Quantum state continuity at a finite Hamiltonian step ($|\psi (0^{+})⟩=|\psi (0^{-})⟩$)

For a fixed Hilbert space, a bounded self-adjoint Hamiltonian operator with a finite step $H_i\to H_f$ at $t=0$ does not deliver an instantaneous rotation of the quantum state. Integrating the Schrödinger equation across a short interval givesFor a normalized quantum state and $\Vert H(t)\Vert \le M$ in operator norm, the difference has norm at most $2ϵM/\hslash$, which tends to zero. The time derivative can jump even though the quantum state is continuous. Here $\hslash$ is the reduced Planck constant. An instantaneous replacement of the eigenbasis changes coordinates, not the vector being represented.

For a real switch occupying $0\le t\le \tau$, remove any real scalar contribution $a(t)I$ that produces only a global phase, and write $|\tilde{\psi }(t)⟩=e^{i{\int }_0^{t}a(s)ds/\hslash }|\psi (t)⟩$. The same integral argument gives the sufficient bounded-model estimateThus $\epsilon \ll 1$ controls the sudden approximation up to a global phase. For any normalized final energy eigenstate, its probability amplitude differs from the instantaneous-switch value by at most $\epsilon$, and its Born rule probability differs by at most $2\epsilon$. These are sufficient error bounds, not necessary criteria. For the fixed-gap two-level sudden quench, any switching path $H(t)=-\Delta \mathbf{n}(t)\cdot \mathbf{\sigma }/2$ with a unit vector $\mathbf{n}(t)$ has operator norm $\Delta /2$, giving $\epsilon \le \Delta \tau /(2\hslash )$.

This estimate is deliberately for bounded Hamiltonian operators. An unbounded energy spectrum does not supply a finite global $M$; relevant quantum states, operator domains and excited energy eigenvalues require separate control. The adiabatic theorem also requires hypotheses about the path and its spectral gaps; there is no universal switch-duration criterion covering all spectra. The cold-atom collapse and revival after a lattice quench illustrates how a single switch can be sudden for one degree of freedom and slow for another.

An impulsive quantum kick is the limit of a pulse with finite integrated Hamiltonian operator, rather than a bounded finite step. For a bounded self-adjoint operator $K$ with units of action, a rectangular pulse $H_{\tau }=K/\tau$ lasting $\tau$ has exact unitary time evolution $e^{-iK/\hslash }$ for every $\tau$. In the limiting notation $H(t)=K\delta (t)$, $\delta$ is the Dirac delta function andThe integral estimate for quantum state continuity at a finite Hamiltonian step no longer tends to zero: the pulse height diverges. A bounded background Hamiltonian operator contributes vanishing evolution during this limit. A more general pulse with noncommuting generators can require time ordering; integrated area alone then need not specify the result.

For example, $K=\hslash \varphi {\sigma }_y/2$, with a $y$ Pauli matrix, sends a qubit in $|0⟩$ to $\cos (\varphi /2)|0⟩+\sin (\varphi /2)|1⟩$. At $\varphi =\pi$, the Born rule predicts $|1⟩$ with probability one, however short the regularized pulse is. Its finite rotation is precisely what a bounded finite step cannot accomplish at the switching instant.

A Hermitian impulse $U\delta (t)$ with zero diagonal has kick operator $e^{-iU/\hslash }$. If $u=U_{21}$, its transition probability is ${\sin }^2(|u|/\hslash )$, reducing to $|u{|}^2/{\hslash }^2$ for a weak impulse. The integrated operator has action units, and a pulse integrated across its full support has no extra half-delta factor.