Lanczos–Krylov spectra and dynamics
Construct normalized Krylov recurrences and distinguish eigenpair convergence from controlled time propagation.
On this page
The question for today
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 , the m-dimensional Krylov space spans . 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.
Project spectra and check actual residuals
Collect the basis vectors as columns of Q and form . 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.
Time propagation introduces a second convergence variable
For a state initially equal to , 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.
Work it through
A two-step Lanczos calculation
Use and , with . Construct Lanczos through termination and find the Ritz energies.
The first expectation is . Subtracting it from leaves , so and .
The second expectation is . Subtracting both the current diagonal component and leaves zero, hence . The two-dimensional Krylov space is now invariant.
The projected matrix equals the original one. Its characteristic equation is , giving in units of . Both lifted eigenvectors have zero recurrence residual.
Lanczos terminates after two directions and gives exact finite-matrix energies (2 ± √2)E_*.
The living notebook
Explore the model
Interactive lab · vectors
Orthogonality as a geometric prerequisite
Reset and set angle , magnitude , then repeat from Reset with both controls , . 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.
- x component
- 1.393
- y component
- 0.975
- Magnitude
- 1.7
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.
Pause & reason
Check your understanding
Your turn
Try it yourself
Let and start from . A one-dimensional Krylov approximation has . Compare its prediction with exact evolution at , 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 , 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 , and including the second direction makes this two-state evolution exact.
Keep the idea
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.
Read further
Original sources and deeper explanations to follow at your own pace.
Optional progress, saved in this browser when available. Revisit any lesson at any time.
Lesson not marked complete.