6.2 The radial equation and the effective potential

The last section compressed the hydrogen atom into an ordinary differential equation for $R(r)$ , containing a term of mysterious origin: $l(l+1)\hbar^2/2\mu r^2$ . In this section we first tidy the equation into the shape we know best, then look that term squarely in the face.

A familiar phenomenon: why satellites don’t fall down

Warm up with classical mechanics. Earth’s gravity pulls relentlessly downward — why doesn’t a satellite fall? Because it has angular momentum. At fixed angular momentum $L$ , the rotational kinetic energy $L^2/2\mu r^2$ climbs steeply as you approach the centre: to get closer you must spin faster, and the kinetic-energy cost outweighs what gravity pays back in potential energy. Book that kinetic energy on the “potential” side of the ledger and you get the effective potential of classical orbital mechanics:

V_\text{eff}(r)=V(r)+\frac{L^2}{2\mu r^2}\tag{6.2.1}

The second term is the centrifugal term (in plain words: a fictitious outward-pushing barrier produced by rotation — really kinetic energy, just booked differently). A planet’s perihelion and aphelion are the two turning points of one-dimensional motion on this curve.

Exactly the same structure will reappear in quantum mechanics — with $L^2$ replaced by its eigenvalue $l(l+1)\hbar^2$ .

Cosmetic surgery: u = rR

The radial equation

-\frac{\hbar^2}{2\mu}\frac{1}{r^2}\frac{\dd}{\dd r}\!\left(r^2\frac{\dd R}{\dd r}\right) +\left[V(r)+\frac{l(l+1)\hbar^2}{2\mu r^2}\right]R=ER\tag{6.2.2}

has a first-derivative term $\frac{2}{r}R'$ that keeps it from looking like the one-dimensional equations we are used to solving. One substitution removes it.

The substitution u(r) = rR(r): the radial equation in standard formbasic~6 min

Step 1: change the function. Set $u(r)=rR(r)$ , i.e. $R=u/r$ . Differentiate term by term:

\frac{\dd R}{\dd r}=\frac{u'}{r}-\frac{u}{r^2} \quad\Longrightarrow\quad r^2\frac{\dd R}{\dd r}=ru'-u\tag{6.2.3}

\frac{\dd}{\dd r}\!\left(r^2\frac{\dd R}{\dd r}\right)=u'+ru''-u'=ru''\tag{6.2.4}

The first-derivative term cancels exactly. That is the entire motivation for choosing $u=rR$ .

Step 2: substitute back. The first term of the original equation becomes $-\frac{\hbar^2}{2\mu}\frac{ru''}{r^2}=-\frac{\hbar^2}{2\mu}\frac{u''}{r}$ ; multiply through by $r$ :

-\frac{\hbar^2}{2\mu}\frac{\dd^2u}{\dd r^2} +\underbrace{\left[V(r)+\frac{l(l+1)\hbar^2}{2\mu r^2}\right]}_{V_\text{eff}(r)}u=Eu\tag{6.2.5}

This has the face of a one-dimensional stationary Schrödinger equation — everything we learned in Chapters 2 and 4 (the node theorem, bound-state criteria, asymptotic analysis) carries over directly. Only two differences remain:

the variable lives on the half-line $r\ge0$ , not the full line;
the potential is replaced by the effective potential $V_\text{eff}$ .

Step 3: boundary conditions. The normalisation condition tidies up too. The volume element is $\dd^3r=r^2\dd r\,\dd\Omega$ , and with the angular part normalised:

\int_0^\infty |R|^2r^2\dd r=\int_0^\infty|u|^2\dd r=1\tag{6.2.6}

The normalisation of $u$ carries no $r^2$ weight — exactly like a one-dimensional wavefunction. At the origin we must require

u(0)=0\tag{6.2.7}

Reason: if $u(0)\ne0$ , then $R\sim u(0)/r$ diverges at the origin, and $\nabla^2(1/r)=-4\pi\delta^3(\vec r)$ would plant a $\delta$ -function source in the equation with nothing to cancel it. So as far as $u$ is concerned, the origin is an infinitely high wall — a one-dimensional problem on the half-line, pinned to zero at the left end, the exact twin of the infinite well’s left wall.

The centrifugal barrier: angular momentum's rent

The quantum effective potential corresponds word for word to the classical one:

V_\text{eff}(r)=-\frac{e^2}{4\pi\varepsilon_0 r}+\frac{l(l+1)\hbar^2}{2\mu r^2}\tag{6.2.8}

As $r\to0$ the centrifugal term diverges like $1/r^2$ , faster than the Coulomb attraction’s $1/r$ — for any $l\ge1$ , the net effect near the origin is repulsive. An electron carrying angular momentum is barred from the nucleus by its own rotation.

Put in numbers to get a feel: for $l=1$ , the two terms balance at $r=2a_0\approx0.106$ nm, the bottom of the well; any farther in, the barrier climbs fast. That is why the probability distributions of p and d electrons are squeezed to zero near the nucleus, while an $l=0$ s electron, facing no such wall, can press right up against it — a “nucleus-hugging” talent that will star twice more in this chapter (the Darwin term and the 21 cm line are both s-state exclusives).

Reading the plot: where the bound states live

The picture

Draw the $V_\text{eff}$ curves in your head.

$l=0$ : just the bare Coulomb potential — a single curve rising from $-\infty$ toward 0, an “infinitely deep” needle of a well, but with no inner wall.
$l\ge1$ : at small $r$ the centrifugal barrier props the curve up to $+\infty$ , while at large $r$ it still approaches 0 from below — leaving a finite-depth bowl in between. The larger $l$ is, the shallower the bowl and the farther out it sits.

Bound states live in the bowl: solutions with negative energy $E$ whose wavefunctions decay on both sides. Solutions with $E>0$ are scattering states (the electron swings past the proton, bends, and leaves), with a continuum of energies.

That the bowl grows shallower with $l$ means: at the same energy, higher- $l$ states “live” farther out; and when $l$ is too large the bowl becomes too shallow to hold a state below a given energy — which will explain the constraint $l\le n-1$ appearing in the next section.

The mathematics

Position and depth of the bowl’s bottom (for the Coulomb potential, set $V_\text{eff}'=0$ ):

r_\text{min}=l(l+1)\,a_0,\qquad V_\text{eff}(r_\text{min})=-\frac{13.6\ \text{eV}}{l(l+1)}\tag{6.2.9}

where $a_0=4\pi\varepsilon_0\hbar^2/\mu e^2=0.0529$ nm.

$l$	Bowl bottom	Depth
0	none (plunges to the origin)	$-\infty$
1	$2a_0$	$-6.8$ eV
2	$6a_0$	$-2.27$ eV
3	$12a_0$	$-1.13$ eV

Behaviour of the solution as $r\to0$ : the equation reduces to $u''=\frac{l(l+1)}{r^2}u$ ; trying $u\sim r^s$ gives $s(s-1)=l(l+1)$ , i.e. $s=l+1$ or $s=-l$ . The latter violates $u(0)=0$ (or is unnormalisable), so

u(r)\xrightarrow{r\to0} C\,r^{\,l+1}\tag{6.2.10}

The larger $l$ is, the flatter the wavefunction is pressed near the origin — the quantitative version of the centrifugal barrier.

The Coulomb potential’s privilege: infinitely many bound states

Chapter 4’s finite well had only finitely many bound states. The Coulomb potential is different.

Why a 1/r potential holds infinitely many levelsadvanced~6 min

The criterion can be read from the “width of the bowl”. Take an energy $E=-\varepsilon$ (with $\varepsilon$ a small positive number). The outer turning point of the classically allowed region sits at

\frac{e^2}{4\pi\varepsilon_0 r_\text{out}}=\varepsilon \quad\Longrightarrow\quad r_\text{out}=\frac{e^2}{4\pi\varepsilon_0\,\varepsilon}\xrightarrow{\ \varepsilon\to0\ }\infty\tag{6.2.11}

The closer the energy gets to zero, the wider the allowed region — its width diverges. And the number of standing waves that fit in an allowed region of width $L$ at depth of order $\varepsilon$ grows with $L$ (recall the infinite well: half-wavelengths $\lambda/2$ must fit inside $L$ ). The width diverges faster than the wavelength grows, so below $E=0$ you can pack infinitely many ever-more-crowded levels — piling up without limit toward $E=0$ . This is exactly the pattern we met in the Bohr model: $E_n=-13.6/n^2$ eV, with $n$ arbitrarily large and level spacings $\to0$ .

Contrast: a short-range potential (the Yukawa potential $\ee^{-r/r_0}/r$ , or anything essentially zero beyond a finite distance) has an allowed-region width that saturates at $\sim r_0$ as $\varepsilon\to0$ — too small for infinitely many states. Bound states are finite in number, possibly zero. Hydrogen’s infinity of levels is rooted in the long tail of the Coulomb $1/r$ : it decays so slowly that even at the ends of the earth it still tugs on the electron just a little.

A startling corollary comes free: since $n$ can be arbitrarily large, an atom can sit in a state with $n\sim100$ or higher. Such Rydberg atoms have radii $\sim n^2a_0$ reaching the micrometre scale — ten thousand times the ground state. They really are made in the laboratory, and radio astronomy has observed interstellar hydrogen transitions with $n$ in the hundreds.

Taking stock

After the substitution $u=rR$ , the radial problem is a one-dimensional Schrödinger equation on the half-line, with $u(0)=0$ at the left end and potential $V_\text{eff}$ .
The centrifugal barrier $l(l+1)\hbar^2/2\mu r^2$ keeps $l\ge1$ electrons away from the nucleus; near the origin $u\sim r^{l+1}$ .
The Coulomb potential’s long tail guarantees infinitely many bound states, piling up toward $E=0$ .

We spent a whole chapter solving one-dimensional problems; every needed technique is in hand. The next section is the frontal assault: put the Coulomb potential in and solve for $u(r)$ exactly.

Key formulas

Radial equation (standard form)
$-\frac{\hbar^2}{2\mu}\frac{\dd^2u}{\dd r^2}+V_{\text{eff}}(r)\,u=Eu,\qquad u=rR$
A 1D problem on the half-line; normalisation ∫|u|²dr = 1, u(0)=0
Effective potential
$V_{\text{eff}}(r)=-\frac{e^2}{4\pi\varepsilon_0 r}+\frac{l(l+1)\hbar^2}{2\mu r^2}$
The centrifugal term is bookkept rotational kinetic energy, not a new force
Behaviour at the origin
$u(r)\sim r^{\,l+1}\quad(r\to0)$
Larger l presses the wavefunction farther from the origin; only s states hug the nucleus
Bowl bottom
$r_{\min}=l(l+1)\,a_0,\qquad V_{\text{eff}}(r_{\min})=-\frac{13.6\ \text{eV}}{l(l+1)}$
Larger l: shallower well, farther out

Self-check4 questions

What comes next

The equation is now in one-dimensional standard form, and both ends of its behaviour are understood: $u\sim r^{l+1}$ at the origin, exponential decay at infinity. The next section pinches these two ends together — bridging the middle with a power series. You will see that the series must terminate abruptly at some order, or the wavefunction blows up; and the termination condition is precisely the energy quantisation condition. Out of pure mathematics, $-13.6\ \text{eV}/n^2$ will grow all by itself.

The radial equation and the effective potential

Recommended first

After this section you should be able to

A familiar phenomenon: why satellites don’t fall down

Cosmetic surgery: u = rR

Reading the plot: where the bound states live

The picture

The mathematics

The Coulomb potential’s privilege: infinitely many bound states

Taking stock

Key formulas

Self-check4 questions

What comes next

A familiar phenomenon: why satellites don’t fall down

Cosmetic surgery: u = rR

Reading the plot: where the bound states live

◑The picture

∑The mathematics

The Coulomb potential’s privilege: infinitely many bound states

Taking stock

∎Key formulas

What comes next

The picture

The mathematics

Key formulas