advanced30 min studyLesson 937 in Physics

Time-dependent basis truncation and convergence

Control representation error with moving-basis connections, state residuals and bounded propagator approximations.

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What you’ll understand

  • Distinguish accurate propagation inside a basis from small error against the reference system.
  • Include the connection term of an orthonormal moving basis.
  • State the Hermitian spectral and residual domains behind Krylov and Chebyshev checks.

A finite basis is a physical approximation as well as a storage choice. An integrator may converge to machine precision within that basis while missing population transfer into excluded states. A time-dependent basis introduces an additional geometric term even before truncation error is considered. Reliable dynamics therefore checks both how well the represented coefficients are propagated and how well their reconstructed state satisfies the chosen reference Schrödinger equation.

Measure the equation defect in a specified reference space

For a fixed orthogonal projector P, replacing the reference Hamiltonian by P H P gives a Hermitian evolution within the retained subspace. That restricted evolution preserves its norm even when the full Hamiltonian couples retained and omitted states. Small final population near a selected cutoff is useful evidence but can miss virtual transitions or earlier excursions. Vary the basis and examine the target observables over the whole time interval rather than relying on normalization alone.

A reconstructed approximate state has a residual equal to i hbar times its derivative minus the reference H acting on it. For a finite Hermitian reference Hamiltonian, variation of constants and unitarity bound state error by initial error plus the integrated residual norm divided by hbar. In a continuum problem this reasoning additionally needs a self-adjoint operator, a common valid domain and a well-defined integrable residual. For a general non-Hermitian reference generator, the estimate needs the corresponding propagation-norm factor.

r(t)=iℏψ˙N−HψN,∥ψ(t)−ψN(t)∥≤∥ψ(0)−ψN(0)∥+1ℏ∫0t∥r(s)∥ dsr(t)=i\hbar\dot\psi_N-H\psi_N,\qquad \|\psi(t)-\psi_N(t)\|\le\|\psi(0)-\psi_N(0)\|+\frac1\hbar\int_0^t\|r(s)\|\,ds
The stated bound uses unitary reference propagation under the finite-Hermitian or qualified self-adjoint domain assumptions. The residual has energy units for dimensionless state coefficients.↔ Scroll wide equations horizontally.

Retain the connection of a moving orthonormal basis

Let columns of U be orthonormal retained basis vectors that vary with time, and write the reconstructed state as U times a coefficient vector. Substituting that representation into the Schrödinger equation and projecting gives the basis Hamiltonian plus minus i hbar times U-adjoint times its derivative. This connection term accounts for motion of the basis itself. Dropping it mistakes a coordinate rotation in state space for additional physical dynamics.

Differentiating orthonormality shows that U-adjoint times its derivative is anti-Hermitian, so the connection contribution is Hermitian. The resulting projected coefficient dynamics can preserve norm, while residual components orthogonal to the moving subspace still measure missing physics. Nonorthogonal bases require an overlap matrix and related derivative terms. A small coefficient integration tolerance cannot diagnose an incorrectly omitted connection or certify that the moving subspace follows all relevant transitions.

ψN=Uc,U†U=I,iℏc˙=(U†HU−iℏU†U˙)c\psi_N=Uc,\quad U^\dagger U=I,\qquad i\hbar\dot c=\left(U^\dagger HU-i\hbar U^\dagger\dot U\right)c
The moving-basis connection has energy units. For a truncated basis this is a projected equation, whose full-space residual need not vanish.↔ Scroll wide equations horizontally.

Check propagator subspaces and polynomial spectra

Krylov propagation approximates the action of an exponential in the span of the initial vector and successive Hamiltonian applications. Hermitian Lanczos reduction gives a small tridiagonal problem, with accuracy depending on subspace size, orthogonality and a suitable defect or error estimate. If the generated subspace closes exactly under H, the reduced action is exact for that initial vector. Otherwise norm preservation inside the reduced problem does not bound the omitted full-space dynamics.

Chebyshev propagation instead expands the exponential in polynomials of a shifted, scaled Hermitian Hamiltonian. Verified energy bounds must put its spectrum inside minus one to one; only then does the familiar polynomial norm bound support an omitted-coefficient-tail estimate. An underestimated spectral interval can create large polynomial growth. A single small last coefficient is not automatically a rigorous bound on the whole tail. Driven Hamiltonians also require treatment of their time dependence, separate from polynomial or Krylov approximation of a frozen operator.

H~=H−EcId,spec⁡(H~)⊆[−1,1], d>0 ⟹ ∥Tn(H~)∥≤1(H=H†)\widetilde H=\frac{H-E_cI}{d},\quad \operatorname{spec}(\widetilde H)\subseteq[-1,1],\ d>0\ \Longrightarrow\ \|T_n(\widetilde H)\|\le1\quad(H=H^\dagger)
E-c and d are energy center and half-width, not a spatial spacing. For this Hermitian spectral domain, a Chebyshev truncation error can be bounded using the full omitted coefficient tail.↔ Scroll wide equations horizontally.

An exactly closed two-vector Krylov space

Use H=ϵ(010101010)H=\epsilon\begin{pmatrix}0&1&0\\1&0&1\\0&1&0\end{pmatrix} and initial middle state (0,1,0)(0,1,0). Build its first two Krylov directions, show closure, and find probabilities when ϵt/ℏ=π/(22)\epsilon t/\hbar=\pi/(2\sqrt2).

  1. Applying H to the middle state gives epsilon times (1,0,1)(1,0,1). Normalize this second direction as v=(1,0,1)/2v=(1,0,1)/\sqrt2, orthogonal to the middle state.

  2. Applying H to v gives square-root-two epsilon times the middle state. The two directions therefore form an invariant subspace with reduced Hamiltonian 2ϵσx\sqrt2\epsilon\sigma_x.

  3. At the stated time its rotation angle in the exponential is pi over two, so the state is −iv=−i(1,0,1)/2-iv=-i(1,0,1)/\sqrt2. The end probabilities are one half each and middle probability zero. Closure makes this reduction exact for the specified initial state, not a general promise for every truncated Krylov calculation.

The result

The two-direction invariant Krylov space gives the exact three-state result: equal end populations and zero middle population at the specified time.

Explore the model

Interactive lab

Interactive lab · spectral

Adding retained modes changes a represented quantity

Reset and set mode count two and coefficient decay one; at time zero predict the sum at pi over two is one. Compare a fresh Reset run at mode count three and decay one; predict two thirds. The instrument is the fixed sine sum with common time phase, not a quantum basis, Krylov reduction, Chebyshev propagator or physical spectral flow.

Adding retained modes changes a represented quantityA finite sum of sin(nx − t)/nᵖ. Playback changes t; increasing the cutoff includes finer spatial scales. This is not a proof about a continuum limit or renormalization. Current readings: Sum at x = π/2: 0.867; Smallest wavelength: 2π / 5; Highest-mode coefficient: 0.2. All parameters are adjustable with the labeled controls below.01.573.144.716.28−2.56−1.2801.282.56x · radiansFinite sine sum
Finite sumHighest included mode
Sum at x = π/2
0.867
Smallest wavelength
2π / 5
Highest-mode coefficient
0.2
t = 0 s

A finite sum of sin(nx − t)/nᵖ. Playback changes t; increasing the cutoff includes finer spatial scales. This is not a proof about a continuum limit or renormalization. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forRepresentation size can change a computed quantity even when its finite formula is evaluated exactly.

Check your understanding

When does the displayed Chebyshev polynomial norm bound apply?

Try it yourself

Let the finite reference Hamiltonian be H=ϵσxH=\epsilon\sigma_x with initial state ∣0⟩|0\rangle. A one-state truncation keeps only this vector and predicts the stationary approximate state ∣0⟩|0\rangle. Find its residual norm and the bound at ϵt/ℏ=0.2\epsilon t/\hbar=0.2. Compare with exact state error and missed excited probability.

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

The reference Hamiltonian sends zero into one, even though its retained diagonal matrix element vanishes. Use the exact two-level cosine and sine.

Compare with the solution

The residual is minus epsilon times state one, with norm epsilon. Initial error is zero, so the integrated bound is 0.2. Exact evolution is cos⁡(0.2)∣0⟩−isin⁡(0.2)∣1⟩\cos(0.2)|0\rangle-i\sin(0.2)|1\rangle, whose distance from the approximate zero basis state is 2sin⁡(0.1)≈0.19972\sin(0.1)\approx0.1997. The omitted excited probability is sin⁡2(0.2)≈0.0395\sin^2(0.2)\approx0.0395. Both states have norm one; the reference-space residual detects the missing coupling that normalization cannot.

Take it with you

  • A basis-converged time integrator can still approximate the wrong reference dynamics.
  • Moving bases require their connection term.
  • Residual and polynomial bounds have explicit Hermitian, spectral and domain assumptions.

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