Chapter 4

The Hydrogen Atom

4.1Introduction

The hydrogen atom is the only atom whose Schrödinger equation can be solved exactly in closed form. This chapter is therefore the analytical foundation of the course: every atom and molecule that follows is described by saying how it differs from this solution.

We will carry the calculation out completely. The two-body problem is reduced to one body, the Laplacian is written in spherical coordinates, the variables are separated, both resulting ordinary differential equations are solved, and the quantization of the energy is obtained from the requirement that the radial solution remain finite. The complete normalized wave functions and the energy spectrum are collected in Sec. 4.7.

4.2The Two-Body Problem and the Reduced Mass

A hydrogen atom is an electron of mass $m_{e}$ and charge $- e$ at position ${\vec{r}}_{e}$ , and a proton of mass $m_{p}$ and charge $+ e$ at ${\vec{r}}_{p}$ . The Hamiltonian is

\begin{matrix} \hat{H} = - \frac{ℏ^{2}}{2 m_{e}} \nabla_{e}^{2} - \frac{ℏ^{2}}{2 m_{p}} \nabla_{p}^{2} - \frac{e^{2}}{4 π ϵ_{0} | {\vec{r}}_{e} - {\vec{r}}_{p} |} . \end{matrix}

Introduce center-of-mass and relative coordinates,

\begin{matrix} \vec{R} = \frac{m_{e} {\vec{r}}_{e} + m_{p} {\vec{r}}_{p}}{m_{e} + m_{p}}, \vec{r} = {\vec{r}}_{e} - {\vec{r}}_{p}, \end{matrix}

and define the total and reduced masses

\begin{matrix} M = m_{e} + m_{p}, μ = \frac{m_{e} m_{p}}{m_{e} + m_{p}} \end{matrix}

In these coordinates the kinetic energy separates exactly,

\begin{matrix} - \frac{ℏ^{2}}{2 m_{e}} \nabla_{e}^{2} - \frac{ℏ^{2}}{2 m_{p}} \nabla_{p}^{2} = - \frac{ℏ^{2}}{2 M} \nabla_{R}^{2} - \frac{ℏ^{2}}{2 μ} \nabla_{r}^{2}, \end{matrix}

and the potential depends on $\vec{r}$ alone. Writing $Ψ = Φ (\vec{R}) ψ (\vec{r})$ therefore separates the problem: $Φ$ is a free particle of mass $M$ (the atom as a whole drifting through space, which we discard) and $ψ$ obeys a one-body equation.

\begin{matrix} \hat{H} = - \frac{ℏ^{2}}{2 μ} \nabla^{2} + V (r), V (r) = - \frac{Z e^{2}}{4 π ϵ_{0} r} \end{matrix}

where we have allowed a nuclear charge $Z e$ so that the result covers He $^{+}$ , Li $^{2 +}$ and so on; for hydrogen $Z = 1$ .

Numerically, $m_{p} / m_{e} = 1836.15$ , so

\begin{matrix} μ = \frac{m_{e}}{1 + m_{e} / m_{p}} = m_{e} (1 - \frac{1}{1837.15}) = 0.999456 m_{e} . \end{matrix}

The correction is $5.4 \times 10^{- 4}$ — small, but large enough to be measured, and it is the origin of the isotope shift between hydrogen and deuterium.

4.3The Time-Independent Schrödinger Equation

We solve $\hat{H} ψ = E ψ$ , that is

\begin{matrix} [- \frac{ℏ^{2}}{2 μ} \nabla^{2} - \frac{Z e^{2}}{4 π ϵ_{0} r}] ψ (r, θ, φ) = E ψ (r, θ, φ) . \end{matrix}

Because $V$ depends only on $r$ , spherical coordinates are the natural choice. The Laplacian in spherical coordinates is

\begin{matrix} \nabla^{2} = \frac{1}{r^{2}} \frac{\partial}{\partial r} (r^{2} \frac{\partial}{\partial r}) + \frac{1}{r^{2} \sin θ} \frac{\partial}{\partial θ} (\sin θ \frac{\partial}{\partial θ}) + \frac{1}{r^{2} \sin^{2} θ} \frac{\partial^{2}}{\partial φ^{2}} . \end{matrix}

Substituting Eq. (4.8) into Eq. (4.7) gives the equation in full:

\begin{matrix} \begin{aligned} - \frac{ℏ^{2}}{2 μ} [ & \frac{1}{r^{2}} \frac{\partial}{\partial r} (r^{2} \frac{\partial ψ}{\partial r}) + \frac{1}{r^{2} \sin θ} \frac{\partial}{\partial θ} (\sin θ \frac{\partial ψ}{\partial θ}) \\ + \frac{1}{r^{2} \sin^{2} θ} \frac{\partial^{2} ψ}{\partial φ^{2}}] - \frac{Z e^{2}}{4 π ϵ_{0} r} ψ = E ψ . \end{aligned} \end{matrix}

Comparing Eq. (4.8) with the operator ${\hat{L}}^{2}$ of Chapter 3, Eq. (3.31), the two angular terms are precisely $- {\hat{L}}^{2} / ℏ^{2} r^{2}$ . So Eq. (4.8) can be written compactly as

\begin{matrix} \nabla^{2} = \frac{1}{r^{2}} \frac{\partial}{\partial r} (r^{2} \frac{\partial}{\partial r}) - \frac{{\hat{L}}^{2}}{ℏ^{2} r^{2}}, \end{matrix}

which is the form that makes the separation transparent.

4.4Separation of Variables

Assume a product solution,

\begin{matrix} ψ (r, θ, φ) = R (r) Y (θ, φ) . \end{matrix}

Substitute Eq. (4.11) into Eq. (4.9). The radial derivative acts only on $R$ and the angular derivatives only on $Y$ , so

\begin{matrix} - \frac{ℏ^{2}}{2 μ} [\frac{Y}{r^{2}} \frac{d}{d r} (r^{2} \frac{d R}{d r}) + \frac{R}{r^{2}} Λ Y] - \frac{Z e^{2}}{4 π ϵ_{0} r} R Y = E R Y, \end{matrix}

where for brevity

\begin{matrix} Λ Y \equiv \frac{1}{\sin θ} \frac{\partial}{\partial θ} (\sin θ \frac{\partial Y}{\partial θ}) + \frac{1}{\sin^{2} θ} \frac{\partial^{2} Y}{\partial φ^{2}} . \end{matrix}

Now multiply Eq. (4.12) through by $- \frac{2 μ r^{2}}{ℏ^{2} R Y}$ and rearrange so that all $r$ dependence is on the left and all angular dependence on the right:

\begin{matrix} \underset{function of r only}{\underset{⏟}{\frac{1}{R} \frac{d}{d r} (r^{2} \frac{d R}{d r}) + \frac{2 μ r^{2}}{ℏ^{2}} [E + \frac{Z e^{2}}{4 π ϵ_{0} r}]}} = \underset{function of θ, φ only}{\underset{⏟}{- \frac{Λ Y}{Y}}} . \end{matrix}

A function of $r$ can equal a function of $(θ, φ)$ for all values of all three variables only if both are the same constant. Call that separation constant $l (l + 1)$ — a choice of name, justified in Sec. 4.5 when the constant turns out to be forced into exactly that form. Equation (4.14) then splits into two ordinary differential equations.

The angular equation

\begin{matrix} \frac{1}{\sin θ} \frac{\partial}{\partial θ} (\sin θ \frac{\partial Y}{\partial θ}) + \frac{1}{\sin^{2} θ} \frac{\partial^{2} Y}{\partial φ^{2}} + l (l + 1) Y = 0 . \end{matrix}

The radial equation

\begin{matrix} \frac{1}{r^{2}} \frac{d}{d r} (r^{2} \frac{d R}{d r}) + [\frac{2 μ}{ℏ^{2}} (E - V (r)) - \frac{l (l + 1)}{r^{2}}] R = 0 . \end{matrix}

Notice what the separation has achieved: the potential appears only in the radial equation. Equation (4.15) contains no reference to $V$ , so its solutions are the same for every central potential. They are solved once, in Sec. 4.5, and reused for every atom in this book.

4.4.3Reduction to One-Dimensional Form

The radial equation (4.16) has a first-derivative term, which is inconvenient. It is removed by the substitution

\begin{matrix} U (r) = r R (r) \end{matrix}

Let us verify this explicitly, since the manipulation recurs throughout the book. From $R = U / r$ ,

\begin{matrix} \frac{d R}{d r} = \frac{1}{r} \frac{d U}{d r} - \frac{U}{r^{2}}, so r^{2} \frac{d R}{d r} = r \frac{d U}{d r} - U . \end{matrix}

Differentiating again,

\begin{matrix} \frac{d}{d r} (r^{2} \frac{d R}{d r}) = \frac{d U}{d r} + r \frac{d^{2} U}{d r^{2}} - \frac{d U}{d r} = r \frac{d^{2} U}{d r^{2}} . \end{matrix}

Substituting Eq. (4.19) into Eq. (4.16) and multiplying by $r$ gives the result:

\begin{matrix} - \frac{ℏ^{2}}{2 μ} \frac{d^{2} U}{d r^{2}} + [V (r) + \frac{ℏ^{2} l (l + 1)}{2 μ r^{2}}] U = E U \end{matrix}

This has exactly the form of the one-dimensional Schrödinger equation of Chapter 2, on the half-line $0 < r < \infty$ , with an effective potential

\begin{matrix} V_{eff} (r) = V (r) + \frac{ℏ^{2} l (l + 1)}{2 μ r^{2}} . \end{matrix}

The added term is the centrifugal barrier: repulsive, and divergent as $r \to 0$ for every $l > 0$ .

The boundary condition is $U (0) = 0$ , which follows from requiring $R = U / r$ to remain finite at the origin.

4.5Solving the Angular Equation

Equation (4.15) separates once more. Write

\begin{matrix} Y (θ, φ) = Θ (θ) Φ (φ), \end{matrix}

substitute, and multiply by $\sin^{2} θ / Θ Φ$ :

\begin{matrix} \underset{θ only}{\underset{⏟}{\frac{\sin θ}{Θ} \frac{d}{d θ} (\sin θ \frac{d Θ}{d θ}) + l (l + 1) \sin^{2} θ}} = \underset{φ only}{\underset{⏟}{- \frac{1}{Φ} \frac{d^{2} Φ}{d φ^{2}}}} \equiv m^{2}, \end{matrix}

introducing a second separation constant, written $m^{2}$ in anticipation.

The azimuthal equation

\begin{matrix} \frac{d^{2} Φ}{d φ^{2}} = - m^{2} Φ ⟹ Φ (φ) = \frac{1}{\sqrt{2 π}} e^{i m φ} . \end{matrix}

The physical requirement is that $Φ$ be single-valued: rotating by $2 π$ returns to the same point in space, so $Φ (φ + 2 π) = Φ (φ)$ . This requires $e^{2 π i m} = 1$ , hence

\begin{matrix} m = 0, \pm 1, \pm 2, \dots \end{matrix}

This is where quantization first enters, and it comes from a boundary condition, not from any postulate.

The polar equation

With $x = \cos θ$ , Eq. (4.23) becomes the associated Legendre equation

\begin{matrix} \frac{d}{d x} [(1 - x^{2}) \frac{d Θ}{d x}] + [l (l + 1) - \frac{m^{2}}{1 - x^{2}}] Θ = 0 . \end{matrix}

Solving this by power series — the same technique applied to the radial equation in Sec. 4.6 — gives solutions that remain finite at $x = \pm 1$ (that is, at the poles $θ = 0, π$ ) only if

\begin{matrix} l = 0, 1, 2, \dots and | m | \leq l . \end{matrix}

This is why the separation constant was written $l (l + 1)$ : the condition for a finite solution forces it into that form with integer $l$ . The solutions are the associated Legendre polynomials $P_{l}^{m} (x)$ , given by

\begin{matrix} P_{l}^{m} (x) = (1 - x^{2})^{| m | / 2} {(\frac{d}{d x})}^{| m |} P_{l} (x), P_{l} (x) = \frac{1}{2^{l} l!} {(\frac{d}{d x})}^{l} (x^{2} - 1)^{l}, \end{matrix}

the second expression being Rodrigues' formula for the ordinary Legendre polynomials.

The spherical harmonics

Combining and normalizing on the unit sphere gives the standard result.

\begin{matrix} Y_{l}^{m} (θ, φ) = \sqrt{\frac{(2 l + 1)}{4 π} \frac{(l - | m |)!}{(l + | m |)!}} P_{l}^{m} (\cos θ) e^{i m φ} \end{matrix}

orthonormal in the sense

\begin{matrix} \int_{0}^{2 π} \int_{0}^{π} Y_{l^{'}}^{m^{'} *} Y_{l}^{m} \sin θ d θ d φ = δ_{l l^{'}} δ_{m m^{'}} . \end{matrix}

The lowest few, written out:

\begin{matrix} \begin{aligned} Y_{0}^{0} & = \frac{1}{\sqrt{4 π}}, & Y_{1}^{0} & = \sqrt{\frac{3}{4 π}} \cos θ, & Y_{1}^{\pm 1} & = \mp \sqrt{\frac{3}{8 π}} \sin θ e^{\pm i φ}, \\ Y_{2}^{0} & = \sqrt{\frac{5}{16 π}} (3 \cos^{2} θ - 1), & Y_{2}^{\pm 1} & = \mp \sqrt{\frac{15}{8 π}} \sin θ \cos θ e^{\pm i φ}, & Y_{2}^{\pm 2} & = \sqrt{\frac{15}{32 π}} \sin^{2} θ e^{\pm 2 i φ} . \end{aligned} \end{matrix}

Comparing Eq. (4.15) with Eq. (3.31) of Chapter 3 identifies what has been solved: the angular equation is the eigenvalue equation for ${\hat{L}}^{2}$ , and Eq. (4.27) is the spectrum obtained there by algebra alone. The two routes agree, as they must.

4.6Solving the Radial Equation

Now put the Coulomb potential into Eq. (4.20):

\begin{matrix} - \frac{ℏ^{2}}{2 μ} \frac{d^{2} U}{d r^{2}} + [- \frac{Z e^{2}}{4 π ϵ_{0} r} + \frac{ℏ^{2} l (l + 1)}{2 μ r^{2}}] U = E U . \end{matrix}

We seek bound states, $E < 0$ . Define

\begin{matrix} κ = \frac{\sqrt{- 2 μ E}}{ℏ} (real and positive for E < 0), \end{matrix}

and the dimensionless variable and parameter

\begin{matrix} ρ = κ r, ρ_{0} = \frac{μ Z e^{2}}{2 π ϵ_{0} ℏ^{2} κ} . \end{matrix}

Dividing Eq. (4.32) by $E$ and rewriting in terms of $ρ$ gives the equation in its cleanest form:

\begin{matrix} \frac{d^{2} U}{d ρ^{2}} = [1 - \frac{ρ_{0}}{ρ} + \frac{l (l + 1)}{ρ^{2}}] U . \end{matrix}

Step 1: the asymptotic behavior

As $ρ \to \infty$ the bracket tends to $1$ , so $d^{2} U / d ρ^{2} \to U$ , giving $U \sim e^{\pm ρ}$ . The growing solution is not normalizable, so

\begin{matrix} U ⟶ e^{- ρ} (ρ \to \infty) . \end{matrix}

As $ρ \to 0$ the $l (l + 1) / ρ^{2}$ term dominates, so $d^{2} U / d ρ^{2} ≃ l (l + 1) U / ρ^{2}$ , whose solutions are $U \sim ρ^{l + 1}$ and $U \sim ρ^{- l}$ . The second violates $U (0) = 0$ , so

\begin{matrix} U ⟶ ρ^{l + 1} (ρ \to 0) . \end{matrix}

Step 2: strip out both limits

Write

\begin{matrix} U (ρ) = ρ^{l + 1} e^{- ρ} v (ρ), \end{matrix}

which builds in both asymptotic forms and leaves $v$ to be found. Substituting into Eq. (4.35) and simplifying gives

\begin{matrix} ρ \frac{d^{2} v}{d ρ^{2}} + 2 (l + 1 - ρ) \frac{d v}{d ρ} + [ρ_{0} - 2 (l + 1)] v = 0 . \end{matrix}

Step 3: power series and the recursion relation

Expand

\begin{matrix} v (ρ) = \sum_{j = 0}^{\infty} c_{j} ρ^{j} . \end{matrix}

Substituting into Eq. (4.39) and collecting the coefficient of $ρ^{j}$ gives the recursion relation

\begin{matrix} c_{j + 1} = \frac{2 (j + l + 1) - ρ_{0}}{(j + 1) (j + 2 l + 2)} c_{j} \end{matrix}

Step 4: the series must terminate

For large $j$ , Eq. (4.41) behaves as $c_{j + 1} ≃ (2 / j) c_{j}$ , which sums to $v \sim e^{2 ρ}$ and hence $U \sim e^{+ ρ}$ — the divergent solution we rejected. The only escape is for the series to stop: there must be a largest power $j_{max}$ with $c_{j_{max}} \neq 0$ and $c_{j_{max} + 1} = 0$ . From Eq. (4.41) that requires

\begin{matrix} ρ_{0} = 2 (j_{max} + l + 1) . \end{matrix}

Define the principal quantum number

\begin{matrix} n \equiv j_{max} + l + 1, so ρ_{0} = 2 n . \end{matrix}

Since $j_{max} \geq 0$ and $l \geq 0$ are integers, $n$ is a positive integer, and

\begin{matrix} n = 1, 2, 3, \dots and l = 0, 1, \dots, n - 1 \end{matrix}

The restriction $l \leq n - 1$ is not imposed — it is forced by $j_{max} = n - l - 1 \geq 0$ .

Step 5: the energy

Combining $ρ_{0} = 2 n$ with the definitions in Eqs. (4.33) and (4.34),

\begin{matrix} \frac{μ Z e^{2}}{2 π ϵ_{0} ℏ^{2} κ} = 2 n ⟹ κ = \frac{μ Z e^{2}}{4 π ϵ_{0} ℏ^{2} n} = \frac{Z}{n} \cdot \frac{μ}{m_{e}} \cdot \frac{1}{a_{0}}, \end{matrix}

where the Bohr radius is

\begin{matrix} a_{0} \equiv \frac{4 π ϵ_{0} ℏ^{2}}{m_{e} e^{2}} = 0.529 177 Å \end{matrix}

Remark 4.6.1

m_{e}

μ

The length appearing naturally in the derivation above is $4 π ϵ_{0} ℏ^{2} / μ e^{2}$ , with the reduced mass. The Bohr radius $a_{0}$ is defined by convention with $m_{e}$ , as in Eq. (4.46), and $0.529 177$ Å is that value. The two differ by

\frac{4 π ϵ_{0} ℏ^{2} / μ e^{2}}{a_{0}} = \frac{m_{e}}{μ} = 1.000545,

using Eq. (4.6) — five parts in ten thousand. We write $a_{0}$ throughout and carry $μ$ explicitly where it matters, which is exactly the isotope shift noted in Sec. 4.2. Be aware that many texts silently set $μ = m_{e}$ from the outset; the distinction only becomes visible in high-precision spectroscopy.

Then from $E = - ℏ^{2} κ^{2} / 2 μ$ ,

\begin{matrix} E_{n} = - \frac{μ Z^{2} e^{4}}{2 (4 π ϵ_{0})^{2} ℏ^{2}} \frac{1}{n^{2}} = - \frac{μ}{m_{e}} \frac{ℏ^{2} Z^{2}}{2 m_{e} a_{0}^{2} n^{2}} \end{matrix}

Setting $μ \to m_{e}$ — the limit of infinite nuclear mass — and $Z = 1$ gives the familiar

\begin{matrix} E_{n} = - \frac{ℰ_{Ryd}}{n^{2}}, ℰ_{Ryd} = 13.6057 eV \end{matrix}

$ℰ_{Ryd}$ is the Rydberg energy, and it is an infinite-mass quantity: for real hydrogen the factor $μ / m_{e} = 0.999456$ of Eq. (4.6) reduces it to $13.598$ eV, which is the measured ionization energy. The two agree to four significant figures, and we use $13.606$ eV throughout except where the difference is the point. The levels crowd toward $E = 0$ as $n \to \infty$ .

Step 6: the radial functions

The terminating series $v (ρ)$ of Eq. (4.40) is, up to normalization, an associated Laguerre polynomial $L_{n - l - 1}^{2 l + 1} (2 ρ)$ , defined by

\begin{matrix} L_{q}^{p} (x) = \frac{e^{x} x^{- p}}{q!} {(\frac{d}{d x})}^{q} (e^{- x} x^{p + q}) . \end{matrix}

Carrying out the normalization $\int_{0}^{\infty} | R_{n l} |^{2} r^{2} d r = 1$ gives the complete result:

\begin{matrix} R_{n l} (r) = \sqrt{{(\frac{2 Z}{n a_{0}})}^{3} \frac{(n - l - 1)!}{2 n (n + l)!}} e^{- Z r / n a_{0}} {(\frac{2 Z r}{n a_{0}})}^{l} L_{n - l - 1}^{2 l + 1} (\frac{2 Z r}{n a_{0}}) \end{matrix}

For $Z = 1$ the lowest few, written out explicitly, are

\begin{matrix} \begin{aligned} R_{10} & = \frac{2}{a_{0}^{3 / 2}} e^{- r / a_{0}}, \\ R_{20} & = \frac{1}{2 \sqrt{2} a_{0}^{3 / 2}} (2 - \frac{r}{a_{0}}) e^{- r / 2 a_{0}}, & R_{21} & = \frac{1}{2 \sqrt{6} a_{0}^{3 / 2}} \frac{r}{a_{0}} e^{- r / 2 a_{0}}, \\ R_{30} & = \frac{2}{81 \sqrt{3} a_{0}^{3 / 2}} (27 - \frac{18 r}{a_{0}} + \frac{2 r^{2}}{a_{0}^{2}}) e^{- r / 3 a_{0}}, & R_{31} & = \frac{4}{81 \sqrt{6} a_{0}^{3 / 2}} (\frac{6 r}{a_{0}} - \frac{r^{2}}{a_{0}^{2}}) e^{- r / 3 a_{0}}, \\ R_{32} & = \frac{4}{81 \sqrt{30} a_{0}^{3 / 2}} \frac{r^{2}}{a_{0}^{2}} e^{- r / 3 a_{0}} . \end{aligned} \end{matrix}

The polynomial factor in $R_{n l}$ has degree $n - l - 1$ , so $R_{n l}$ has

\begin{matrix} n - l - 1 radial nodes, \end{matrix}

and the total number of nodes, radial plus angular, is $n - 1$ .

4.7The Complete Solution

Collecting Eqs. (4.29) and (4.50), the normalized eigenfunctions of the hydrogen atom are

\begin{matrix} ψ_{n l m} (r, θ, φ) = R_{n l} (r) Y_{l}^{m} (θ, φ) \end{matrix}

or, written out in full,

\begin{matrix} \begin{aligned} ψ_{n l m} (r, θ, φ) = & \sqrt{{(\frac{2 Z}{n a_{0}})}^{3} \frac{(n - l - 1)!}{2 n (n + l)!}} e^{- Z r / n a_{0}} {(\frac{2 Z r}{n a_{0}})}^{l} L_{n - l - 1}^{2 l + 1} (\frac{2 Z r}{n a_{0}}) \\ \times \sqrt{\frac{(2 l + 1)}{4 π} \frac{(l - | m |)!}{(l + | m |)!}} P_{l}^{m} (\cos θ) e^{i m φ}, \end{aligned} \end{matrix}

with energies given by Eq. (4.47) and quantum numbers

\begin{matrix} n = 1, 2, 3, \dots; l = 0, 1, \dots, n - 1; m = - l, - l + 1, \dots, l - 1, l . \end{matrix}

The three quantum numbers came from three separate places, and it is worth recording which:

Number	Arises from	Physical meaning
$m$	single-valuedness in $φ$	$L_{z} = m ℏ$
$l$	finiteness of $Θ$ at the poles	$L^{2} = l (l + 1) ℏ^{2}$
$n$	termination of the radial series	energy $E_{n}$

Each is a boundary condition, not a postulate. Quantization in quantum mechanics is always of this kind.

4.7.1Degeneracy

The energy in Eq. (4.47) depends on $n$ alone. Counting the states that share it, using Eq. (4.55),

\begin{matrix} \sum_{l = 0}^{n - 1} (2 l + 1) = 2 \cdot \frac{(n - 1) n}{2} + n = n^{2} - n + n = n^{2}, \end{matrix}

so level $n$ is $n^{2}$ -fold degenerate, or $2 n^{2}$ counting the two spin states of Chapter 3.

The $m$ -degeneracy is expected: it follows from rotational symmetry, and any central potential has it. The $l$ -degeneracy is not. Chapter 3 showed that different $l$ means a different centrifugal barrier and hence a physically different radial equation — Eq. (4.20) literally changes — yet the energies coincide. This is special to the exact $1 / r$ potential. It is called the accidental degeneracy, and Chapter 5 shows it breaking as soon as a second electron is present.

4.7.2Expectation Values and the Size of an Orbital

The probability of finding the electron between $r$ and $r + d r$ , integrated over all angles, is

\begin{matrix} P (r) d r = | R_{n l} (r) |^{2} r^{2} d r = | U_{n l} (r) |^{2} d r, \end{matrix}

which is why $U = r R$ is the natural object: $| U |^{2}$ is the radial probability density directly.

Example 4.7.1The most probable radius of the ground state

Find where the $1 s$ electron is most likely to be found, and compare with $⟨ r ⟩$ .

Solution.

From Eqs. (4.51) and (4.57) with $Z = 1$ ,

P (r) = \frac{4}{a_{0}^{3}} r^{2} e^{- 2 r / a_{0}} .

Differentiate and set to zero:

\frac{d P}{d r} = \frac{4}{a_{0}^{3}} (2 r - \frac{2 r^{2}}{a_{0}}) e^{- 2 r / a_{0}} = 0 ⟹ r = a_{0} .

So the most probable radius is exactly the Bohr radius — the quantum-mechanical solution reproduces the radius of Bohr's circular orbit, though as the peak of a distribution rather than a trajectory.

The mean radius is different. Using $\int_{0}^{\infty} r^{n} e^{- α r} d r = n! / α^{n + 1}$ ,

⟨ r ⟩ = \int_{0}^{\infty} r P (r) d r = \frac{4}{a_{0}^{3}} \int_{0}^{\infty} r^{3} e^{- 2 r / a_{0}} d r = \frac{4}{a_{0}^{3}} \cdot \frac{3!}{(2 / a_{0})^{4}} = \frac{4}{a_{0}^{3}} \cdot \frac{6 a_{0}^{4}}{16} = \frac{3 a_{0}}{2} .

The mean exceeds the mode because the distribution has a long tail to large $r$ . Both numbers are useful, and they are not the same number — a distinction worth keeping straight whenever an “atomic size” is quoted.

The general results, quoted, are

\begin{matrix} {⟨ r ⟩}_{n l} = \frac{a_{0}}{2 Z} [3 n^{2} - l (l + 1)], {⟨ \frac{1}{r} ⟩}_{n l} = \frac{Z}{n^{2} a_{0}}, {⟨ r^{2} ⟩}_{n l} = \frac{n^{2} a_{0}^{2}}{2 Z^{2}} [5 n^{2} + 1 - 3 l (l + 1)] . \end{matrix}

The middle one is worth noting: $⟨ 1 / r ⟩$ depends only on $n$ , which combined with $⟨ V ⟩ = - Z e^{2} ⟨ 1 / r ⟩ / 4 π ϵ_{0} = 2 E_{n}$ is the quantum version of the virial theorem for a $1 / r$ potential.

Orbitals therefore grow as $n^{2}$ : $1.5 a_{0}$ for the ground state, about $150 a_{0} \approx 8$ nm at $n = 10$ , and larger than many molecules by $n = 30$ .

4.8Atomic Units

The constants in Eqs. (4.47) and (4.50) are cumbersome, and research papers eliminate them by a choice of units. They state this once, usually in a single parenthetical sentence, and everything after it is unreadable if the sentence is skipped.

Definition 4.8.1Atomic units

Atomic units are defined by setting

\begin{matrix} ℏ = m_{e} = e = \frac{1}{4 π ϵ_{0}} = 1 . \end{matrix}

Every other unit follows. From Eq. (4.46), the unit of length is $a_{0}$ ; from Eq. (4.47), the unit of energy is $μ e^{4} / (4 π ϵ_{0})^{2} ℏ^{2}$ , called the hartree:

Quantity	Unit	Expression	Value
Length	bohr	$a_{0} = 4 π ϵ_{0} ℏ^{2} / m_{e} e^{2}$	$0.529 177$ Å
Energy	hartree	$E_{h} = m_{e} e^{4} / (4 π ϵ_{0})^{2} ℏ^{2}$	$27.2114$ eV
Energy	rydberg	$\frac{1}{2} E_{h}$	$13.6057$ eV
Time	—	$ℏ / E_{h}$	$24.19$ as
Velocity	—	$α c$	$2.19 \times 10^{6}$ m/s
Field	—	$E_{h} / e a_{0}$	$5.14 \times 10^{11}$ V/m

In these units Eqs. (4.5), (4.20) and (4.47) become simply

\begin{matrix} \hat{H} = - \frac{1}{2} \nabla^{2} - \frac{Z}{r}, - \frac{1}{2} \frac{d^{2} U}{d r^{2}} + [- \frac{Z}{r} + \frac{l (l + 1)}{2 r^{2}}] U = E U, E_{n} = - \frac{Z^{2}}{2 n^{2}} . \end{matrix}

Two entries in the table carry physics rather than convenience. The atomic unit of time is $24$ attoseconds — the natural timescale of electronic motion, and the reason attosecond pulses are the tool of choice for watching electrons. And the atomic unit of field, $5.14 \times 10^{11}$ V/m, is the field the electron feels in the ground state; a laser reaching an appreciable fraction of it competes with the binding of the atom itself, which is what the phrase “strong field” means.

Example 4.8.1Converting a laser specification

A laser has wavelength $800$ nm and peak intensity $1 \times 10^{13}$ W/cm $^{2}$ . Express the photon energy, the optical period and the field amplitude in atomic units.

Solution.

Photon energy. $E = h c / λ$ , and with $h c = 1239.8$ eV nm,

E = \frac{1239.8}{800} = 1.550 eV = \frac{1.550}{27.2114} = 0.0570 a.u.

Period. $T = 2 π / ω$ with $ω = E$ in atomic units, so

T = \frac{2 π}{0.0570} = 110.2 a.u. = 110.2 \times 24.19 as = 2.67 fs,

which is the correct optical period for $800$ nm light.

Field. From $I = \frac{1}{2} ϵ_{0} c E_{0}^{2}$ with $I = 1 \times 10^{17}$ W/m $^{2}$ ,

E_{0} = \sqrt{\frac{2 I}{ϵ_{0} c}} = \sqrt{\frac{2 \times 10^{17}}{8.854 \times 10^{- 12} \times 3 \times 10^{8}}} = 8.68 \times 10^{9} V / m,

so in atomic units

E_{0} = \frac{8.68 \times 10^{9}}{5.14 \times 10^{11}} = 0.0169 a.u.

This is about $1.7 %$ of the field binding the ground state — a genuine perturbation for hydrogen, but note that a Rydberg state at $n = 10$ is bound by $1 / n^{2} = 1 %$ of the ground-state energy and feels an internal field smaller by $n^{4}$ , so the same laser is overwhelmingly strong for it. “Strong” is always relative to the state.

4.9Rydberg States

Definition 4.9.1Rydberg state

A Rydberg state is a state of high principal quantum number, in which the electron is far from the nucleus and weakly bound.

Four scalings follow directly from the results above, and all four are used later in this book.

Binding energy $\propto n^{- 2}$ . From Eq. (4.48), the electron is bound by only $13.606 / n^{2}$ eV: at $n = 10$ , that is $0.136$ eV, less than an infrared photon.

Size $\propto n^{2}$ . From Eq. (4.58).

Level spacing $\propto n^{- 3}$ . Differentiating Eq. (4.47),

\begin{matrix} Δ E = E_{n + 1} - E_{n} ≃ \frac{d E_{n}}{d n} = \frac{ℰ_{Ryd} \cdot 2}{n^{3}} = \frac{27.21}{n^{3}} eV, \end{matrix}

so levels crowd toward the ionization threshold and the density of states per unit energy grows as $n^{3}$ .

Probability density near the nucleus $\propto n^{- 3}$ . This is the important one. The wave function is normalized over a region of size $n^{2}$ , so its amplitude in any fixed small region near the origin must shrink. Explicitly, for $l = 0$ and $r ≪ a_{0} n^{2}$ , the exponential and polynomial factors in Eq. (4.50) are both of order unity, and the normalization prefactor $(2 Z / n a_{0})^{3 / 2} \sqrt{(n - l - 1)! / 2 n (n + l)!}$ supplies the entire $n$ dependence, giving

\begin{matrix} | U_{n l} (r) |^{2} \propto n^{- 3} for fixed small r \end{matrix}

Equation (4.62) deserves to be stated as a principle:

Any process requiring the electron to be near the nucleus runs at a rate proportional to $n^{- 3}$ .

A quantity linear in the wave function rather than quadratic — a dipole matrix element, for instance — carries the square root, $n^{- 3 / 2}$ . These two exponents govern the Rydberg physics of Chapters 5, 9 and 10.

Example 4.9.1How classical is a Rydberg state?

Estimate the orbital period of an $n = 20$ electron in hydrogen and compare it with the beat period between adjacent levels.

Solution.

For a circular Bohr orbit of radius $r_{n} = n^{2} a_{0}$ , the classical period in atomic units is $T = 2 π n^{3}$ . For $n = 20$ ,

T = 2 π (8000) = 5.03 \times 10^{4} a.u. = 5.03 \times 10^{4} \times 24.19 as = 1.22 ps .

The spacing to the neighboring level, from Eq. (4.61), is $Δ E = 1 / n^{3}$ a.u. $= 1.25 \times 10^{- 4}$ a.u. $= 3.4$ meV, and a superposition of two levels beats at

T_{beat} = \frac{2 π}{Δ E} = 2 π n^{3},

which is the same number.

This is the correspondence principle made quantitative: a superposition of a few neighboring Rydberg levels forms a localized packet that orbits the nucleus on the classical Kepler period and stays recognisable for many orbits. It is also the mechanism behind the quantum beats measured in Chapter 11 — there, between molecular Rydberg states, and on a timescale of tens of femtoseconds rather than a picosecond.

4.10Summary

The two-body problem reduces to one body of reduced mass $μ = m_{e} m_{p} / (m_{e} + m_{p}) = 0.99946 m_{e}$ in the potential $V = - Z e^{2} / 4 π ϵ_{0} r$ .
Separating $ψ = R (r) Y (θ, φ)$ with separation constant $l (l + 1)$ gives the angular equation (4.15), which contains no potential and so is solved once for all central problems, and the radial equation (4.16).
The substitution $U = r R$ removes the first derivative and yields the one-dimensional form (4.20) with the centrifugal barrier $ℏ^{2} l (l + 1) / 2 μ r^{2}$ and boundary condition $U (0) = 0$ .
The angular solutions are the spherical harmonics $Y_{l}^{m}$ , Eq. (4.29), built from associated Legendre polynomials; single-valuedness quantizes $m$ and finiteness at the poles quantizes $l$ , with $| m | \leq l$ .
The radial solution is obtained by stripping out the asymptotic forms $e^{- ρ}$ and $ρ^{l + 1}$ , expanding in a power series, and requiring the series to terminate. Termination gives $ρ_{0} = 2 n$ with $n = j_{max} + l + 1$ , hence $l \leq n - 1$ , and fixes the energy.
Results: $E_{n} = - μ Z^{2} e^{4} / 2 (4 π ϵ_{0})^{2} ℏ^{2} n^{2} = - 13.606 Z^{2} / n^{2}$ eV, radial functions Eq. (4.50) with $n - l - 1$ nodes, and complete wave functions $ψ_{n l m} = R_{n l} Y_{l}^{m}$ .
Degeneracy $n^{2}$ (or $2 n^{2}$ with spin). The $m$ -degeneracy follows from rotational symmetry; the $l$ -degeneracy is accidental, special to $1 / r$ , and breaks in every other atom.
$⟨ r ⟩ = (a_{0} / 2 Z) [3 n^{2} - l (l + 1)]$ ; the most probable radius of the $1 s$ state is exactly $a_{0}$ while $⟨ r ⟩ = 1.5 a_{0}$ .
Atomic units set $ℏ = m_{e} = e = 1 / 4 π ϵ_{0} = 1$ : $a_{0}$ , $27.21$ eV, $24$ as, $5.14 \times 10^{11}$ V/m.
Rydberg scalings: binding $n^{- 2}$ , size $n^{2}$ , spacing $n^{- 3}$ , and probability density near the nucleus $n^{- 3}$ — the last governing every core-sensitive rate, with dipole matrix elements carrying $n^{- 3 / 2}$ .

4.11Exercises

The substitution, verified. Carry out the steps of Eqs. (4.18)–(4.19) yourself, and confirm that Eq. (4.16) becomes Eq. (4.20). State where the boundary condition $U (0) = 0$ comes from.
Checking a wave function. Take $R_{10} = 2 a_{0}^{- 3 / 2} e^{- r / a_{0}}$ from Eq. (4.51), with $Z = 1$ .
1. Verify that $U_{10} = r R_{10}$ satisfies Eq. (4.20) with $l = 0$ and $E = - 13.606$ eV, by direct substitution.
2. Verify the normalization $\int_{0}^{\infty} | R_{10} |^{2} r^{2} d r = 1$ .
3. Repeat the normalization check for $R_{21}$ .
The recursion relation. Substitute the series (4.40) into Eq. (4.39) and derive the recursion (4.41). Then, for $n = 2$ and $l = 0$ , use it to find $c_{1} / c_{0}$ and confirm that $c_{2} = 0$ , so the polynomial is $1 - ρ / 2$ — which reproduces the $(2 - r / a_{0})$ factor of $R_{20}$ in Eq. (4.51) up to normalization.
Degeneracy and expectation values.
1. Prove Eq. (4.56) step by step.
2. Using Eq. (4.58), compute $⟨ r ⟩$ for the $3 s$ , $3 p$ and $3 d$ states of hydrogen in units of $a_{0}$ . They have the same energy: explain in one sentence how they can have different mean radii.
3. Verify $⟨ V ⟩ = 2 E_{n}$ for the ground state using $⟨ 1 / r ⟩ = Z / n^{2} a_{0}$ .
Atomic units. Redo Example 4.8.1 for an XUV pulse of photon energy $17.6$ eV: give the energy in atomic units, the wavelength in nm, and the optical period in attoseconds. Then confirm the claim in Sec. 4.8 that the dipole approximation is safe by comparing the wavelength with $⟨ r ⟩$ for the ground state.

4.12Project: Hydrogen, Numerically and Exactly

The problem. Solve Eq. (4.20) on a computer and confirm every analytic result of this chapter. The same solver, with the potential changed, handles the screened atoms of Chapter 5 and the molecular vibrations of Chapter 7, so build it carefully.

Work in atomic units, where the equation is the middle expression of Eq. (4.60).

On paper. Discretize the equation. On a uniform grid $r_{i} = i h$ , the second derivative is $U^{″} (r_{i}) ≃ (U_{i + 1} - 2 U_{i} + U_{i - 1}) / h^{2}$ . Write down the resulting matrix eigenvalue problem $𝐇 \vec{U} = E \vec{U}$ , giving the diagonal and off-diagonal entries explicitly in terms of $h$ , $r_{i}$ and $l$ .
On the computer. Diagonalize it for $l = 0$ and report the four lowest eigenvalues. Compare with $- 1 / 2 n^{2}$ . State the $r_{max}$ and $h$ you needed for four-figure agreement, and explain why too small an $r_{max}$ fails while too small an $h$ merely costs time.
On the computer. Repeat for $l = 1$ and $l = 2$ . Confirm the accidental degeneracy: the lowest $l = 1$ level should equal the second $l = 0$ level to your numerical accuracy. Report the actual discrepancy.
On the computer. Plot $| U_{n l} (r) |^{2}$ for $(n, l) = (1, 0), (2, 0), (2, 1), (3, 0), (3, 2)$ , overlaying the analytic forms from Eq. (4.51). Count the nodes and check Eq. (4.52).
On the computer. Compute $⟨ r ⟩$ numerically for each state in (d) and compare with Eq. (4.58). Also locate the maximum of $| U_{10} |^{2}$ and confirm it sits at $r = a_{0}$ , as Example 4.7.1 predicts.
On the computer. Test Eq. (4.62). For $l = 0$ and $n = 1$ to $8$ , compute $\int_{0}^{a_{0}} | U_{n 0} |^{2} d r$ and plot it against $n$ on log–log axes. Measure the slope; you should find $- 3$ .
What has to move. One animation: $| U_{n 0} (r) |^{2}$ as $n$ steps from $1$ to $8$ , on axes wide enough to show the state spreading out, with the region $r < a_{0}$ shaded so the shrinking amplitude inside it is visible. That shrinkage is the $n^{- 3}$ law.

The check. Part (f) is the one that matters and the one most likely to go wrong. A slope of $- 2$ instead of $- 3$ almost always means the wave functions were normalized on a grid too short to hold the larger states, so their normalization is wrong. Fix the grid, not the exponent.

Be ready to answer. Your part (c) energies came out independent of $l$ , as Eq. (4.47) says. But the equation you actually diagonalized, Eq. (4.20), is visibly different for each $l$ — the centrifugal term changes the matrix. How can three different matrices share eigenvalues, and what single change to the potential in your code would destroy the coincidence?