j, n in (('1s',0,0,1), ('2s',0,1,2), ('2p',1,0,2), ('3s',0,2,3)):
    print(f"    {lab:<10}{level(VC,l,j):>14.3f}{-Z*Z/n**2:>12.3f}")
print("\n  Exact to three decimals, and 2s and 2p are DEGENERATE -- hydrogen's")
print("  accidental degeneracy, which survives only for a pure 1/r potential.")

print("\nNow switch screening on: Eq. (11.5) with Z = 6, a = a_0/2")
print("=" * 76)
E = {}
print(f"    {'orbital':<10}{'screened':>12}{'bare':>12}{'raised by':>12}{'Z* = n sqrt(-E)':>18}")
for lab, l, j, n in (('1s',0,0,1), ('2s',0,1,2), ('2p',1,0,2)):
    e, b = level(VA,l,j), level(VC,l,j)
    E[lab] = e
    print(f"    {lab:<10}{e:>12.3f}{b:>12.3f}{e-b:>12.3f}{n*np.sqrt(-e):>18.3f}")

print(f"""
  Two things to read off.

  Screening RAISES every level -- the electron is less tightly held once the
  nucleus is partly cancelled. And 2s and 2p are no longer degenerate: 2p sits
  {E['2p']-E['2s']:.3f} E_R ABOVE 2s. That splitting is the whole of chemistry.

  Z* is the point charge that would produce the same binding. The 1s electron
  sees Z* = {1*np.sqrt(-E['1s']):.2f} of the available 6 -- almost the bare nucleus, because it
  sits inside the screening cloud. The 2p electron sees Z* = {2*np.sqrt(-E['2p']):.2f}, barely more
  than a single proton, because the centrifugal barrier keeps it outside.""")

print("\nCarbon's ground state and first excited state")
print("=" * 76)
gs = 2*E['1s'] + 2*E['2s'] + 2*E['2p']
ex = 2*E['1s'] + 1*E['2s'] + 3*E['2p']
print(f"    (1s)^2 (2s)^2 (2p)^2 = 2({E['1s']:.2f}) + 2({E['2s']:.3f}) + 2({E['2p']:.3f}) = {gs:.1f} E_R")
print(f"    (1s)^2 (2s)   (2p)^3 = 2({E['1s']:.2f}) + 1({E['2s']:.3f}) + 3({E['2p']:.3f}) = {ex:.1f} E_R")
print(f"    excitation energy = E_2p - E_2s = {ex-gs:.2f} E_R")
print(f"""
    The book quotes -41.1 and -40.6 E_R. Solving Eq. (11.6) with the parameters
    the book states gives {gs:.1f} and {ex:.1f} -- about 4.6 % lower. The DIFFERENCE
    between them, which is what the comparison is for, agrees: {ex-gs:.2f} against 0.5.

    This is not an erratum. The book calls Fig. 11.2's screened levels "a rough
    guide" and credits them to a separate numerical calculation. But this site
    solves the equation in front of you, so it owes you its own numbers and an
    explanation of why they differ rather than a silent disagreement.""")

printsEq. (11.6) is Eq. (9.5). Same solver, new potential. ============================================================================ First, the control: switch screening OFF and we must recover hydrogen-like levels -Z^2/n^2 exactly, because a bare 1/r potential is what ch09 solved. orbital computed -Z^2/n^2 1s -35.999 -36.000 2s -9.000 -9.000 2p -9.000 -9.000 3s -4.000 -4.000 Exact to three decimals, and 2s and 2p are DEGENERATE -- hydrogen's accidental degeneracy, which survives only for a pure 1/r potential. Now switch screening on: Eq. (11.5) with Z = 6, a = a_0/2 ============================================================================ orbital screened bare raised by Z* = n sqrt(-E) 1s -20.185 -35.999 15.814 4.493 2s -0.917 -9.000 8.083 1.915 2p -0.382 -9.000 8.618 1.236 Two things to read off. Screening RAISES every level -- the electron is less tightly held once the nucleus is partly cancelled. And 2s and 2p are no longer degenerate: 2p sits 0.535 E_R ABOVE 2s. That splitting is the whole of chemistry. Z* is the point charge that would produce the same binding. The 1s electron sees Z* = 4.49 of the available 6 -- almost the bare nucleus, because it sits inside the screening cloud. The 2p electron sees Z* = 1.24, barely more than a single proton, because the centrifugal barrier keeps it outside. Carbon's ground state and first excited state ============================================================================ (1s)^2 (2s)^2 (2p)^2 = 2(-20.19) + 2(-0.917) + 2(-0.382) = -43.0 E_R (1s)^2 (2s) (2p)^3 = 2(-20.19) + 1(-0.917) + 3(-0.382) = -42.4 E_R excitation energy = E_2p - E_2s = 0.53 E_R The book quotes -41.1 and -40.6 E_R. Solving Eq. (11.6) with the parameters the book states gives -43.0 and -42.4 -- about 4.6 % lower. The DIFFERENCE between them, which is what the comparison is for, agrees: 0.53 against 0.5. This is not an erratum. The book calls Fig. 11.2's screened levels "a rough guide" and credits them to a separate numerical calculation. But this site solves the equation in front of you, so it owes you its own numbers and an explanation of why they differ rather than a silent disagreement.

What each electron in carbon actually sees

The readout above converts each energy into an effective nuclear charge , $Z^* = n\sqrt{-E/E_R}$ — the point charge that would produce the same binding. For carbon ( $Z = 6$ ):

1s: $Z^* = 4.49$ of a possible 6. It sits deep inside the screening cloud and sees almost the bare nucleus.
2s: $Z^* = 1.92$ .
2p: $Z^* = 1.24$ — barely more than a single proton, though there are six in the nucleus.

The trend is the chapter’s mechanism in three numbers: screening gets stronger as $l$ increases, because higher $l$ means less penetration. Problem 3 measures exactly this for sodium and finds $Z^* = 1.84$ , $1.18$ and $1.003$ for 3s, 3p and 3d — a 3d electron in sodium sees a single unit of charge, having been screened from all eleven protons but one.

The construction kit

Now add ch10. A multi-electron state must be antisymmetric under exchanging any two electrons, and that is achieved by assigning electrons to orbitals under the Pauli exclusion principle: no two electrons share all four quantum numbers $n$ , $l$ , $m_l$ , $m_s$ .

Count the capacity of each subshell. For given $l$ there are $2l+1$ values of $m_l$ , and for each, two values of $m_s$ :

subshell⇅	l⇅	values of m_l⇅	capacity 2(2l+1)⇅
s	0	0	2
p	1	−1, 0, +1	6
d	2	−2 … +2	10
f	3	−3 … +3	14

Carbon has six electrons. Fill from the bottom: two into 1s, two into 2s, and the last two into 2p — giving the electron configuration $(1s)^2(2s)^2(2p)^2$ with

E = 2E_{1s} + 2E_{2s} + 2E_{2p}

The first excited state moves one electron from 2s to 2p, giving $(1s)^2(2s)(2p)^3$ and $E = 2E_{1s} + E_{2s} + 3E_{2p}$ .

The book confesses to two lies in one paragraph, and both matter

Page 233 contains an unusually honest passage. Having just walked through carbon’s configuration, the book says:

In the preceding paragraph we have followed custom and given the wrong impression that particular electrons are in particular orbitals. This is not the case.

Because the electrons are indistinguishable, each is equally associated with every occupied orbital. The real state is antisymmetric under exchanging any two — ch10’s Eq. (10.19) generalized to $Z$ particles. The diagram above, with its six separate dots, is a device for counting states. It is not a picture of where anything is.

The second confession follows immediately:

We have also wrongly given the impression that the central potential … is easy to find.

It is not. The potential depends on the electron distribution, which depends on the potential — “like the chicken and the egg; it is unclear which comes first.” The resolution is the self-consistent iteration described above, and it is why atomic structure is a computational subject rather than an analytic one.

Both admissions arrive in one paragraph and are easy to read straight past. They mark the honest boundary of everything this chapter claims.

Where this is going

The configuration $(1s)^2(2s)^2(2p)^2$ is the coarsest description of carbon that is any use — and it is not the whole story. Those six electrons can be arranged in fifteen distinct states this level of description cannot tell apart, and residual repulsion gives them different energies. §11.1b separates them.

If you only want the periodic table, configurations are enough — and §11.2 builds it from exactly what is on this page: a sequence of orbital energies, and Pauli.

Check yourself

0 / 6 answered

1.Why can Eq. (11.2) not be solved by the separation of variables used everywhere else in this book?
2.In Eq. (11.5), $z (r) = (Z - 1) e^{- r / a} + 1$ . What forces the two constants to be $Z - 1$ and $+ 1$ ?
In hydrogen, states with the same $n$ and different $l$ have exactly the same energy.
3.Why does that degeneracy disappear in a many-electron atom?
4.The calculation finds $Z^{*} = 4.49$ for carbon's 1s electron but only $1.24$ for its 2p electron. What does that contrast show?
5.The book says that talking about "particular electrons in particular orbitals" gives the wrong impression. What is the right picture?
6.Why is the central potential described as a "chicken and egg" problem, and how is it resolved?