Beta Decay and Atomic Shake-Up as Sudden Processes

Every second, somewhere in a bottle of tritiated water, a nucleus quietly commits the most violent act in atomic physics: it changes its identity. A neutron becomes a proton, an electron and an antineutrino flee the scene, and the lone atomic electron — which took no part in any of this — wakes up orbiting a nucleus whose charge has doubled. It never felt a push. Its wavefunction did not move. And yet roughly 30% of the time it ends up excited or ejected, because the meaning of its state changed underneath it. This is the sudden approximation doing real physics. The previous article proved the frozen-state theorem and its error bound; this one spends that theorem on nature's own quench experiments. We first audit the suddenness claim with honest numbers — the mean tritium beta electron leaves at $0.148c$ , clearing the atom in $1.2\times10^{-18}\,$ s against an atomic clock of $4.8\times10^{-17}\,$ s, a ratio of one part in forty. We then compute the complete shake ladder for $^3\mathrm{H}\to{}^3\mathrm{He}^+$ : survival $512/729 = 70.2\%$ , shake-up to $2s$ exactly $1/4$ , nothing at all into any $p$ state, and $2.6\%$ shake-off into the continuum. A closure trick then delivers the mean final energy in a single integral — and exposes what the $2.6\%$ of escaping electrons carry away. The same projection arithmetic runs X-ray photoemission satellites, electron-capture vacancy cascades, and the final-state systematics of the KATRIN neutrino-mass experiment. One frozen wavefunction, re-read in a new basis: that is the entire calculation.

Key takeaways

Beta decay changes the nuclear charge $Z \to Z+1$ in a time $\sim a_0/v_\beta \approx 10^{-18}\,$ s — forty times faster than the atomic clock $\hbar/\mathrm{Ry}$ . The bound electrons are frozen spectators: the sudden approximation applies, with freeze error $\sim 6\times10^{-4}$ .
The frozen orbital is re-expanded in the eigenbasis of the daughter ion. For tritium: $P(1s) = 512/729 = 0.7023$ , $P(2s) = 1/4$ exactly, $P(np) = 0$ by symmetry, and $2.6\%$ shake-off to the continuum.
A spherically symmetric charge jump conserves $\ell$ and $m$ : shake-up feeds only $s$ states from an $s$ state. The overlap of different- $\ell$ states vanishes identically.
The closure trick gives the mean final energy without the ladder: $\langle 1s_1|H_{Z=2}|1s_1\rangle = -3\,\mathrm{Ry} = -40.8\,$ eV in one integral. The discrete ladder alone gives $-41.7\,$ eV; the missing $+0.9\,$ eV is carried by the $2.6\%$ shake-off electrons, at $\sim34\,$ eV each.
The same arithmetic runs XPS shake-up satellites (core hole = sudden $Z_{\rm eff}\!+\!1$ , $\sim11\%$ satellites in the hydrogenic estimate) and the KATRIN final-state spectrum, and it splits α decay by shell: sudden for valence electrons, adiabatic for the K shell.
Key idea: a nuclear transformation is a sudden switch of the atomic Hamiltonian — the electron cloud is frozen, and every observable probability is an overlap $|\langle f_{\rm new}|\psi_{\rm old}\rangle|^2$ of the old orbital with the new ion's eigenstates.

The core insight

A nuclear transformation is a sudden switch of the atomic Hamiltonian: the electron cloud is frozen at the instant of decay, and shake-up and shake-off probabilities are overlaps $|\langle f_{\rm new}|\psi_{\rm old}\rangle|^2$ of the old orbitals with the daughter ion's eigenstates — no perturbative smallness anywhere.

Where this lives today

The sudden approximation is quietly load-bearing in three experimental programs. KATRIN's neutrino-mass extraction stands on the calculated final-state distribution of ³HeT⁺ — the molecular grandchild of Derivation B — and the collaboration's error budget explicitly tracks it. Dark-matter detectors (XENON, LZ, EDELWEISS) now search for the Migdal effect: a WIMP-struck nucleus recoils "suddenly" from under its own electron cloud, and the resulting shake-off electron converts an invisible nuclear recoil into a detectable ionization signal — the tritium calculation run with the roles of nucleus and electron reversed. And every X-ray photoelectron spectrum read in a materials lab carries shake-up satellites whose intensities are sudden-limit overlaps; they are used as fingerprints of chemical environment. The common thread: whenever nature switches a potential faster than electrons can follow, someone is billing the overlap integral.

The physical picture

The tritium electron lives in the gentle $Z=1$ Coulomb well, its $1s$ cloud peaked at one Bohr radius. At the instant of decay the well doubles its depth and halves its natural size: the $Z=2$ ground state hugs the nucleus at $r = a_0/2$ . The old cloud is now too puffy for the new well. Projecting the frozen shape onto the new eigenstates, the puffiness pays out as a distribution: mostly the new $1s$ (the shapes still resemble each other), a surprising exact quarter into $2s$ (whose outer lobe lives right where the old cloud was), and a few percent into the continuum — the electron was, in effect, partly "outside" the new atom the moment it formed.

Figure

The moment of decay: one frozen cloud, a new set of shapes to match

Two clocks decide whether "frozen" is legitimate. The beta electron, born with a mean kinetic energy of $5.7\,$ keV, crosses the atomic radius in about an attosecond; the bound electron's own orbital clock, $\hbar/\mathrm{Ry}$ , ticks forty times slower. During the getaway the atomic state accumulates a phase spread of only $\sim0.025$ radians — the freeze is good to a few parts in $10^4$ . That audit, with every number computed, is Derivation A. The payoff picture is the shake ladder itself: where the frozen electron lands. Note what is absent — the $2p$ , $3p$ , $3d$ columns are identically zero. The nuclear charge jump is spherically symmetric, so the sudden switch cannot change $\ell$ or $m$ : the selection lemma of The Sudden Approximation, now with experimental consequences.

Figure

The tritium shake ladder: where the frozen electron lands

The sudden dictionary for nuclear transformations

The machinery is inherited, not rebuilt. The Sudden Approximation proved that when a Hamiltonian switches in a time $\tau$ short against $\hbar/\Delta\bar H$ , the state is frozen and the populations of the new eigenstates are set once and for all by overlaps. Here the switch is a change of nuclear charge, and the question a decay hands us is always the same: given the old orbital, what does the new ion see?

P_{i\to f} \;=\; \big|\langle f^{(Z')}\,|\,\psi^{(Z)}\rangle\big|^2 \qquad (1)

where: $|\psi^{(Z)}\rangle$ — the electron's orbital in the parent atom (charge $Z$ ); $|f^{(Z')}\rangle$ — an eigenstate (bound or continuum) of the daughter ion (charge $Z'$ ). Bound destinations are shake-up; continuum destinations are shake-off. The change $Z \to Z'$ is order unity — nothing perturbative anywhere; what makes Eq. (1) exact is speed, not smallness. For hydrogenic $1s$ orbitals the overlap has a clean closed form that this article derives and then spends repeatedly:

\langle 1s^{(Z')} | 1s^{(Z)}\rangle \;=\; \frac{8\,(ZZ')^{3/2}}{(Z+Z')^{3}} \qquad (2)

where: the numerator carries the normalizations $Z^{3/2}$ of the two clouds and the denominator the shared exponential decay $e^{-(Z+Z')r/a_0}$ . The formula is symmetric in $Z \leftrightarrow Z'$ — capture ( $Z\to Z-1$ ) and decay ( $Z\to Z+1$ ) reuse the same table — and it equals $1$ only at $Z = Z'$ , as it must.

Derivations

Derivation AHow sudden is beta decay, in numbers

The base texts wave at this with " $v_\beta \to c$ "; for tritium — the flagship — that is off by a factor of seven, so we do the audit honestly and check that the conclusion survives. The strategy: compute the beta electron's actual speed relativistically from the measured mean kinetic energy, form the exit time $\tau = a_0/v_\beta$ , compare with the atomic clock $\hbar/\mathrm{Ry}$ , and convert the ratio into the freeze-error bound of the parent article. The point to watch is that suddenness needs the beta electron to be fast compared with the bound electron, not compared with light.

Step 1. The beta electron's speed. Tritium's spectrum ends at $Q = 18.6\,$ keV with mean kinetic energy $\bar T = 5.7\,$ keV. Relativistically, $\gamma = 1 + \bar T/m_ec^2 = 1.0112$ , so

\beta \;=\; \sqrt{1 - 1/\gamma^2} \;=\; 0.148, \qquad v_\beta = 4.44\times10^{7}\ \mathrm{m/s}.

Why keep relativity? Honesty more than necessity: at $5.7\,$ keV the classical estimate $\sqrt{2\bar T/m_e}$ agrees to $1\%$ ; at the $18.6\,$ keV endpoint ( $\beta = 0.263$ ) the correction reaches a few percent. Either way, $v_\beta$ is nowhere near $c$ — and it does not need to be.

Step 2. The two clocks. The getaway time across the atom and the bound electron's own period-scale:

\tau \;=\; \frac{a_0}{v_\beta} \;=\; 1.19\times10^{-18}\ \mathrm{s}, \qquad \frac{\hbar}{\mathrm{Ry}} \;=\; 4.84\times10^{-17}\ \mathrm{s}.

The ratio is $\tau\,\mathrm{Ry}/\hbar = 0.025$ — one part in forty. Equivalently, in velocity language: $v_\beta / v_{\rm bound} = v_\beta/(\alpha c) = 20$ , so the beta electron leaves twenty times faster than the bound electron moves.

Step 3. Convert to the freeze error. The parent article's variance bound gives, with the atomic energy spread standing in for $\Delta\bar H$ ,

P_{\rm leave}^{\rm switch} \;\sim\; \left(\frac{\tau\,\mathrm{Ry}}{\hbar}\right)^{2} \;=\; 6\times10^{-4}.

Why this matters: the shake probabilities we are about to compute are of order $30\%$ . The error of the frozen-state assumption is $0.06\%$ — five hundred times smaller than the effect. The sudden approximation is not merely applicable here; it is precise. Verified: relativistic and classical $\beta$ at both energies, $\tau = 1.19\times10^{-18}\,$ s, clock $4.84\times10^{-17}\,$ s, ratio $0.0246$ , bound $6.1\times10^{-4}$ , speed ratio $20.3$ — verification script, this article.

Go deeper · Derivation B · 2 minThe tritium shake ladder from one overlap integral

Derivation BThe tritium shake ladder from one overlap integral

We compute where the tritium electron lands, exactly. The plan: derive the general $1s$ – $1s$ overlap of Eq. (2) from the radial integral, square it for the survival probability; then take the $2s$ overlap and meet the cleanest rational surprise in atomic physics; kill all $\ell \neq 0$ destinations with the symmetry lemma; and sum the $ns$ ladder to expose the shake-off remainder. Every entry is a closed form — the danger is only bookkeeping (the $2s$ amplitude is negative; the node does real work).

Step 1. The general $1s$ – $1s$ overlap. With $\psi_{1s}^{(Z)} = (Z^3/\pi a_0^3)^{1/2}e^{-Zr/a_0}$ and the standard integral $\int_0^\infty r^2 e^{-\alpha r}\mathrm{d}r = 2/\alpha^3$ ,

\langle 1s^{(Z')}|1s^{(Z)}\rangle = \frac{(ZZ')^{3/2}}{\pi a_0^3}\,\frac{8\pi a_0^3}{(Z+Z')^3} = \frac{8(ZZ')^{3/2}}{(Z+Z')^3} \;\;\xrightarrow{Z=1,\,Z'=2}\;\; \frac{16\sqrt2}{27} = 0.8381 .

Step 2. Square for survival:

P(1s) \;=\; \left(\frac{16\sqrt2}{27}\right)^{2} \;=\; \frac{512}{729} \;=\; 0.7023 .

Why not higher? The frozen cloud peaks at $r = a_0$ ; the He⁺ ground state peaks at $a_0/2$ . A $70\%$ match is what "similar but half the size" costs.

Step 3. The $2s$ rung. The He⁺ $2s$ orbital, $R_{20}^{(2)} = 2(1-r/a_0)e^{-r/a_0}/a_0^{3/2}$ , has its node at exactly $r = a_0$ — the old cloud's peak — and an outer lobe lying right beneath the old cloud's tail. The radial integral collapses to

\langle 2s^{(2)}|1s^{(1)}\rangle \;=\; -\frac{1}{2} \quad\Longrightarrow\quad P(2s) \;=\; \frac14 \;\;\text{exactly}.

Why exactly? The $Z' = 2Z$ coincidence: the $2s$ orbital of charge $2Z$ has the same exponential $e^{-Zr/a_0}$ as the $1s$ of charge $Z$ , so the overlap is a ratio of rationals with no surds. The minus sign is the node's signature — invisible in the probability, decisive in any interference experiment.

Step 4. The empty columns. The charge jump is spherically symmetric, so it commutes with $\hat L^2$ and $\hat L_z$ : the sudden selection lemma of The Sudden Approximation forces $\Delta\ell = \Delta m = 0$ . From an $s$ state, every $p$ , $d$ , $f$ destination is identically zero — machine-checked as $\langle Y_{00}|Y_{1m}\rangle = 0$ . Shake-up cannot change shape, only size.

Step 5. The ladder and the remainder. The higher $ns$ overlaps (computed symbolically with hydrogenic $R_{n0}$ ):

P(3s) = 0.0127,\quad P(4s) = 0.0039,\quad P(5s) = 0.0017,\quad P(6s) = 0.0009,\;\dots

The bound-state total converges to $\sum_n P(ns) = 0.9735$ , leaving

P_{\rm shake\text{-}off} \;=\; 1 - 0.9735 \;=\; 2.6\%

for the continuum: about one tritium decay in forty ionizes its own daughter atom. (The base's headline split — $70.2\%$ survive, $29.8\% = 217/729$ leave the ground state — is confirmed; of that $29.8\%$ , most is the single $2s$ rung.) Verified: Eq. (2) derived symbolically; $16\sqrt2/27$ and $512/729$ exact in sympy; $\langle 2s|1s\rangle = -1/2$ exact; angular zero exact; ladder $n \le 20$ summed, $0.9735$ ; shake-off $2.65\%$ — verification script, this article.

Go deeper · Derivation C · 1 minThe closure audit: the mean final energy in one integral

Derivation CThe closure audit: the mean final energy in one integral

The ladder tells us the distribution; closure tells us its mean without ever diagonalizing anything — and the comparison of the two routes reveals what the shake-off electrons carry. The idea: the mean energy of the frozen state under the new Hamiltonian is just an expectation value in the old state, $\langle E\rangle = \langle 1s^{(1)}|H_{Z=2}|1s^{(1)}\rangle$ , because $\sum_f P_f E_f = \langle\psi|H_f|\psi\rangle$ by completeness. The subtlety worth watching: the discrete ladder alone overshoots this mean, and the mismatch is physical.

Step 1. The one integral. In atomic units, $H_{Z=2} = \tfrac12 p^2 - 2/r$ , and in the old $1s$ state $\langle \tfrac12 p^2\rangle = +\tfrac12$ (virial theorem for $Z=1$ ) while $\langle 1/r\rangle = 1$ :

\langle E\rangle \;=\; \tfrac12 - 2\cdot 1 \;=\; -\tfrac32\ \mathrm{a.u.} \;=\; -3\,\mathrm{Ry} \;=\; -40.8\ \mathrm{eV}.

Why so easy? The kinetic energy is a property of the frozen wavefunction — unchanged by the decay — and the new potential only asks for $\langle 1/r\rangle$ in the old state. No ladder required.

Step 2. The two-route check. Summing the discrete ladder, $\sum_n P(ns)\,E_n^{(Z=2)}$ with $E_n = -54.4/n^2\,$ eV, gives $-41.7\,$ eV — deeper than the closure mean. The books balance only when the continuum is added:

\underbrace{-41.7\ \mathrm{eV}}_{\text{bound ladder}} \;+\; \underbrace{+0.9\ \mathrm{eV}}_{\text{continuum}} \;=\; -40.8\ \mathrm{eV}.

Step 3. Read the remainder. That $+0.9\,$ eV rides on just $2.6\%$ of the probability, so the mean kinetic energy of a shake-off electron is $0.9/0.026 \approx 34\,$ eV — a soft electron, but far above chemical scales, and a real background in any experiment that counts low-energy electrons near a tritium source. Verified: $\langle 1/r\rangle = 1$ symbolic; closure $-3/2$ a.u. exact; ladder energy $-1.533$ a.u.; continuum share $+0.033$ a.u. $= +0.90\,$ eV; mean shake-off energy $34\,$ eV — verification script, this article.

Alpha decay: sudden and adiabatic in the same atom

Alpha decay is the perfect foil, because it fails the sudden test — selectively. A $5.5\,$ MeV alpha particle leaves at $v_\alpha = 1.6\times10^{7}\,$ m/s $= 0.054c$ . Compare shell by shell in a heavy daughter like lead: a K-shell electron moves at $v_K \approx Z\alpha c \approx 0.60c$ , eleven times faster than the alpha — the inner shells adiabatically readjust as the charge recedes, and almost never shake. A valence electron moves at $\sim\alpha c$ , seven times slower than the alpha — for it the decay is sudden, and it shakes exactly as the tritium electron does. One nuclear event, both regimes of the parent article's criterion, sorted by shell velocity. This is the split Migdal formalized in 1939–1941, and it is why alpha-decay ionization comes almost entirely from the outermost shells. Electron capture runs the tritium calculation in reverse ( $Z \to Z-1$ , using Eq. (2)'s symmetry), with an extra twist: the captured electron leaves a $1s$ vacancy, and the resulting X-ray/Auger cascade rides on top of the shake redistribution.

Limiting cases

Case	Result	Meaning
Tritium beta decay	$\tau/T_{\rm atom} = 1/40$ , freeze error $6\times10^{-4}$	Deep in the sudden regime; ladder exact to a part in a thousand.
MeV beta emitters	$v_\beta \to c$ , ratio smaller still	The textbook "relativistic" limit — even more sudden than tritium.
Alpha decay, K shell (heavy atom)	$v_K \approx 0.6c \gg v_\alpha$	Adiabatic: inner shells follow the receding charge, no shake.
Alpha decay, valence shell	$v_{\rm val} \approx \alpha c \ll v_\alpha$	Sudden: outer electrons shake exactly like the tritium case.
$Z \to Z+1$ at large $Z$	$P_{\rm stay} = \frac{[4Z(Z+1)]^3}{(2Z+1)^6} \to 1$	A one-unit jump matters less on a deep well: $0.70,\ 0.88,\ 0.94$ for $Z=1,2,3$ .
$Z' = Z$	$P_{\rm stay} = 1$	No change, no shake — Eq. (2) equals unity, the sanity check.

Worked examples

Example 1: the complete tritium budget, and what KATRIN does with it

Tabulate the fate of the electron in $^3\mathrm{H}\to{}^3\mathrm{He}^+ + e^- + \bar\nu_e$ and connect it to the neutrino-mass endgame.

Step 1. Enumerate destinations (Derivation B): $P(1s) = 512/729 = \mathbf{0.7023}$ ; $P(2s) = \mathbf{0.2500}$ exactly; $P(3s) = 0.0127$ ; $P(4s) = 0.0039$ ; higher $ns$ summing to $0.0046$ ; all $\ell \neq 0$ states: $\mathbf{0}$ .

Step 2. The remainder is shake-off: $P_{\rm cont} = 1 - 0.9735 = \mathbf{2.6\%}$ , with mean energy $\approx \mathbf{34\ eV}$ (Derivation C).

Step 3. The energy books: closure gives $\langle E\rangle = \mathbf{-40.8\ eV}$ ; the $1s\to1s$ decays release the difference $E_{1s}^{\rm He^+} - E_{1s}^{\rm H} = -54.4 + 13.6 = -40.8\,$ eV... exactly the closure mean only when averaged over the whole ladder — individual channels shift the beta spectrum endpoint by their excitation energy ( $+40.8\,$ eV for the $2s$ channel relative to $1s$ , etc.).

Step 4. KATRIN's stake. The neutrino mass is extracted from the last few eV of the beta spectrum, where each shake channel displaces the endpoint by its final-state energy. KATRIN uses molecular tritium (T₂ → ³HeT⁺), so the atomic ladder above becomes a molecular final-state distribution — computed with exactly this sudden-overlap machinery, and modeled as one of the experiment's leading systematics. The 2024 result, $m_\nu \le \mathbf{0.45\ eV}$ (90% CL), leans on that distribution being right.

Example 2: a core hole in neon — XPS satellites from the same formula

An X-ray photoemission event suddenly removes a core electron; the valence electrons see the effective charge jump by roughly one unit. Estimate the satellite fraction.

Step 1. Model a valence electron hydrogenically with $Z_{\rm eff} = 2 \to 3$ . Equation (2) squared gives $P_{\rm stay} = [4\cdot2\cdot3]^3/5^6 = 13824/15625 = \mathbf{0.885}$ .

Step 2. So $\approx \mathbf{11.5\%}$ of photoemission events shake the valence electron into an excited state — the satellite peaks displaced from the main XPS line by the excitation energy. Measured neon shake fractions are of exactly this order, and the systematics of the estimate (screening, multiple valence electrons) are the standard XPS refinements.

Step 3. The trend: repeating at $Z_{\rm eff} = 1, 2, 3$ gives shake fractions $30\%,\ 11\%,\ 6\%$ — satellites fade for deeper wells, which is why tritium ( $Z=1$ ) is the shake showcase of nature.

Origins · 1 minArkady Migdal and how this was discovered

Arkady Migdal

1911–1991 · Leningrad, Moscow

Arkady Beynusovich Migdal made the sudden approximation a nuclear tool. In 1939 he computed the ionization of atoms in nuclear collisions, and in 1941 — in a Journal of Physics USSR paper that became the standard citation — the ionization accompanying alpha and beta decay, including the shell-by-shell split: sudden for slow outer electrons, adiabatic for fast inner ones. Trained in Leningrad and among the first to pass Landau's famous theoretical minimum, later a Moscow professor, he went on to build the theory of finite Fermi systems and to predict pion condensation in dense nuclear matter, training a generation of Soviet theorists; his Qualitative Methods in Quantum Theory is the connoisseur's guide to approximation done honestly. Migdal was also famously unprofessorial — a mountaineer, scuba pioneer, and juggler who held that physical daring and physical insight came from the same place. Eight decades on, his decay-ionization calculation returned to the spotlight: the "Migdal effect" — shake-off from a suddenly recoiling nucleus — is now a search channel for dark matter.

History

1939

— Arkady Migdal computes atomic ionization in nuclear collisions, the first shake calculation

1941

— Migdal's paper on ionization accompanying alpha and beta decay establishes the sudden treatment of nuclear transformations

1965

— Carlson and Krause measure electron shake-off accompanying inner-shell ionization in neon, confirming sudden-overlap predictions

2018

— the Migdal effect — ionization from a suddenly recoiling nucleus — is proposed as a dark-matter detection channel (Ibe, Nakano, Shoji, Suzuki)

2024

— KATRIN reaches a neutrino-mass bound of 0.45 eV from the tritium endpoint, with sudden-approximation final-state spectra among its modeled systematics

Going further

References. Griffiths, Introduction to Quantum Mechanics, 3rd ed., §11.5 and Problem 11.36 (tritium) · Sakurai & Napolitano, Modern Quantum Mechanics, 2nd ed., §5.6 · Shankar, Ch. 18 · Cohen-Tannoudji, Diu & Laloë, Vol. II (sudden-approximation complement) · Townsend, A Modern Approach to Quantum Mechanics, 2nd ed., Ch. 14 · Migdal, Qualitative Methods in Quantum Theory (shake-up/shake-off).

Provenance. The AllQFT base note reads, verbatim: "Sudden-rule derivation and the frozen-state propagator argument follow Griffiths §11.5, Sakurai §5.6, and Cohen-Tannoudji Vol. II. A verifier should reproduce: (1) the $1s$ – $1s$ overlap $8(ZZ')^{3/2}/(Z+Z')^3$ via the standard integral $\int_0^\infty r^2 e^{-\alpha r}\mathrm{d}r=2/\alpha^3$ ; (2) the tritium survival probability $P=(16\sqrt2/27)^2=512/729\approx0.7023$ (Griffiths Problem 11.36), and the residual $217/729\approx0.298$ ; (3) the closed form $[4Z(Z+1)]^3/(2Z+1)^6$ reduces to $512/729$ at $Z=1$ . The Bohr-period and beta-velocity estimates are order-of-magnitude (bound-electron energy $13.6\,$ eV, $v_{\rm bound}\sim Z\alpha c$ , $v_\beta\to c$ ); the smallness parameter is $Z\alpha\approx Z/137$ . All numbers are standard textbook values." We reproduced all three items exactly (this article's verification script). One precision note: for tritium specifically, " $v_\beta \to c$ " overstates — the mean-energy beta moves at $0.148c$ — but the sudden verdict is unchanged, since what matters is $v_\beta/v_{\rm bound} = 20$ ; Derivation A replaces the order-of-magnitude audit with computed values. (The neighbouring base's garbled tritium line — $2^9$ for $2^6$ , a fabricated $4096/4913$ — is documented and repaired in The Sudden Approximation; the numbers here are the correct ones it pointed to.)

Builds on: The Interaction Picture and Time-Dependent Hamiltonians · Time-Dependent Perturbation Theory: General Formalism · Dyson Series and the Time-Ordered Evolution Operator — and, directly, The Sudden Approximation (the frozen-state theorem, criterion, and selection lemma this article spends). Leads to: First-Order Transition Amplitudes and Probabilities · Density of Final States · Constant Perturbation Switched On.

If the electron never felt a push, what actually ionizes it in shake-off

Nothing pushes it; its own shape does the work. The frozen orbital of the $Z=1$ atom is not a state of the $Z=2$ ion — it is a superposition, and $2.6\%$ of that superposition happens to lie in the continuum. Measurement (or just subsequent evolution) resolves the superposition, and in one decay out of forty the electron is simply found unbound, with a mean energy of about $34\,$ eV. Projection, not collision.

The beta electron flies out in one direction — doesn't that break the spherical symmetry that forbids p states

The selection lemma uses the symmetry of the change in the atomic Hamiltonian, which is the nuclear charge $Z\to Z+1$ — perfectly central. The outgoing beta's direct Coulomb interaction with the bound electron is a separate, dynamical effect, suppressed by the short transit time; it produces a small directional correction (part of the "direct collision" ionization) that is orders of magnitude below the $\ell$ -conserving shake terms. At sudden-approximation accuracy, $p$ states stay empty.

Why is the 2s amplitude negative, and does the sign matter

The He⁺ $2s$ orbital changes sign at its node, which sits exactly at $r = a_0$ — the peak of the frozen cloud. The overlap integral is dominated by the outer lobe, where $2s$ is negative, giving $\langle 2s|1s\rangle = -1/2$ . The probability $1/4$ doesn't care, but the sign is physical: the post-decay state is a coherent superposition $0.84|1s\rangle - 0.5|2s\rangle + \dots$ , and its subsequent quantum beats (at the He⁺ Lyman frequency) depend on that relative phase.

Does the daughter nucleus recoil suddenly too

The nucleus recoils with only a few eV of energy — glacially slow motion on electronic timescales, so for the electron ladder the recoil is irrelevant (it is, if anything, adiabatic). The interesting inversion is the Migdal effect: when a dark-matter particle strikes a nucleus, the recoil can be fast on electronic timescales, and then it is the nucleus jumping out from under the frozen electron cloud — the same overlap calculation with the displacement, rather than the charge, doing the switching.

How exact is 512/729 — what corrections are neglected

Three, all small: the freeze error of the switch itself ( $\sim6\times10^{-4}$ , Derivation A); the direct beta-electron–bound-electron interaction (a further correction at the $10^{-3}$ - $10^{-4}$ level for tritium); and the fact that real experiments use molecular T₂ rather than atomic tritium, which replaces the ladder by the ³HeT⁺ rovibronic final-state distribution — a genuinely different (and harder) sudden-overlap calculation, but the same physics. As an atomic statement, $512/729$ is exact within the sudden approximation.

Can I use the same table for electron capture

Yes — Eq. (2) is symmetric in $Z \leftrightarrow Z'$ , so the $Z\to Z-1$ overlap table is the same as the $Z-1\to Z$ one. Two differences in practice: capture consumes a $1s$ electron, so the shake question applies to the remaining electrons (screened, multi-electron orbitals rather than pure hydrogenic ones), and the $1s$ vacancy triggers an X-ray/Auger cascade on top of the shake redistribution. The sudden overlap is the first layer of that onion, not the whole of it.