research28 min studyLesson 562 in Physics

Lanczos–Krylov spectra and dynamics

Construct normalized Krylov recurrences and distinguish eigenpair convergence from controlled time propagation.

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What you’ll understand

  • Derive the normalized Lanczos three-term recurrence.
  • Evaluate a Ritz-pair residual from the final recurrence coefficient.
  • Explain why spectral broadening and time-propagation convergence need their own checks.

A sparse Hamiltonian becomes useful when an algorithm can extract physical information through repeated actions on a vector. Lanczos builds an orthonormal basis from such actions, replacing a large Hermitian problem by a small tridiagonal projection. Its early success for extremal energies does not imply that the entire spectrum or every long-time observable has converged. Starting-vector overlap, numerical orthogonality, projection dimension, and time interval each control a different part of the calculation.

Normalize every new Krylov direction

Starting from a unit vector q1q_1, the m-dimensional Krylov space spans q1,Hq1,…,Hm−1q1q_1,Hq_1,\ldots,H^{m-1}q_1. Those powers usually become nearly parallel, so Lanczos constructs orthonormal directions instead. At step j, subtract components along the current and previous vectors, then normalize the remainder. For Hermitian H, exact arithmetic makes the remaining earlier overlaps vanish automatically. The diagonal coefficient is a Rayleigh expectation, while the nonnegative off-diagonal coefficient is a norm.

A zero remainder means the generated space is invariant under H, so recurrence should terminate rather than divide by zero. A small floating-point remainder can instead amplify roundoff when normalized. Loss of orthogonality can produce repeated approximations to an already found eigenvalue, sometimes called ghosts. Reorthogonalization and independent residual checks help distinguish such numerical repetitions from a genuine degenerate eigenspace. One starting vector samples only its projection into a degenerate subspace and cannot reveal every independent vector there.

rj=Hqj−αjqj−βj−1qj−1,αj=qj†Hqj,βj=∥rj∥2,qj+1=rj/βjr_j=Hq_j-\alpha_jq_j-\beta_{j-1}q_{j-1},\quad \alpha_j=q_j^\dagger Hq_j,\quad \beta_j=\|r_j\|_2,\quad q_{j+1}=r_j/\beta_j
Set beta_0 = 0. The q vectors have unit norm and alpha, beta have energy units. The normalized update is used only when beta_j is nonzero.↔ Scroll wide equations horizontally.

Project spectra and check actual residuals

Collect the basis vectors as columns of Q and form Tm=Q†HQT_m=Q^\dagger HQ. Its eigenvalues are Ritz values, and lifting a normalized small eigenvector y gives the approximation Qy. In exact orthogonality, the final Lanczos coefficient and the last component of y determine the full-space residual norm. In finite precision, applying H again to the recovered vector is the safer verification. A lowest Ritz value remains variationally above the true lowest energy in the searched invariant sector.

A spectral function uses an operator-excited starting vector, whose norm carries the total spectral weight. Renormalizing that vector for the recurrence must not erase its prefactor from the final correlation function. Finite lattices have discrete spectral poles. Adding an imaginary broadening parameter makes a smooth plot but does not create a physical lifetime. Converge Krylov dimension first, then examine system size and artificial broadening to decide which features are supported by the calculation.

HQm=QmTm+βmqm+1emT,∥H(Qmy)−θQmy∥2=βm∣emTy∣HQ_m=Q_mT_m+\beta_mq_{m+1}e_m^T,\qquad \|H(Q_my)-\theta Q_my\|_2=\beta_m|e_m^Ty|
The residual identity assumes orthonormal Q_m and a normalized Ritz eigenvector y of T_m with eigenvalue theta. e_m selects its final component; the residual has energy units.↔ Scroll wide equations horizontally.

Time propagation introduces a second convergence variable

For a state initially equal to q1q_1, projected real-time propagation evaluates a small matrix exponential and maps its coefficients back to the original space. This preserves norm in an orthonormal basis because the projected Hamiltonian is Hermitian. Norm preservation is necessary but insufficient: a tiny Krylov space can evolve unitarily within itself while missing important physical amplitudes outside it. Increasing the subspace dimension or reducing the time interval tests that error.

For long calculations, rebuild the Krylov space after short propagation intervals and control accumulated error, rather than relying on one fixed space indefinitely. Real time uses an imaginary exponential and rotates phases. Imaginary time uses a decaying exponential and, after normalization, filters higher-energy contributions if the initial state overlaps the desired ground sector. These evolutions have different numerical and physical roles. A geometric vector display can illustrate orthogonality but cannot generate a Krylov basis, a spectral function, or either many-body exponential.

∣ψ(t)⟩≈Qme−iTmt/ℏe1,∣ψ(τ)⟩∝e−Hτ∣ψ(0)⟩|\psi(t)\rangle\approx Q_m e^{-iT_mt/\hbar}e_1,\qquad |\psi(\tau)\rangle\propto e^{-H\tau}|\psi(0)\rangle
Real time t is in seconds if T_m is in joules. Imaginary parameter tau is defined here in inverse-energy units; the second expression must be normalized when used as a state.↔ Scroll wide equations horizontally.

A two-step Lanczos calculation

Use H=E∗(1113)H=E_*\begin{pmatrix}1&1\\1&3\end{pmatrix} and q1=(1,0)Tq_1=(1,0)^T, with E∗>0E_*>0. Construct Lanczos through termination and find the Ritz energies.

  1. The first expectation is α1=E∗\alpha_1=E_*. Subtracting it from Hq1Hq_1 leaves r1=E∗(0,1)Tr_1=E_*(0,1)^T, so β1=E∗\beta_1=E_* and q2=(0,1)Tq_2=(0,1)^T.

  2. The second expectation is α2=3E∗\alpha_2=3E_*. Subtracting both the current diagonal component and β1q1\beta_1q_1 leaves zero, hence β2=0\beta_2=0. The two-dimensional Krylov space is now invariant.

  3. The projected matrix equals the original one. Its characteristic equation is (1−x)(3−x)−1=0(1-x)(3-x)-1=0, giving x=2±2x=2\pm\sqrt2 in units of E∗E_*. Both lifted eigenvectors have zero recurrence residual.

The result

Lanczos terminates after two directions and gives exact finite-matrix energies (2 ± √2)E_*.

Explore the model

Interactive lab

Interactive lab · vectors

Orthogonality as a geometric prerequisite

Reset and set angle 0∘0^\circ, magnitude 11, then repeat from Reset with both controls 90∘90^\circ, 11. Predict x components one and zero, respectively. The two unit directions are orthogonal. This two-dimensional projection instrument does not orthogonalize Krylov vectors or diagonalize a Hamiltonian.

Orthogonality as a geometric prerequisitePlayback traces a fixed vector and its projections. Angle and magnitude set the geometry; the moving marker does not change them. Current readings: x component: 1.393; y component: 0.975; Magnitude: 1.7. All parameters are adjustable with the labeled controls below.−7.46−3.7303.737.46−3.5−1.7501.753.5xyvvₓ = 1.39
Vectorx / y projectionsTrace marker
x component
1.393
y component
0.975
Magnitude
1.7
t = 0 s

Playback traces a fixed vector and its projections. Angle and magnitude set the geometry; the moving marker does not change them. 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 forAn orthonormal basis is a computational structure that must be maintained, not assumed from a small residual alone.

Check your understanding

In exact arithmetic, what does a zero Lanczos remainder at step m imply?

Try it yourself

Let H=E∗σxH=E_*\sigma_x and start from (1,0)T(1,0)^T. A one-dimensional Krylov approximation has T1=0T_1=0. Compare its prediction with exact evolution at E∗t/ℏ=π/4E_*t/\hbar=\pi/4, including populations and norm.

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

Give me a hint

Use the exact exponential of a Pauli matrix and distinguish norm preservation from propagation accuracy.

Compare with the solution

The one-dimensional approximation leaves the state unchanged, predicting populations one and zero. Exact evolution is (cos⁡(π/4),−isin⁡(π/4))T(\cos(\pi/4),-i\sin(\pi/4))^T, so both populations are one half. Both states have unit norm, demonstrating that norm conservation alone does not certify accuracy. The omitted first recurrence coupling is β1=E∗\beta_1=E_*, and including the second direction makes this two-state evolution exact.

Take it with you

  • Lanczos coefficients require normalized orthogonal directions and nonzero-remainder checks.
  • A Ritz residual concerns the finite Hamiltonian and must survive direct recomputation.
  • Time propagation, spectral weights, and artificial broadening each need convergence beyond a low-energy Ritz value.

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