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Dynamical Decoupling in Quantum Circuits

Dynamical decoupling (DD) uses control pulses to reduce unwanted evolution while a qubit stores or processes information. It is particularly useful when noise changes slowly enough for the control to reverse its accumulated effect. Real pulses also introduce errors, so the best sequence is the one that preserves the desired quantum operation under the available timing, calibration and noise conditions.

This page applies DD to idle intervals and computation: how to preserve the intended operation, fit pulses into a schedule and compare the resulting performance. The general dynamical-decoupling lesson develops the sequence and filter theory. DD changes the physical channel before measurement; it does not infer an ideal answer from noisy records or perform syndrome-based recovery.

Required background. Noise and mitigation distinguishes physical suppression from estimation. Dynamical decoupling introduces toggling frames and refocusing.

Helpful background. Pulse sequences describes sequence families; average Hamiltonian theory explains controlled evolution; variance and covariance supports the paired comparison below.

An idle means identity in a specified rotating or logical frame. Laboratory-frame precession may be intentional, and a virtual frame update changes the axes of later pulses. Always compare the entire controlled interval with its intended operation in the same frame.

Write an ideal rotation as Rx(θ)=e−iθX/2R_x(\theta)=e^{-i\theta X/2}. If pulses P1,…,PNP_1,\ldots,P_N occur in chronological order, their product acts right-to-left. For a protected idle, the pulse-only requirement is

PN⋯P1=eiϕI.P_N\cdots P_1=e^{i\phi}I.

Four Xπ=−iXX_\pi=-iX pulses give identity; two give −I-I. One uncompensated XπX_\pi pulse does not preserve an arbitrary input, even when it refocuses dephasing. An echo experiment can account for that known rotation in its readout or frame convention. The phase above is harmless only if it is global over the entire relevant space; different phases on leakage states or coherently controlled branches can matter.

The pulse product alone does not test evolution between or during pulses. For a gate, the complete product must implement the target gate, rather than merely end with an identity pulse cycle. A simple warning is a desired static ZZ rotation: the same balanced echo that removes a nuisance ZZ detuning also removes that rotation. Preserving useful dynamics while canceling errors requires appropriately designed control, not arbitrary insertion of an idle sequence.

The group-averaging formulation of Viola, Knill and Lloyd (1999) explains how control can suppress selected interaction terms. Higher-order effects and finite pulses require additional analysis; there is no universal sequence that cancels arbitrary noise without changing the task.

A pulse with center tjt_j and duration τj\tau_j occupies [tj−τj/2,tj+τj/2][t_j-\tau_j/2,t_j+\tau_j/2]. Check those edges against neighboring operations, waveform tails, minimum free gaps and shared drive or coupler resources. A center inside an idle is insufficient if part of its pulse lies outside.

For a concrete example, take a 1200 ns1200\,\mathrm{ns} idle, four requested centers (150,450,750,1050) ns(150,450,750,1050)\,\mathrm{ns}, a nearest 8 ns8\,\mathrm{ns} timing grid and 40 ns40\,\mathrm{ns} pulses:

Pulse Realized center (ns) Occupied interval (ns)
1 152 132–172
2 448 428–468
3 752 732–772
4 1048 1028–1068

The free gaps are (132,256,264,256,132) ns(132,256,264,256,132)\,\mathrm{ns}. They satisfy a 100 ns100\,\mathrm{ns} boundary guard and 250 ns250\,\mathrm{ns} minimum internal gap. The pulse duty fraction is 160/1200≃13.3%160/1200\simeq13.3\%. These geometric checks say that the schedule fits; they do not establish noise suppression.

In an instantaneous-pulse approximation, y(t)=(−1)n(t)y(t)=(-1)^{n(t)} changes sign at each pulse center. For static detuning ξ0\xi_0, the remaining phase is ξ0Y(0)\xi_0Y(0), where Y(0)=∫0Ty(t) dtY(0)=\int_0^T y(t)\,dt. The requested times give zero, but the rounded schedule gives

Y(0)=152−296+304−296+152=16 ns.\begin{aligned} Y(0)&=152-296+304-296+152\\ &=16\,\mathrm{ns}. \end{aligned}

Thus its normalized static residual is about 1.33%1.33\%. The timing grid has changed the cancellation even though the ideal pulse product and spacing tests still pass. Use the emitted times, including any inserted frame updates, when predicting behavior. Pulse-level control explains the schedule representation.

During a finite pulse, Uc†(t)ZUc(t)U_c^\dagger(t)ZU_c(t) generally rotates continuously and has several Pauli components. Replacing it by y(t)Zy(t)Z throughout the pulse is not exact. Integrate the control Hamiltonian, or state and justify a strong-control or negligible-pulse-duration approximation. Leakage and spectator disturbance need a larger model than this two-level one.

For Hnoise(t)=ξ(t)Z/2H_{\rm noise}(t)=\xi(t)Z/2 in units with ℏ=1\hbar=1, a zero-mean stationary classical Gaussian fluctuation gives coherence C=e−χC=e^{-\chi} under ideal sign modulation. Fix the spectral convention by

⟨ξ(t)ξ(0)⟩=∫−∞∞dω2πSξξ(ω)e−iωt,Y(ω,T)=∫0Ty(t)eiωt dt,χ=12∫−∞∞dω2πSξξ(ω)∣Y(ω,T)∣2.\begin{aligned} \langle\xi(t)\xi(0)\rangle &=\int_{-\infty}^{\infty}\frac{d\omega}{2\pi} S_{\xi\xi}(\omega)e^{-i\omega t},\\ Y(\omega,T)&=\int_0^T y(t)e^{i\omega t}\,dt,\\ \chi&=\frac12\int_{-\infty}^{\infty}\frac{d\omega}{2\pi} S_{\xi\xi}(\omega)|Y(\omega,T)|^2. \end{aligned}

Here ξ\xi has units s−1\mathrm{s}^{-1}, YY has units s\mathrm{s}, and SξξS_{\xi\xi} has units s−1\mathrm{s}^{-1}, making χ\chi dimensionless. The factor 1/21/2 comes from the Gaussian characteristic function. Non-Gaussian noise generally needs more than its two-point spectrum. Cywiński et al. (2008) study both Gaussian descriptions and strongly coupled fluctuators.

A low-frequency zero suppresses only the spectral contributions it actually overlaps. Moving weight away from low frequencies can move it toward another noise band. A high-order zero of a filter is also distinct from cancellation order in a Magnus expansion; Paz-Silva and Viola (2014) develop this distinction. Neither theoretical order alone ranks experimental performance.

If ⟨ξ(t)ξ(t′)⟩=2γδ(t−t′)\langle\xi(t)\xi(t')\rangle=2\gamma\delta(t-t'), then

χ=γ∫0Ty(t)2dt=γT.\chi=\gamma\int_0^T y(t)^2dt=\gamma T.

Ideal sign reversals cannot reduce this white-noise exposure. Equivalently, changing ZZ to −Z-Z leaves the Markovian dissipator unchanged. This statement assumes ideal instantaneous modulation and that same noise model; it does not cover every possible driven open system.

Unitary pulses also cannot reverse an unknown amplitude-damping jump. They may change the coupling or effective relaxation in a structured environment, but that needs an explicit physical model. To illustrate a remaining relaxation cost, suppose a phenomenological channel has transverse contraction ηC\sqrt\eta C and longitudinal contraction η=e−T/T1\eta=e^{-T/T_1}. Its average input-state fidelity is

Favg=3+η+2ηC6.F_{\rm avg}=\frac{3+\eta+2\sqrt\eta C}{6}.

At T=20 μsT=20\,\mu\mathrm{s} and T1=60 μsT_1=60\,\mu\mathrm{s}, changing CC from e−20/12e^{-20/12} to e−20/45e^{-20/45} raises this fidelity from .673.673 to .800.800. Even C=1C=1 gives only .902.902. This comparison assumes unchanged amplitude damping; it is not the channel automatically generated by applying XX pulses during relaxation. Such pulses can rotate damping into excitation in the toggling frame.

More pulses can improve environmental refocusing while worsening control error. Consider four instantaneous pulses, no environmental evolution, perfect axes and the same angle error ϵ\epsilon on every pulse. Four xx pulses implement a residual rotation by 4ϵ4\epsilon after removing their ideal identity. Their average infidelity is

rXXXX=1−cos⁡(4ϵ)3.r_{XXXX}=\frac{1-\cos(4\epsilon)}3.

For chronological axes x,y,x,yx,y,x,y, the product is

θ=π+ϵ,UXYXY=[Ry(θ)Rx(θ)]2.\begin{aligned} \theta&=\pi+\epsilon,\\ U_{XYXY}&=[R_y(\theta)R_x(\theta)]^2. \end{aligned}

It is −I-I at ϵ=0\epsilon=0. Let q=sin⁡(ϵ/2)q=\sin(\epsilon/2). Direct multiplication gives

rXYXY=83q4(1−q4).r_{XYXY}=\frac83q^4(1-q^4).

The leading errors are 8ϵ2/38\epsilon^2/3 and ϵ4/6\epsilon^4/6, respectively. At ϵ=.1\epsilon=.1 radians they are about .0263.0263 and 1.66×10−51.66\times10^{-5}. Both sequences have four pulses, but their coherent errors combine differently.

Average infidelity versus shared pulse-angle error, comparing four same-axis pulses with alternating x and y pulses on a logarithmic vertical scale

Exact two-level pulse products in the stated angle-error model. Alternating axes cancels the leading coherent error; the vertical axis is logarithmic. Environmental noise, detuning errors and finite pulse widths are absent, so the plot does not establish a universal sequence ranking.

A state aligned with the residual rotation axis can look perfectly preserved while other states fail. This is why a single favorable survival probability does not establish arbitrary-state protection. Souza, Álvarez and Suter (2011) and Wang et al. (2012) analyze pulse-error robustness in more realistic settings. Ezzell et al. (2023) also show why pulse spacing and device context matter in experimental comparisons.

Measure the quantity the application needs: coherence of a particular state, average or worst-state memory fidelity, or an algorithmic observable. A longer fitted T2T_2 is useful but does not establish all of these. Include leakage, loss and spectator effects; conditioning on surviving computational states can hide increased failure.

Interleave baseline and DD measurements in matched time blocks to reduce drift confounding. Select the sequence on separate data, then evaluate the chosen sequence on fresh states or circuits. With matched quality scores bi,dib_i,d_i, calculate differences δi=di−bi\delta_i=d_i-b_i and use their sample variance:

δ‾=1m∑iδi,SE⁡(δ‾)=sδm.\begin{aligned} \overline\delta&=\frac1m\sum_i\delta_i,\\ \operatorname{SE}(\overline\delta)&=\frac{s_\delta}{\sqrt m}. \end{aligned}

For baseline scores (.78,.80,.76,.81,.77,.79)(.78,.80,.76,.81,.77,.79) and DD scores (.88,.87,.83,.89,.80,.86)(.88,.87,.83,.89,.80,.86), the mean gain is .070.070 and its paired standard error is .00931.00931. Treating the records as independent gives .0158.0158. A Student-tt interval is approximately [.0461,.0939][.0461,.0939] if the six block differences are independent and approximately normally distributed. Pairing does not justify that distributional assumption or remove correlations between blocks.

Count calibration and tuning runs as well as final shots. If pulses lengthen the circuit, either compare with an equal-duration baseline or include the extra time in the resource comparison. Added drive power, occupied hardware and recalibration can outweigh a small gain. Keeping the original idle can be the better choice.

Combining DD with extrapolation or probabilistic cancellation changes the noise those estimators encounter. Check the actual combined circuits: scaling durations can change DD timing, and a cancellation model for an unprotected gate need not apply after protection. A readout response need only be recalibrated if it changes or its transfer is uncertain; physical suppression does not automatically alter every downstream detector parameter.

Compute the products of two and four ideal XπX_\pi pulses and of chronological Xπ,Yπ,Xπ,YπX_\pi,Y_\pi,X_\pi,Y_\pi. Why is a single echo pulse insufficient for an arbitrary-state identity?

Solution

The first two products are −I-I and II. The alternating product is YXYX=−IYXYX=-I, including the four factors of −i-i. Each is identity up to a global phase. A single pulse is −iX-iX, which swaps ∣0⟩|0\rangle and ∣1⟩|1\rangle; its known action must be compensated or incorporated in the intended final frame or measurement.

Find the free gaps, duty fraction and static moment from the timing table. Does symmetric placement around the midpoint force zero static moment?

Solution

Subtract successive pulse edges and the window boundaries to get (132,256,264,256,132) ns(132,256,264,256,132)\,\mathrm{ns}. Pulse occupancy is 4(40)=160 ns4(40)=160\,\mathrm{ns}, hence duty 2/152/15. Instantaneous signed center intervals give Y(0)=16 nsY(0)=16\,\mathrm{ns}. Reflection symmetry alone does not balance the total positive and negative durations for this even-pulse pattern.

A one-sided ordinary-frequency PSD obeys ⟨ξ2⟩=∫0∞Sf(1)(f)df\langle\xi^2\rangle=\int_0^\infty S_f^{(1)}(f)df. Express the two-sided angular-frequency PSD used above.

Solution

For a real stationary process,

Sξξ(ω)=12Sf(1)(∣ω∣/(2π)).S_{\xi\xi}(\omega)=\tfrac12S_f^{(1)}(|\omega|/(2\pi)).

The half accounts for positive and negative frequencies. Since dω/(2π)=dfd\omega/(2\pi)=df, no extra 2π2\pi factor appears in this convention. A PSD integrated with dωd\omega alone would have a different normalization.

Use Favg(U,I)=(∣Tr⁡U∣2+2)/6F_{\rm avg}(U,I)=(|\operatorname{Tr}U|^2+2)/6 to derive both curves. Explain why multiplying individual pulse fidelities fails.

Solution

The same-axis product has trace 2cos⁡(2ϵ)2\cos(2\epsilon), giving r=(1−cos⁡4ϵ)/3r=(1-\cos4\epsilon)/3. For the alternating product, let c=cos⁡((π+ϵ)/2)=−qc=\cos((\pi+\epsilon)/2)=-q; its trace is 2(2c4−1)2(2c^4-1). Substitution gives r=(8/3)q4(1−q4)r=(8/3)q^4(1-q^4). Coherent rotations compose as amplitudes with signs and axes, not independent failure probabilities. For NN equal same-axis small errors, the leading infidelity is N2ϵ2/6N^2\epsilon^2/6, rather than Nϵ2/6N\epsilon^2/6.

5. Distinguish dephasing suppression from relaxation

Section titled “5. Distinguish dephasing suppression from relaxation”

Derive the average fidelity of the phenomenological channel above. What prevents the value from reaching one when C=1C=1?

Solution

For Bloch action r↦Ar+c\boldsymbol r\mapsto A\boldsymbol r+\boldsymbol c, average pure-input fidelity is (3+Tr⁡A)/6(3+\operatorname{Tr}A)/6: spherical averaging gives ⟨rirj⟩=δij/3\langle r_ir_j\rangle=\delta_{ij}/3 and ⟨r⟩=0\langle\boldsymbol r\rangle=0. Here AA has diagonal entries (ηC,ηC,η)(\sqrt\eta C,\sqrt\eta C,\eta). Even with C=1C=1, η<1\eta<1 contracts the Bloch sphere. Removing pure dephasing has not undone damping.

Recompute the gain and standard error from the six pairs. What additional assumption enters the stated confidence interval?

Solution

The differences are (.10,.07,.07,.08,.03,.07)(.10,.07,.07,.08,.03,.07), with mean .07.07 and sample variance .00052.00052. Thus the standard error is .00052/6=.00931\sqrt{.00052/6}=.00931. Multiplying by t.975,5≃2.571t_{.975,5}\simeq2.571 gives half-width .0239.0239. The exact Student-tt construction assumes independent normal differences; six pairs alone do not establish normality. Correlated blocks or same-data sequence selection require another analysis or new measurements.

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