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The sudden approximation treats a rapid, finite change of a Hamiltonian operator as leaving the quantum state approximately unchanged during the switch. The energy measurement can nevertheless change immediately: the same quantum state has different probability amplitudes in the new eigenbasis. Compute their inner products with the new energy eigenstates, then apply the Born rule. The two-level sudden quench below makes this distinction quantitative without assuming the change is small.

An ideal finite step has exact quantum state continuity at a finite Hamiltonian step; a finite-duration implementation has a small evolution error under the bounded-model conditions below. An impulsive quantum kick is different: its Hamiltonian operator grows as its duration shrinks, retaining a finite integrated action and a finite change of quantum state. A sudden approximation therefore concerns short evolution during switching, not conservation of energy during externally driven change. The subsequent unitary time evolution can be substantial.^[ref]

The drive can supply or extract work while the quantum state stays approximately fixed. For a differentiable bounded Hamiltonian operator, the Schrödinger equation givesThe state-evolution terms cancel, leaving the external drive's power. For a finite ideal step the change in mean energy is $⟨\psi |H_f-H_i|\psi ⟩$. Thus conservation of energy for a closed system with a time-independent Hamiltonian operator does not assert conservation of the driven subsystem's energy during switching.

Specify work operationally by two energy measurements: one before the switch and one after. For the prepared initial ground state, the first outcome is definite; the second samples the new Born rule probabilities. This protocol gives a distribution of measured energy differences, rather than generally the measurement distribution of the single operator $H_f-H_i$.^[ref]and henceFor ${\Delta }_i={\Delta }_f=\Delta$ and $\theta =\pi /3$, the work is $0$ or $\Delta$, with probabilities $3/4$ and $1/4$; its expectation is $\Delta /4$ and its variance is $3{\Delta }^2/16$. The initial mean energy $-\Delta /2$ becomes $-\Delta /4$. Using different initial and final scalar offsets would also shift the measured work; the displayed Hamiltonian operators fix that convention explicitly.