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Tensor networks · A course with experiments

L4 · Chapter 8

Read Figure 3(a) of García-Ripoll's time-evolution paper

Open J. J. García-Ripoll, Time evolution of matrix product states, arXiv:cond-mat/0610210v1, at page 12. Figure 3(a) plots the error of two Trotter methods against the time step on logarithmic axes, for a chain small enough that nothing is truncated. The error is a squared norm, so each curve rises at twice the order of its method. An independent full-vector reproduction matches the second-order curve within 0.18 decade along its straight part. Along its straight part, the printed fourth-order curve lies about 3.7 decades below the Forest–Ruth formula the text prints, and instead matches an optimized fourth-order split from the paper's reference 28.

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Find the setup before the figure

Start at Section 5, Comparison. García-Ripoll evolves one initial state, every spin pointing along the same direction perpendicular to the z axis, under the spin chain of his Eq. 32: an exchange term weighted by the cosine of an angle and a field along z weighted by its sine. Figure 3 uses eight spins, an angle of 0.35, no z exchange, so the chain is an XY model, and a total time of ten.

Two choices decide how to read every curve. First, the error is the squared norm of the difference between the computed and the exact wave function. It is not the error of a magnetization or an energy. Second, the matrices have size sixteen, which holds every state of eight spins: the middle cut has four spins on each side and sixteen patterns. Nothing is truncated, so Figure 3 isolates the time-step error and the arithmetic of the computer.

H=∑k[cos⁡θ (skxsk+1x+skysk+1y+Δ skzsk+1z)+sin⁡θ skz],ε:=∥ψD(T)−U(T)ψ(0)∥2H=\sum_k\Big[\cos\theta\,\big(s^x_ks^x_{k+1}+s^y_ks^y_{k+1}+\Delta\,s^z_ks^z_{k+1}\big)+\sin\theta\,s^z_k\Big],\qquad \varepsilon:=\big\lVert\psi_D(T)-U(T)\psi(0)\big\rVert^2

Read panel (a): axes, curves and three regions

Both axes are logarithmic. The time step runs from 0.002 to 5 and the error from ten to the minus twenty up to one. The legend sits beside panel (c). In panel (a), blue is the symmetric second-order split of Eq. 21 and red the fourth-order Forest–Ruth formula of Eq. 22. Each colour holds two curves drawn on top of each other: a line with dots, labelled DMRG Trotter, and a plain line, labelled MPS Trotter. The legend tags blue with the time step squared and red with its fourth power. The solid black line is the Arnoldi and Lanczos reference that appears in every panel. A thin dashed curve that the legend does not list runs between blue and red; leave it aside.

Each curve has three regions. At the right, large steps saturate the error near one. In the middle the curve is a straight line, the power law of the method. At the left the error rises again as the step shrinks. The paper attributes this branch to the limited precision of the computer, growing like the inverse square of the step, because more steps accumulate more rounding. Read slopes only in the middle region.

Worked example

Measure the slope, then read the label

Measure the blue curve between steps of 0.01 and 0.1. It rises by about four decades. The label says the time step squared. Both are right, because the label names the order of the method while the plot shows a squared norm: if the error of the state scales as the p-th power of the step, its square scales as the 2p-th power. Vidal's Eq. 24 states the same rule, a neglected correction scaling as the step to the power 2p times the square of the total time.

An independent check reproduces the curve. Multiply the exact two-site gates of Eq. 21 on a full vector of eight spins, which equals the paper's untruncated matrix product state run. The fitted slope between 0.01 and 0.1 is 4.0003, and a separate SciPy rebuild agrees to better than one part in a billion. The figure is stored in the PDF as vector lines, so every printed point can be read from the file instead of from pixels. From steps of 0.01 to 2.5 the printed blue curve lies 0.10 to 0.18 decade below the reproduction, a factor of about 1.36, with the same slope.

∥δψ∥∝Δt p ⇒ ε=∥δψ∥2∝Δt 2p,p=2: slope 4 (measured 4.0003)\lVert\delta\psi\rVert\propto\Delta t^{\,p}\ \Rightarrow\ \varepsilon=\lVert\delta\psi\rVert^2\propto\Delta t^{\,2p},\qquad p=2:\ \text{slope }4\ (\text{measured }4.0003)

Which fourth-order formula the red curve shows

Run Eq. 22 the same way. A least-squares fit over steps of 0.01, 0.02, 0.05 and 0.1 gives a slope of 7.998, so the state error is fourth order, 3.999, as the formula promises. Yet at the five printed points between steps of 0.11 and 0.5 the red curve lies 3.5 to 3.8 decades below this result. The second-order curve comes from the same code, chain, initial state and error measure and agrees within 0.18 decade, so the gap is not in the setup.

Eq. 22 cites references 8 and 28, and reference 28 by Omelyan, Mryglod and Folk is the place to look. They add one pair of exponentials to the Forest–Ruth split and choose its coefficients to make the leading error small. Their Eq. 17 lists the coefficients, and the error constant drops about 42 times. Run their Eq. 9 with those coefficients, odd bonds outermost, on the same chain: it lands within 0.17 decade of all five printed points. García-Ripoll's text prints Eq. 22 and does not say which split made the figure, so this match is our inference, not his statement.

Two parts stay open. Above a step of 0.5 the printed curve climbs 7.2 decades by a step of 2, against 3.8 for Eq. 22 and 4.9 for the optimized split, and we cannot say why. Below 0.08 it rises again as the step shrinks. The paper blames the finite precision of its optimization and gives a worst-case estimate, the square of the number of steps times ten to the minus sixteen; the printed branch sits 1.3 to 1.7 decades under that bound. A full-vector reproduction has no optimization step, so it keeps falling past ten to the minus twenty.

θ=12−21/3≈1.3512,γFRγEFRL≈0.0390.00092≈42\theta=\frac{1}{2-2^{1/3}}\approx 1.3512,\qquad \frac{\gamma_{\mathrm{FR}}}{\gamma_{\mathrm{EFRL}}}\approx\frac{0.039}{0.00092}\approx 42

A checklist for the next error plot

Before reading slopes, find out what the error measures: a norm, a squared norm, a fidelity or one observable. Check whether truncation is active by comparing the bond dimension with the largest Schmidt rank the chain can have. Read slopes only where the curve is straight, and compare them with twice the method order when the error is squared.

Then note where the curve bends up at small steps and what the authors blame for it. If you reproduce a figure, compare absolute values as well as slopes. A curve with the right slope but the wrong height can come from the coefficients of the method, so open the references cited next to the formula. Say which curves you matched and which you did not. Then open the quench Lab, choose the coarse time-step preset, and see the same time-step error appear while the discarded weight stays small.

An experiment to think through

Predict, change, explain

  1. In the quench Lab, predict how the largest local error at a fixed time changes when you halve the time step, given that the Lab uses a second-order split.
  2. Run the coarse time-step preset, then lower the time step in two halvings.
  3. The Lab error is a difference of local magnetizations. Explain why it falls by a factor of four each time the step is halved, while the squared norm in Figure 3(a) falls by sixteen.
Follow a quench until the simulation stops being trusted

The Labs simulate open spin-1/2 chains with energies in units of the coupling and time in units of the reduced Planck constant divided by the coupling. Each Lab states its model and limits in its notes and can export the data behind its readouts.

Try explaining it yourself

1. Two fourth-order splits differ by a factor of 42 in their leading error constant. Roughly how far apart should their curves sit in Figure 3(a), and what slope should both have?

Read an explained answer

The state error scales with the constant, and the plotted squared norm with its square, so about 1800 times or 3.2 decades. Both rise at twice the order, eight decades per decade of step. On this chain the measured gap is 3.7 decades, because the constant of Omelyan and coauthors is a norm of the coefficients of several nested commutators, and this chain does not weight those commutators equally.

2. Why does a matrix size of sixteen mean no truncation for eight spins?

Read an explained answer

The largest Schmidt rank of any cut is the number of patterns on the smaller side. The middle cut has four spins per side, sixteen patterns, so sixteen states per link hold every state exactly.

3. Can Figure 3 tell you how truncation errors grow during a long run?

Read an explained answer

No. Truncation is switched off by construction. The paper turns to sixteen spins with several matrix sizes in Figure 4 to show that effect.

Sources to read alongside

  1. G. Vidal, Efficient simulation of one-dimensional quantum many-body systems, Physical Review Letters 93, 040502 (2004)

    Introduces the time-evolving block decimation: split the Hamiltonian into two-site gates, apply them and keep the largest Schmidt coefficients after each gate.

  2. J. J. García-Ripoll, Time evolution of matrix product states, New Journal of Physics 8, 305 (2006)

    Compares several ways of evolving matrix product states in time. Figure 3 plots the error of Trotter-type integrators against the time step on log–log axes.

  3. I. P. Omelyan, I. M. Mryglod and R. Folk, Optimized Forest–Ruth- and Suzuki-like algorithms for integration of motion in many-body systems, Computer Physics Communications 146, 188 (2002)

    Reference 28 of García-Ripoll. Adds one pair of exponentials to the Forest–Ruth split and chooses its coefficients to shrink the leading error. Equation 17 gives the coefficients; the error constant falls about 42 times.

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