7.8 The sudden approximation

The last section finished off the “infinitely slow” end: in adiabatic evolution the system never strays a step from the instantaneous eigenstate. To close the chapter, look at the other end of the scale — the Hamiltonian switching in an instant. Surprisingly, this end is even simpler than the adiabatic one, and it comes with a textbook-grade real case whose answer is almost too beautiful to be true.

The case is tritium. Tritium ( $^3$ H) is hydrogen’s radioactive isotope: one proton, two neutrons, with one electron in orbit whose wavefunction is our old friend, the hydrogen 1s state. The tritium nucleus undergoes beta decay:

^3\text{H}\ \longrightarrow\ ^3\text{He}^+ + e^-_{\beta} + \bar\nu\tag{7.8.1}

A neutron turns into a proton, ejecting a high-energy beta electron (up to 18.6 keV) and an antineutrino. For the orbital electron left behind, the sky falls in a very specific way: the nuclear charge jumps from $Z=1$ to $Z=2$ in an instant. The “hydrogen atom” it lived in is gone; underfoot is now a “helium ion” — a whole new ladder of levels with completely different spacings.

The question: which level of the new system does this electron land on, and with what probabilities?

First, be clear why none of the old tools apply. Perturbation theory? The potential doubles outright — no small parameter. The adiabatic theorem? It requires the change to be far slower than the electron’s orbital period, and here it is exactly the opposite — the beta electron bursts out of the atom at nearly the speed of light (crossing $a_0$ takes about $10^{-19}$ seconds), while the orbital electron’s intrinsic period is $\sim10^{-16}$ seconds: the change is a thousand times faster than the system’s beat.

Too fast to react — which makes it easy

The Schrödinger equation says the wavefunction’s rate of change is $\partial_t\Psi=-\ii\hat H\Psi/\hbar$ — a finite number. A finite rate of change times a switching time tending to zero gives zero change:

\ket{\Psi(0^+)}=\ket{\Psi(0^-)}\tag{7.8.2}

At the instant of the switch, the wavefunction is untouched. That is the entire content of the sudden approximation — what it approximates is not the wavefunction (the wavefunction is exactly continuous) but the idealisation that “the switching time is strictly zero”.

What changes is the wavefunction’s identity. Before the switch it was an eigenstate of the old Hamiltonian, sitting pretty; after the switch, the very same wavefunction finds itself on the new Hamiltonian’s turf, where it is no eigenstate at all but a superposition of the new eigenstates. The script from here on we have rehearsed countless times in section 3.8: expand it in the new basis, and the squared moduli of the expansion coefficients are the probabilities of finding each new level.

\ket{\Psi}=\sum_n c_n\ket{n^{\text{new}}},\qquad c_n=\braket{n^{\text{new}}}{\Psi},\qquad P_n=|c_n|^2\tag{7.8.3}

Set the criterion first. “Instantaneous” is always relative: if the switch takes a time $\tau$ , and the system’s intrinsic beat is $\hbar/\Delta E$ ( $\Delta E$ being the spacing of the relevant levels — the timescale on which the wavefunction’s components drift out of phase), then

\tau\ \ll\ \frac{\hbar}{\Delta E} \quad\Longleftrightarrow\quad\text{the sudden approximation holds}\tag{7.8.4}

Tritium’s ledger: $\tau\sim10^{-19}$ s, $\hbar/\Delta E\sim10^{-16}$ s (level spacings of tens of eV) — the criterion is met with three orders of magnitude to spare.

The picture

Two faces of the same coin. Adiabatic and sudden are the two limits of the same ratio $\tau\Delta E/\hbar$ :

Far greater than 1: adiabatic — the wavefunction follows step for step, staying an instantaneous eigenstate; populations unchanged, only phases accumulate (last section);
Far less than 1: sudden — the wavefunction stands stock-still while its identity is reshuffled; populations redistributed.

Slow change means following in place, fast change means freezing in place — both ends are simple. The genuinely hard part is the middle ground, $\tau\Delta E/\hbar\sim1$ — there you can neither freeze nor follow, and must honestly solve the time-dependent equation (or fall back on section 7.5’s time-dependent perturbation theory, if the change is small in amplitude).

The mathematics

Same initial state $\ket{n^{\text{old}}}$ , same change from $\hat H_{\text{old}}$ to $\hat H_{\text{new}}$ :

\text{Adiabatic:}\quad P_{m}=\delta_{mn}\qquad\text{(follows into the corresponding new state)}\tag{7.8.5}

\text{Sudden:}\quad P_{m}=\bigl|\braket{m^{\text{new}}}{n^{\text{old}}}\bigr|^2\tag{7.8.6}

In the sudden limit the probabilities are settled once and for all by the overlap of old and new eigenstates — no evolution equation to solve; the entire dynamical problem collapses into a single integral.

A worked example: where did tritium’s electron go?

Level populations of the daughter ion after tritium decaybasic~8 min

The frozen old state: the $Z=1$ 1s wavefunction

\psi_{1s}^{(Z=1)}=\frac{1}{\sqrt{\pi a_0^3}}\,\ee^{-r/a_0}\tag{7.8.7}

The new basis: the $Z=2$ hydrogen-like wavefunctions (take the hydrogen solutions and replace $a_0\to a_0/Z$ ). The new 1s state:

\psi_{1s}^{(Z=2)}=\sqrt{\frac{8}{\pi a_0^3}}\,\ee^{-2r/a_0}\tag{7.8.8}

Clear the field with symmetry first: the old state is spherically symmetric ( $\ell=0$ ), and the charge jump involves no direction whatsoever — angular momentum is strictly conserved across the switch. So the electron can only land in the s states of the new system: $P(2p)=P(3p)=P(3d)=\cdots=0$ , not a scrap of probability. All that remains is the overlap with each $n$ s state.

Overlap with the new 1s (the angular integral gives $4\pi$ ; the radial one is elementary):

c_{1s}=\int\psi_{1s}^{(Z=2)}\,\psi_{1s}^{(Z=1)}\,\dd^3r =\frac{2\sqrt2\cdot4\pi}{\pi a_0^3}\int_0^\infty r^2\ee^{-3r/a_0}\,\dd r =8\sqrt2\cdot\frac{2}{27}=\frac{16\sqrt2}{27}\tag{7.8.9}

(using $\int_0^\infty r^2\ee^{-\lambda r}\dd r=2/\lambda^3$ ). Hence

P(1s)=\left(\frac{16\sqrt2}{27}\right)^2=\frac{512}{729}\approx0.702\tag{7.8.10}

Overlap with the new 2s: insert $\psi_{2s}^{(Z=2)}=\sqrt{\dfrac{1}{\pi a_0^3}}\left(1-\dfrac{r}{a_0}\right)\ee^{-r/a_0}$ ; both pieces are the same kind of integral, and the result is suspiciously clean:

c_{2s}=-\frac12,\qquad P(2s)=\frac14=0.250\tag{7.8.11}

The tally:

Final state	Probability
He⁺ 1s	70.2%
He⁺ 2s	25.0%
All p and d states	Strictly 0
Higher ns + ionisation	4.8%

Seven in ten electrons settle quietly into the new ground state, a quarter get lifted to 2s (falling back later with a flash of light), and about 3% are ionised outright and fly away — one decay, with a miniature “chemical reaction” thrown in for free. These numbers are no paper game: precision measurements of the tritium decay spectrum (today the heart of the KATRIN neutrino-mass experiment) must subtract exactly these electron-side energy allocations term by term, using precisely the overlap integrals computed here (carried, of course, to far higher accuracy).

All the dynamics, collapsed into a geometry problem

The sudden approximation turns a time-dependent problem into pure geometry: how long are the projections of the old-state vector onto the directions of the new basis? No evolution equation, no time integral — only angles in Hilbert space. The intuitive reading of P(1s)=70%: the $Z=2$ orbital radius shrinks to half, so the old wavefunction is “too fat” and matches the new ground state’s shape only 70% of the way.

This also explains why the answer cares nothing for the beta electron’s energy or the neutrino’s direction — once the change is “fast enough”, the details of the fastness are wiped away entirely. The virtue of approximation methods reaches its peak here: not grinding through every detail, but seeing which details never enter the answer at all.

'Sudden' does not mean 'sloppy'

Three pitfalls that are easy to step in:

Using the wrong energy scale in the criterion. $\Delta E$ is the spacing of the levels you care about. The same nuclear decay may no longer count as sudden for the inner-shell electrons (keV scales, $10^{-19}$ -second beats) — in the beta decay of heavy atoms, inner-shell shake-off probabilities must be corrected beyond the zeroth-order sudden approximation.
Do not forget the conservation laws. “The wavefunction is unchanged” automatically respects symmetry (the vanishing p-state probabilities above are its gift), but if the switch itself carries away angular momentum or momentum (recoil, say), that part must be put into the final states explicitly.
The probabilities must balance the books. All the $P_n$ must sum to 1 (continuum included). The 4.8% “remainder” above is not an error but the genuine share of the higher excitations and the ionisation channel — throw it away as error, and the books will not balance.

Key formulas

Sudden approximation
$\ket{\Psi(0^+)}=\ket{\Psi(0^-)},\qquad P_m=\bigl|\braket{m^{\text{new}}}{\Psi(0^-)}\bigr|^2$
The wavefunction freezes; probability = squared overlap of old and new eigenstates
Criterion
$\tau\ll\frac{\hbar}{\Delta E}$
The two limits of the same ratio as the adiabatic criterion; the middle ground needs the full time-dependent equation
Tritium: new ground-state population
$P(1s)=\left(\frac{16\sqrt2}{27}\right)^2=\frac{512}{729}\approx0.702$
2s takes 1/4; p and d states strictly zero by angular-momentum conservation

Self-check4 questions

What comes next

Chapter 7’s inventory is complete: perturbation theory (static, degenerate, time-dependent), the variational method, WKB, the golden rule, adiabatic and sudden — in a world without exact solutions, these seven tools carry nearly every practical calculation.

But there is one account this chapter never dared examine closely. Helium appeared three times (perturbation in 7.1, variation in 7.3), and we kept calling its two electrons “electron 1” and “electron 2”, writing the wavefunction as $\psi(\vec r_1,\vec r_2)$ as if one wore a red hat and the other a blue one. But electrons have no hats — swap the two electrons, and no observable can possibly change. This seemingly harmless “indistinguishability” will react back on the mathematical structure of the wavefunction with astonishing force: it splits all many-particle states into two great camps, symmetric and antisymmetric, giving birth to the Pauli exclusion principle, the periodic table, the degeneracy pressure that holds up white dwarfs — and it will circle back to settle the final 1.5 eV this chapter still owes on the helium atom. Next chapter: identical particles.

The sudden approximation

Recommended first

After this section you should be able to

Too fast to react — which makes it easy

The picture

The mathematics

A worked example: where did tritium’s electron go?

Key formulas

Self-check4 questions

What comes next

Too fast to react — which makes it easy

◑The picture

∑The mathematics

A worked example: where did tritium’s electron go?

∎Key formulas

What comes next

The picture

The mathematics

Key formulas