Radial Probability Distribution and Expectation Values

Ask the most natural question in atomic physics — how far from the nucleus is the electron? — and quantum mechanics appears to give two contradictory answers. The 3D probability density $|\psi_{100}|^2$ of the hydrogen ground state is largest at $r = 0$ , right on top of the proton. Yet if you measure the electron's distance many times, the most common answer is not zero but exactly one Bohr radius. Both statements are true, and the object that reconciles them is this article's subject: the radial probability distribution $P(r) = r^2|R_{n\ell}(r)|^2$ , the probability per unit radius of finding the electron in a thin spherical shell. The factor $r^2$ is not decoration — it is the surface area of the shell doing the asking. Near the origin the density is high but the shells are tiny; far out the shells are huge but the fog is thin. $P(r)$ scores the competition, and for the ground state the contest peaks precisely at $a_0$ : Bohr's orbit reborn as the mode of a distribution. From $P(r)$ flow the numbers that run atomic physics: the moments $\langle r^k \rangle$ . The mean radius sets atomic size, $\langle 1/r \rangle$ fixes the Coulomb energy, $\langle 1/r^2 \rangle$ prices the centrifugal term, and $\langle 1/r^3 \rangle$ controls fine structure. For hydrogen they obey a remarkable recursion — the Kramers relation — that generates the whole family with almost no integration. This article builds $P(r)$ , computes its landmarks honestly, and hands you the moment machinery.

Key takeaways

The probability of finding the electron between $r$ and $r + dr$ is $P(r)\,dr = r^2|R_{n\ell}(r)|^2\,dr$ — the $r^2$ is the shell-volume factor from $d^3r$ .
In the language $u = rR$ of the effective 1D problem, $P(r) = |u(r)|^2$ : an ordinary one-dimensional probability density on the half-line.
Density and shell probability answer different questions: $|R_{10}|^2$ peaks at $r = 0$ while $P_{10}(r)$ peaks at $a_0$ . For 1s: mode $a_0$ < median $1.337\,a_0$ < mean $1.5\,a_0$ .
Hydrogen moments: $\langle r \rangle = \tfrac{a_0}{2}[3n^2 - \ell(\ell+1)]$ , $\langle 1/r \rangle = 1/(n^2a_0)$ ( $\ell$ -independent!), $\langle 1/r^2 \rangle = 1/(n^3(\ell+\tfrac12)a_0^2)$ , $\langle 1/r^3 \rangle = 1/(n^3\ell(\ell+\tfrac12)(\ell+1)a_0^3)$ .
The Kramers–Pasternack recursion links $\langle r^s \rangle$ , $\langle r^{s-1} \rangle$ , $\langle r^{s-2} \rangle$ — the whole moment family from two seeds, no integrals.
Each moment samples different physics: positive powers measure size and polarizability, negative powers price binding, centrifugal, and spin–orbit energies.
Key idea: one 1D distribution, $P(r) = |u|^2$ , carries every radial question about the atom — and its moments are the currency of atomic physics.

The core insight

The radial probability density is $P(r) = r^2|R_{n\ell}(r)|^2 = |u_{n\ell}(r)|^2$ with $u = rR$ ; expectation values of radial quantities are moments of this single one-dimensional distribution, $\langle f(r) \rangle = \int_0^\infty f(r)\,P(r)\,dr$ .

Where this lives today

Radial moments are working numbers across modern physics. Every fine-structure and hyperfine calculation begins with $\langle 1/r^3 \rangle$ ; every atomic-clock systematic budget prices field shifts through $\langle r^2 \rangle$ -type polarizabilities; every Rydberg-atom quantum-computing proposal leans on the $n^2$ growth of $\langle r \rangle$ to make two atoms talk across micrometers. The proton-radius puzzle of the 2010s was, at bottom, a radial-distribution story: muonic hydrogen's Lamb shift samples the s-state contact region $10^6$ times more strongly than electronic hydrogen (the $1/a^3$ scaling of $|u(0^+)|^2$ -adjacent physics), turning a femtometer nuclear-size question into a measurable spectroscopic one. And the distributions themselves are no longer textbook abstractions: photoionization microscopy (2013) imaged the nodal structure of hydrogen stationary states directly, and quantum-gas microscopes routinely reconstruct $P(r)$ -analogues for atoms in traps. The century-old question "where is the electron?" is now answered one detection at a time, and the histograms converge on exactly the curves drawn above.

The physical picture

Figure

Density versus shell probability — the two answers reconciled

Picture the electron cloud as fog. The fog is thickest at the nucleus — that is the red curve, the density $|R|^2$ . But "how far out is the electron?" sums the fog over an entire shell of radius $r$ , and shell area grows as $r^2$ : a thin fog spread over a big sphere can hold more total probability than a dense wisp over a vanishing one. The blue curve is that competition scored at every radius. At the origin it must start at zero — not because the density vanishes (for s-states it does not) but because the shell does. There is a deeper reason the blue curve is the honest 1D object. The substitution $u = rR$ that turned the radial problem into a half-line Schrödinger problem at Effective Potential and the Centrifugal Barrier also turns the measure flat: its identity $\int|R|^2r^2dr = \int|u|^2dr$ says $P(r) = |u|^2$ exactly. The radial distribution is not a bolt-on definition — it is the ordinary $|\text{wavefunction}|^2$ of the effective one-dimensional particle.

Figure

2s versus 2p — nodes sculpt the distribution, and the landmarks disagree

The $n = 2$ pair shows how much structure one energy level can hide. The 2p distribution is a clean single hump peaking at $4a_0$ . The 2s carries its radial node at $2a_0$ : a small inner hump, a forbidden radius, and a dominant outer hump at $5.24\,a_0$ . Which state is "bigger"? By the mode, 2s ( $5.24 > 4$ ); by the mean, also 2s ( $6a_0 > 5a_0$ ) — but tails, humps, and landmarks must be kept straight, and the worked examples below do the accounting honestly. The general anatomy is inherited from the wavefunction catalogue of Hydrogenic Wavefunctions and Quantum Numbers: near the origin $P \sim r^{2\ell+2}$ (the centrifugal wall squared, plus the shell factor), at large $r$ the tail $P \sim r^{2n}e^{-2r/na_0}$ , and in between $n - \ell - 1$ zeros where the radial nodes forbid the electron entirely.

Building the distribution

Start from the full 3D statement — the Born rule with the spherical volume element — and integrate out what you do not care about. The probability of finding the electron in $d^3r$ is $|\psi_{n\ell m}|^2\,r^2\sin\theta\,dr\,d\theta\,d\phi$ ; asking only about $r$ means integrating over all angles. Because the states factor as $R_{n\ell}Y_\ell^m$ (the separation established at Separation of Variables in Spherical Coordinates) and the spherical harmonics are unit-normalized on the sphere, the angular integral collapses to one:

P(r)\,dr = \left(\int|Y_\ell^m|^2\,d\Omega\right)r^2|R_{n\ell}(r)|^2\,dr = r^2|R_{n\ell}(r)|^2\,dr \qquad (1)

where: $P(r)$ — probability per unit radius, dimensions of 1/length; the harmonics' normalization is what makes every trace of $m$ and the angular shape drop out. Any factored state — any central potential, not just hydrogen — has a radial distribution of exactly this form. In the reduced language the same object is cleaner still:

u_{n\ell}(r) \equiv r\,R_{n\ell}(r), \qquad P(r) = |u_{n\ell}(r)|^2, \qquad \int_0^\infty |u_{n\ell}|^2\,dr = 1 \qquad (2)

where: $u$ — the half-line wavefunction of Effective Potential and the Centrifugal Barrier, whose measure identity this equation cashes; the normalization statement is the doctrine of Boundary Conditions and Normalization of Radial Wavefunctions. Read (2) as a license: every intuition you own about 1D probability densities — normalization, modes, spreads — applies verbatim to the radial problem. Radial questions then become moments of one distribution:

\langle f(r) \rangle = \int_0^\infty f(r)\,P(r)\,dr \qquad (3)

where: $f$ — any function of radius alone. Angular observables never enter: for those, the harmonics' catalogue does the work. The art is in which $f$ to ask for — and hydrogen's answer sheet is short enough to memorize.

Derivation AThe most probable radius — Bohr's orbit as a mode

The goal is the peak of the ground-state distribution: show that $P_{10}(r) = (4/a_0^3)r^2e^{-2r/a_0}$ is maximized at exactly $r = a_0$ , the Bohr radius — the cleanest meeting point between wave mechanics and the old planetary picture. The strategy is elementary calculus on the product $r^2 \times$ decaying exponential; the instructive part is why there is a peak at all: the shell factor grows, the density falls, and the maximum sits where their logarithmic slopes cancel. We work at $Z = 1$ ; the hydrogenic generalization is a one-line rescaling.

Step 1. Form the distribution. With $R_{10} = 2a_0^{-3/2}e^{-r/a_0}$ from the catalogue of Hydrogenic Wavefunctions and Quantum Numbers,

P_{10}(r) = r^2|R_{10}|^2 = \frac{4}{a_0^3}\,r^2e^{-2r/a_0}.

Why: the density $|R_{10}|^2$ is monotonically falling — every bit of structure in $P$ comes from the fight with $r^2$ .

Step 2. Differentiate and set to zero:

\frac{dP}{dr} = \frac{4}{a_0^3}\left(2r - \frac{2r^2}{a_0}\right)e^{-2r/a_0} = \frac{8r}{a_0^3}\left(1 - \frac{r}{a_0}\right)e^{-2r/a_0} = 0.

Why: the root $r = 0$ is the shell-area zero (a minimum); the balance point is the nonzero root.

Step 3. Read off the maximum: $r_{\rm mp} = a_0$ . In logarithmic terms, $d\ln P/dr = 2/r - 2/a_0$ vanishes where the shell growth rate $2/r$ equals the density decay rate $2/a_0$ — the two-force picture made quantitative.

Step 4. Generalize. For a circular state ( $\ell = n-1$ ) of any $n$ , $P \propto r^{2n}e^{-2r/na_0}$ and the same cancellation gives $r_{\rm mp} = n^2a_0$ — the Bohr-orbit ladder, recovered as the modes of node-free distributions. For $Z \ne 1$ , every length shrinks by $1/Z$ (the scaling law of Hydrogenic Ions and Scaling with Nuclear Charge). Verified: $r_{\rm mp} = a_0$ and the circular-state peaks $n^2a_0$ for $n = 2, 3, 4$ found symbolically (sympy, verify script sections 2 and 8); reproduces Griffiths §4.2.

The mode is only one landmark, and for skewed distributions the landmarks disagree. The 1s distribution has mode $a_0$ , but its long tail drags the mean out to $\langle r \rangle = \tfrac32 a_0$ (a two-line gamma integral: $\langle r \rangle = \tfrac{4}{a_0^3}\int r^3e^{-2r/a_0}dr = \tfrac{4}{a_0^3}\cdot\tfrac{6a_0^4}{16}$ ), and the median — the radius with a 50/50 chance of finding the electron inside — sits between them at $1.337\,a_0$ , from solving $1 - e^{-2x}(1 + 2x + 2x^2) = \tfrac12$ numerically. Mode, median, mean: $1 < 1.337 < 1.5$ Bohr radii. When someone asks "how big is the hydrogen atom," all three are defensible answers; know which one you are quoting.

The moment machinery

Hydrogen's radial moments have closed forms, and four of them do most of atomic physics:

\langle r \rangle_{n\ell} = \frac{a_0}{2}\left[3n^2 - \ell(\ell+1)\right], \qquad \langle r^2 \rangle_{n\ell} = \frac{a_0^2n^2}{2}\left[5n^2 + 1 - 3\ell(\ell+1)\right] \qquad (4)

where: at fixed $n$ , higher $\ell$ means smaller $\langle r \rangle$ — circular orbits hug their shell radius while penetrating s-states throw long tails both inward and outward. The leading behavior $\langle r \rangle \approx \tfrac32 n^2a_0$ is the $n^2$ growth that makes Rydberg atoms micron-sized.

\Big\langle \frac{1}{r} \Big\rangle_{n\ell} = \frac{1}{n^2a_0}, \qquad \Big\langle \frac{1}{r^2} \Big\rangle_{n\ell} = \frac{1}{n^3(\ell+\tfrac12)a_0^2}, \qquad \Big\langle \frac{1}{r^3} \Big\rangle_{n\ell} = \frac{1}{n^3\,\ell(\ell+\tfrac12)(\ell+1)\,a_0^3} \quad (\ell \ne 0) \qquad (5)

where: the first is startlingly clean — no $\ell$ -dependence at all — and that is not an accident but a fingerprint: $\langle 1/r \rangle$ fixes $\langle V \rangle$ , and the energy's $\ell$ -blindness is exactly hydrogen's accidental degeneracy (its explanation lives at Accidental Degeneracy and the SO(4) Symmetry). The $\langle 1/r^3 \rangle$ formula prices spin–orbit coupling and diverges for $\ell = 0$ : since $P \sim r^{2\ell+2}$ , the integrand of $\langle r^{-3} \rangle$ goes as $1/r$ near the origin for s-states — logarithmically divergent, which is why s-state fine structure needs a separate (Darwin) term, a story for the fine-structure articles. Note what $\langle 1/r \rangle \ne 1/\langle r \rangle$ already teaches: for 1s, $\langle 1/r \rangle = 1/a_0$ but $1/\langle r \rangle = 2/3a_0$ . Moments do not commute with reciprocals, and each physical effect chooses its own.

Go deeper · Derivation B · 1 minThe l-blind moment — ⟨1/r⟩ from the virial theorem, no integrals

Go deeper · Derivation C · 2 minThe Kramers–Pasternack recursion — the moment factory

Derivation CThe Kramers–Pasternack recursion — the moment factory

The goal is the three-term recursion that generates all hydrogen radial moments — $\tfrac{s+1}{n^2}\langle r^s \rangle - (2s+1)a_0\langle r^{s-1} \rangle + \tfrac{s}{4}[(2\ell+1)^2 - s^2]a_0^2\langle r^{s-2} \rangle = 0$ — and an honest demonstration of how it is used. The full proof multiplies the radial equation for $u(r)$ by $r^su'$ and by $r^{s-1}u$ , integrates by parts repeatedly, and discards boundary terms (legal precisely when the integrals converge); we run the strategy, then cash the payoff at $s = 1$ , where the recursion delivers $\langle r \rangle$ from nothing but $\langle 1/r \rangle$ and normalization. Watch the seeds: the factory needs exactly two inputs, and both are already on the shelf.

Step 1. Set up the two ingredients. The radial equation in $u$ -form (from The Radial Equation, in units $a_0 = 1$ , $E_n = -1/2n^2$ hartree) is $u'' = \left[\ell(\ell+1)/r^2 - 2/r + 1/n^2\right]u$ . Multiply by $r^su$ and integrate: every term becomes a moment, except $\int r^s u u''\,dr$ , which integration by parts trades for $\langle r^{s-2} \rangle$ -type terms and $\int r^{s-1}u'^2$ -type remainders. Why: the recursion is the statement that those remainders cancel between the two weighted integrals — the radial equation eats its own derivative terms.

Step 2. Collect (the algebra is mechanical; Griffiths Problem 4.43 walks it) into the boxed three-term relation. Note its anatomy: the $s$ -th moment is priced by the energy ( $1/n^2$ ), the $(s-1)$ -th by the Coulomb term, the $(s-2)$ -th by the centrifugal constant $(2\ell+1)^2$ .

Step 3. Cash it at $s = 1$ . The recursion reads $\tfrac{2}{n^2}\langle r \rangle - 3a_0 + \tfrac14[(2\ell+1)^2 - 1]a_0^2\langle r^{-1} \rangle = 0$ . Insert Derivation B's $\langle 1/r \rangle = 1/n^2a_0$ and $(2\ell+1)^2 - 1 = 4\ell(\ell+1)$ :

\frac{2}{n^2}\langle r \rangle = 3a_0 - \frac{\ell(\ell+1)}{n^2}a_0 \quad\Longrightarrow\quad \langle r \rangle = \frac{a_0}{2}\left[3n^2 - \ell(\ell+1)\right].

Why: this is equation (4) — obtained without evaluating a single Laguerre integral. Iterating upward ( $s = 2, 3, \dots$ ) generates every positive moment; the one thing the recursion cannot do is descend below $\langle r^{-2} \rangle$ on its own, because at $s = 0$ the $\langle r^{-2} \rangle$ term enters multiplied by zero — fixing $\langle 1/r^2 \rangle$ takes one extra input (the standard route differentiates $E$ with respect to $\ell$ — the Hellmann–Feynman idea, named here but not developed).

Step 4. State the seeds. With $\langle r^0 \rangle = 1$ (normalization) and $\langle r^{-1} \rangle$ (virial), the factory runs upward indefinitely: $s = 2$ delivers $\langle r^2 \rangle$ of equation (4), $s = 3$ delivers $\langle r^3 \rangle$ , and so on. Verified: the recursion checked by direct sympy integration for every $(n, \ell)$ with $n \le 4$ at $s = 1, 2, 3$ , and the $s = 1$ instance solved symbolically to reproduce (4) exactly (verify script section 5).

Limiting cases

Case	Result	Meaning
$r \to 0$	$P(r) \sim r^{2\ell+2}$	centrifugal wall squared, times shell area: only s-states have contact weight (their $P \sim r^2$ )
$r \to \infty$	$P(r) \sim r^{2n}e^{-2r/na_0}$	universal exponential tail; reach grows as $na_0$
1s landmarks	mode $a_0$ · median $1.337a_0$ · mean $1.5a_0$	skewed distribution: three "sizes," one state
Circular states $\ell = n{-}1$	single hump at $r_{\rm mp} = n^2a_0$	Bohr's orbit ladder as distribution modes; relative width shrinks with $n$
Large $n$ (fixed $\ell$ )	$\langle r \rangle \approx \tfrac32 n^2a_0$	Rydberg giants: $n = 50$ means $\sim$ 0.2 μm
$Z \ne 1$	$\langle r^k \rangle \to Z^{-k}\langle r^k \rangle$	lengths shrink $1/Z$ , inverse moments grow $Z^k$ — the scaling article's one rule

Common trap|R|² is not "the probability of being at radius r"

The single most common error in this chapter is reading $|R_{n\ell}(r)|^2$ — or worse, $|\psi(r,\theta,\phi)|^2$ — as the radial probability. $|\psi|^2$ is probability per unit volume; comparing it across radii without the $r^2$ shell factor answers a question nobody asked. The tell-tale symptom: concluding that the 1s electron is "most likely at the nucleus." Per unit volume, yes; per shell — the thing a distance measurement samples — the answer is $a_0$ , and the origin has exactly zero shell probability. The same slip corrupts moments: $\langle r^k \rangle$ computed with $|R|^2\,dr$ instead of $r^2|R|^2\,dr$ is wrong for every state. When in doubt, work with $u = rR$ and $P = |u|^2$ : the 1D language has no shell factor to forget.

Worked examples

1 · Tails and contact — where the 1s electron actually is (confidence builder)

For ground-state hydrogen: (a) the probability of finding the electron beyond $2a_0$ ; (b) the probability of finding it inside the proton (charge radius $r_N = 0.84\ \mathrm{fm}$ ).

Step 1. Set up the tail integral in the scaled variable $x = r/a_0$ : $P(r > 2a_0) = 4\int_2^\infty x^2e^{-2x}dx$ .

Step 2. Integrate by parts (or use the incomplete-gamma shortcut $\int_c^\infty x^2e^{-2x}dx = \tfrac{e^{-2c}}{4}(1 + 2c + 2c^2)\cdot\tfrac{\dots}{\dots}$ — cleanly): the cumulative function is $1 - e^{-2x}(1 + 2x + 2x^2)$ , so the tail is $e^{-4}(1 + 4 + 8) = \mathbf{13e^{-4} = 0.238}$ . Nearly a quarter of all position measurements land beyond twice the Bohr radius — the exponential tail is fat.

Step 3. For the contact probability, note $r_N/a_0 = 1.59\times10^{-5} \equiv x_N$ : the exponential is 1 to ten digits, so $P(r < r_N) \approx 4\int_0^{x_N}x^2dx = \tfrac43 x_N^3$ .

Step 4. Evaluate — carefully. $\tfrac43(1.59\times10^{-5})^3 = \mathbf{5.4\times10^{-15}}$ . (A caution from this article's own verification: evaluating the exact cumulative $1 - e^{-2x}(1+2x+2x^2)$ at $x_N$ in double precision returns garbage in the last digits — the expression cancels to $10^{-15}$ , at the edge of machine epsilon. The series form is the honest route; the high-precision check confirms it.) Tiny as it is, this s-state contact weight is nonzero — the seed of hyperfine structure and electron capture; a 2p electron, with $P \sim r^4$ , has contact probability smaller by another nine orders of magnitude.

2 · The 2s/2p size paradox (exam level)

Which is bigger, 2s or 2p? Compute all the landmarks and adjudicate.

Step 1. Means, from equation (4): $\langle r \rangle_{20} = \tfrac{a_0}{2}[12 - 0] = \mathbf{6a_0}$ ; $\langle r \rangle_{21} = \tfrac{a_0}{2}[12 - 2] = \mathbf{5a_0}$ . By the mean, 2s is larger.

Step 2. Modes. $P_{21} \propto r^4e^{-r/a_0}$ peaks at $r = \mathbf{4a_0}$ (log-derivative: $4/r = 1/a_0$ ). For 2s, $P_{20} \propto x^2(2-x)^2e^{-x}$ (with $x = r/a_0$ ) has critical points where $x(2-x)(x^2 - 6x + 4) = 0$ : the node at $x = 2$ , and humps at $x = 3 \pm \sqrt5$ — an inner bump at $0.76a_0$ and the dominant outer hump at $\mathbf{5.24a_0}$ (the outer maximum is 3.7 times taller).

Step 3. Adjudicate. Every landmark agrees here: 2s is the larger state ( $6 > 5$ , $5.24 > 4$ ) and the more penetrating one — its inner hump puts real probability at small radii where 2p, walled off by $P \sim r^4$ , has almost none. "Bigger" and "closer-in" are not opposites for a two-humped distribution.

Step 4. Extract the physics. Penetration is why, the moment anything screens the nucleus (any multi-electron atom), s-states drop below p-states of the same shell: the inner hump samples the unscreened charge. The 2s/2p split of every alkali atom — invisible in hydrogen by the accidental degeneracy — is this distribution shape, priced in energy.

Origins · 1 minHendrik Anthony Kramers and how this was discovered

Hendrik Anthony Kramers

1894–1952 · Leiden and Utrecht

Kramers appears twice in this track, and the two appearances measure the man's range: the dynamics chapter tells the 1925 story — the dispersion relation that was the old quantum theory's last calculation and the new mechanics' first. This article's Kramers is twenty years on: the master calculator of Leiden, Ehrenfest's successor, whose lectures and 1937 textbook distilled hydrogen's unwieldy Laguerre integrals into the three-term recursion that now bears his name (independently derived by Simon Pasternack, hence the double-barreled title). It is a characteristic Kramers artifact: unglamorous, exact, and the fastest route to an answer — colleagues joked that almost everything he touched ended up carrying someone else's name or none at all, from the WKB method's middle initial to dispersion relations shared with Kronig. He computed the way others theorized: renormalization-flavored subtractions, barrier-penetration factors, and moment recursions, all decades before their fields matured. The recursion's fine print is pure Kramers too — it silently fails for the negative moments that diverge, exactly where the physics (spin–orbit, the Darwin term) also changes character.

History

1913

— Bohr's model fixes the orbit radius a₀; wave mechanics will later demote it to the mode of a distribution.

1926

— Schrödinger's hydrogen solution replaces orbits with wavefunctions; Born's rule makes |ψ|² a probability density.

1930

— Kramers' Leiden lectures systematize hydrogen expectation values; the moment recursion enters the toolkit.

1937

— Pasternack publishes the general ⟨rˢ⟩ recursion; the Kramers–Pasternack relation gets its modern form.

1963

— Hydrogen-maser precision forces sub-percent knowledge of radial moments for the 21 cm hyperfine problem.

Going further

Texts. Griffiths & Schroeter, Introduction to Quantum Mechanics, 3rd ed., §4.2 (and Problem 4.43) · Sakurai & Napolitano, 2nd ed., §3.7 · Shankar, 2nd ed., Ch. 13 · Cohen-Tannoudji, Diu & Laloë, Vol. 1, Ch. VII · Bethe & Salpeter, Quantum Mechanics of One- and Two-Electron Atoms, §3.

Provenance. Moment formulas ( $\langle r\rangle$ , $\langle r^2\rangle$ , $\langle 1/r\rangle$ , $\langle 1/r^2\rangle$ , $\langle 1/r^3\rangle$ ) and the Kramers recursion are standard; a verifier should recompute the 1s integrals via $\int_0^\infty r^n e^{-\beta r}\mathrm{d}r=n!/\beta^{n+1}$ and check $r_{\max}=a_0$ , $\langle r\rangle_{1s}=\tfrac32 a_0$ against Griffiths §4.2 and Bethe–Salpeter §3. The virial derivation of $\langle 1/r\rangle$ and Hellmann–Feynman derivations of $\langle 1/r^2\rangle,\langle 1/r^3\rangle$ appear in Sakurai §3.7 and Shankar Ch. 13. Bohr radius $a_0\approx 0.529\,\text{Å}$ is the CODATA value. [verbatim from the AllQFT export] Edition note (2026-07-08). The base's large- $r$ limit card printed the placeholder $P(r) \sim r^{?}e^{-2r/(na_0)}$ — a template artifact; the correct power, verified symbolically, is $r^{2n}$ (from $R \sim r^{n-1}e^{-r/na_0}$ plus the shell factor), and this article ships it. All moment formulas and the recursion were machine-verified against direct integration before shipping. Radial-function forms are consumed from Hydrogenic Wavefunctions and Quantum Numbers; the virial theorem is consumed from the dynamics chapter's Ehrenfest's Theorem.

Prerequisites. The radial equation · boundary conditions and normalization of radial wavefunctions · separation of variables in spherical coordinates · hydrogenic wavefunctions and quantum numbers · the Bohr radius and characteristic atomic scales. Leads to. Hydrogenic ions and scaling with nuclear charge · the Laplace–Runge–Lenz vector in quantum mechanics · effective potential and the centrifugal barrier.

Is P(r) a different physical postulate from the Born rule?

No — it is the Born rule, marginalized. $|\psi|^2d^3r$ is the full 3D statement; integrating over angles (where the normalized harmonics contribute exactly 1) leaves the probability in a shell. Nothing new is postulated, which is why $P(r) = |u|^2$ holds for any central potential, not just hydrogen. What is genuinely hydrogen-specific is the closed-form moment family.

Why does the most probable radius equal Bohr's orbit exactly — coincidence or necessity?

Semi-necessity. For the node-free circular states ( $1s$ , $2p$ , $3d$ , …) the distribution is a single hump $r^{2n}e^{-2r/na_0}$ whose log-slope cancellation lands at $n^2a_0$ — precisely the Bohr-model radii. The correspondence is structural: Bohr's quantization and the quantum mode both balance Coulomb attraction against an angular-momentum budget. But it holds only for the circular family: the 2s mode ( $5.24a_0$ ) matches no Bohr orbit, and means never match ( $\tfrac32 a_0$ vs $a_0$ for 1s).

How can 2s be both bigger than 2p AND more penetrating?

Because a two-humped distribution refuses one-number summaries. The 2s node at $2a_0$ splits its probability: about 5% sits in the inner hump near $0.76a_0$ (inside where 2p barely reaches, since $P_{2p} \sim r^4$ at small $r$ ), while the dominant outer hump at $5.24a_0$ lies beyond 2p's peak at $4a_0$ . Both statements — larger mean, deeper penetration — are true simultaneously, and the second one runs chemistry: penetrating s-electrons see less-screened nuclear charge, which is why $ns$ fills before $np$ in every multi-electron atom.

⟨1/r⟩ = 1/(n²a₀), so is ⟨r⟩ = n²a₀?

No — $\langle 1/r \rangle \ne 1/\langle r \rangle$ except for zero-width distributions. For 1s: $\langle 1/r \rangle = 1/a_0$ but $\langle r \rangle = 1.5a_0$ . The gap is a theorem (Jensen's inequality: $1/r$ is convex, so $\langle 1/r \rangle \ge 1/\langle r \rangle$ ) and a physics lesson: binding energy samples the inner distribution, size samples the tail. Each observable picks the moment that matches its own $r$ -weighting — that is the entire point of keeping a family of moments rather than one "radius."

Where exactly does the Kramers recursion stop working, and what takes over?

Two places. Descending: the derivation's discarded boundary terms are only legal while the integrals converge, and $\langle r^s \rangle$ exists only for $s > -(2\ell + 3)$ — for s-states, $\langle 1/r^3 \rangle$ is already divergent, matching the recursion's blind spot at $s = 0$ where the $\langle r^{-2} \rangle$ term enters with coefficient zero. Fixing $\langle 1/r^2 \rangle$ needs an independent input — the standard trick treats $\ell$ as a continuous parameter and differentiates the energy (the Hellmann–Feynman idea). Physically the breakdown radii are honest: the very moments the recursion cannot reach are the ones dominated by the near-origin region where relativistic corrections take over anyway.

Does P(r) tell me where the electron IS between measurements?

It tells you the statistics of position measurements on identically prepared atoms — the standard Born-rule reading established back at Born Statistical Interpretation. Between measurements the electron does not have a radius any more than a spin has all three components; a stationary state's $P(r)$ is time-frozen precisely because nothing is orbiting. What is physically real without measurement is the moments' appearance in energies: $\langle 1/r \rangle$ shows up in the binding energy whether or not anyone measures a position, which is why moment machinery, not the full distribution, is what most of atomic physics actually consumes.