advanced30 min studyLesson 933 in Physics

Split-operator Fourier time evolution

Fix FFT signs and periodic modes before composing kinetic and potential propagators.

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What you’ll understand

  • Apply a unitary discrete Fourier convention on a four-point ring.
  • Distinguish spectral kinetic phases from finite-difference dispersion.
  • Explain splitting order, commutator error and aliasing limits.

Fourier propagation alternates between representations where different parts of the Hamiltonian are simple. Position-diagonal potential phases are applied on the spatial grid, while free kinetic phases are applied to Fourier coefficients. The transforms can be fast and each phase exactly norm preserving, but that does not eliminate splitting error, finite momentum bandwidth or periodic wraparound. A trustworthy calculation states its transform convention and tests each of those approximations separately.

Define the periodic discrete Fourier representation

For N equally spaced positions on a ring of length N times a, use a unitary forward transform with a negative exponential and an inverse with a positive exponential. For even N, label modes from zero upward and then use negative indices, assigning the Nyquist entry consistently. The momentum for mode m is two pi hbar m divided by the ring length. Reversing transform signs without relabeling momentum changes propagation and phase interpretation.

The kinetic operator in this Fourier representation uses the continuum quadratic momentum evaluated at the retained modes. It is not the centered finite-difference kinetic matrix from the first lesson, whose dispersion follows a cosine. Both define finite Hermitian models, but substituting one dispersion into the other's benchmark compares different systems. At a fixed grid, sufficiently high-frequency signals also alias into the retained band; increasing time accuracy alone cannot recover missing spatial information.

c~m=1N∑j=0N−1e−2πimj/Ncj,pm=2πℏmNa,c~m↦e−ipm2Δt/(2mparticleℏ)c~m\widetilde c_m=\frac1{\sqrt N}\sum_{j=0}^{N-1}e^{-2\pi imj/N}c_j,\qquad p_m=\frac{2\pi\hbar m}{Na},\qquad \widetilde c_m\mapsto e^{-ip_m^2\Delta t/(2m_{\rm particle}\hbar)}\widetilde c_m
The mode index m differs from the particle mass m-particle. For N four the ordered mode labels are zero, one, minus two, minus one; this sets a periodic spectral kinetic model.↔ Scroll wide equations horizontally.

Compose substeps with a stated order

For an autonomous Hamiltonian T plus V, a symmetric Strang step applies a potential half-step, a kinetic full step and another potential half-step. With finite bounded Hermitian matrices, its local error is third order in timestep and its accumulated error over a fixed smooth evolution interval is second order. Nested commutators determine the leading splitting error. If T and V commute, the split becomes exact apart from arithmetic and representation error.

A time-dependent potential requires an appropriate time-centered prescription to retain the desired order; merely evaluating every factor at the old time can lower it. A product of unitary factors still remains unitary even when it uses a poor approximation to time ordering. Adaptive comparisons should therefore examine the state or observable difference under step subdivision, not just norm drift. A small change in one occupation can also hide a relative-phase error in another observable.

US(Δt)=e−iVΔt/(2ℏ)e−iTΔt/ℏe−iVΔt/(2ℏ),US−e−i(T+V)Δt/ℏ=O(Δt3)U_S(\Delta t)=e^{-iV\Delta t/(2\hbar)}e^{-iT\Delta t/\hbar}e^{-iV\Delta t/(2\hbar)},\qquad U_S-e^{-i(T+V)\Delta t/\hbar}=O(\Delta t^3)
The order statement is for the declared autonomous finite-matrix setting. It is not a uniform error claim for arbitrary unbounded continuum operators or singular potentials.↔ Scroll wide equations horizontally.

Check wraparound, bandwidth and operator error independently

A Fourier grid is periodic even when the physical problem was intended to be an open line. A packet reaching the edge can reappear at the opposite edge. Enlarging the domain and using carefully tested absorption can delay or suppress that artifact, but a generic FFT does not provide an outgoing boundary condition. Potential multiplication can generate unresolved Fourier content, especially for abrupt features, so both bandwidth and the real-space sampling of the potential need convergence checks.

Commutator bounds provide a small-matrix diagnostic of operator splitting. For a first-order Lie product of finite Hermitian pieces, the operator-norm error is bounded by one half timestep squared over hbar squared times the commutator norm. The symmetric method has different nested-commutator constants, so its leading coefficient cannot be inferred from that first-order bound alone. These distinctions are useful when comparing methods rather than labeling every norm-preserving FFT routine exact quantum dynamics.

∥ULie−U∥≤Δt22ℏ2∥[T,V]∥for finite Hermitian T and V\|U_{\rm Lie}-U\|\le\frac{\Delta t^2}{2\hbar^2}\|[T,V]\|\qquad\text{for finite Hermitian T and V}
This is a unitary finite-matrix operator-norm bound. Halving the timestep quarters this local Lie bound; a leading third-order Strang local error instead scales by one eighth.↔ Scroll wide equations horizontally.

A free step on a four-point periodic grid

Take N four with initial coefficients (1,0,0,0)(1,0,0,0) and V zero. Use the declared spectral modes and choose timestep so the kinetic phase at modes plus or minus one is minus i. The Nyquist kinetic energy is four times theirs. Find the final coefficients and probabilities.

  1. The unitary forward transform gives every Fourier coefficient one half. The phases in index order are (1,−i,1,−i)(1,-i,1,-i) because the Nyquist phase is the fourth power of minus i, equal to one.

  2. Apply the positive-exponential inverse transform. The spatial result is ((1−i)/2,0,(1+i)/2,0)((1-i)/2,0,(1+i)/2,0). The two nonzero entries each have squared magnitude one half.

  3. The final probability sum is one, so this exact free spectral step preserves the finite norm. It is evolution on a four-point ring, with very limited momentum content, rather than free propagation on an unbounded line.

The result

The four-point state has probability one half at sites zero and two and zero at the other two sites.

Explore the model

Interactive lab

Interactive lab · modular

Periodic labels wrap around

Reset and set modulus four and additive step one; predict return to residue zero after four Steps. Compare a fresh Reset run at modulus four and additive step two; predict return after two Steps. The instrument advances one addition per half-second Step. This illustrates periodic indexing only; it computes no Fourier phases, wavefunction or kinetic propagator.

Periodic labels wrap aroundAddition modulo m starts at zero: rₙ = na mod m. Playback advances two additions per second. The orbit has period m/gcd(m,a), including period one for a = 0. This is not a cryptographic security test or a primality algorithm. Current readings: Residue r0: 0; Orbit period · m/gcd(m,a): 12; Distinct residues visited: 1 of 12. All parameters are adjustable with the labeled controls below.Repeated addition modulo 12: the current residue is 0, and the orbit period is 12.01234567891011+ 5mod 12An additive orbitn = 0rₙ = 0Period = 12Cycle starts at 0:0 → 5 → 10 → 3 →8 → 1 → 6 → 11 →4 → 9 → 2 → 7 → 0Exact residues · two additions per second
Additive orbitCurrent residue / next edge
Residue r0
0
Orbit period · m/gcd(m,a)
12
Distinct residues visited
1 of 12
t = 0 s

Addition modulo m starts at zero: rₙ = na mod m. Playback advances two additions per second. The orbit has period m/gcd(m,a), including period one for a = 0. This is not a cryptographic security test or a primality algorithm. 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 forA periodic representation has an explicit index domain, while modular addition is not quantum evolution.

Check your understanding

Why can a symmetric split-operator calculation preserve norm while accumulating an incorrect phase?

Try it yourself

Let T=ϵσxT=\epsilon\sigma_x, V=ϵσzV=\epsilon\sigma_z and ϵΔt/ℏ=0.1\epsilon\Delta t/\hbar=0.1. Compute the commutator norm and the displayed local Lie-product error bound. What happens to that bound when timestep is halved, and how would a leading Strang local error scale?

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 Pauli commutator and its unit operator norm. Keep the energy-squared units until multiplying by timestep squared over hbar squared.

Compare with the solution

The commutator is [T,V]=−2iϵ2σy[T,V]=-2i\epsilon^2\sigma_y, with norm 2ϵ22\epsilon^2. The bound is therefore (ϵΔt/ℏ)2=0.01(\epsilon\Delta t/\hbar)^2=0.01. Halving timestep gives bound 0.0025. A leading third-order Strang local error would decrease by a factor eight, provided its asymptotic regime applies. That scaling does not supply its unknown nested-commutator coefficient or address the finite Hilbert-space approximation.

Take it with you

  • FFT signs, mode ordering and periodic boundaries define the spectral model.
  • Unitary splitting can still accumulate commutator and time-order errors.
  • Temporal refinement does not recover aliased or omitted spatial modes.

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