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Device characterization uses experiments to learn a predictive model of a quantum device. A Ramsey trace may reveal detuning; measurements with a neighboring qubit in different states may reveal a coupling. The central question is whether the chosen experiments distinguish the proposed explanations. A precise fit is useful only within a model that predicts other observations.

Required background. Rabi and Ramsey control introduces driven-qubit experiments; quantum noise distinguishes channels and fluctuations. Process tomography explains channel reconstruction and randomized benchmarking explains ensemble performance measurements.

Identifying a frequency from Ramsey measurements

Section titled “Identifying a frequency from Ramsey measurements”

Prepare ∣+⟩|+\rangle, evolve under H=ℏΔZ/2H=\hbar\Delta Z/2, and measure along XX or YY. Here Δ\Delta is angular-frequency detuning relative to the control reference. With ideal axes, a visibility VV, an initial phase ϕ\phi and an exponential envelope, the probabilities of the positive outcomes are

pX(t)=1+Ve−γtcos⁡θ(t)2,pY(t)=1+Ve−γtsin⁡θ(t)2,θ(t)=Δt+ϕ.\begin{aligned} p_X(t)&=\frac{1+Ve^{-\gamma t}\cos\theta(t)}2,\\ p_Y(t)&=\frac{1+Ve^{-\gamma t}\sin\theta(t)}2,\\ \theta(t)&=\Delta t+\phi. \end{aligned}

These signs specify the analysis convention; a pulse-based implementation must translate its outcome labels accordingly. The exponential is an assumption to test, not part of the definition of Ramsey interferometry.

Measuring only XX cannot distinguish (Δ,ϕ)(\Delta,\phi) from (−Δ,−ϕ)(-\Delta,-\phi). Adding YY distinguishes the phase direction. However, one evolution time still determines only Δt+ϕ\Delta t+\phi: measuring more quadratures at that same time cannot separate an unknown detuning from an unknown initial phase.

Multiple times help, but the time grid matters. At t=mτt=m\tau, every integer kk gives the same two-quadrature data for

Δ′=Δ+2πkτ.\Delta' =\Delta+\frac{2\pi k}{\tau}.

This is frequency aliasing. More shots at those same times cannot remove it. A justified prior frequency interval or additional times that distinguish the candidate aliases are needed.

Two Ramsey curves with frequencies differing by one cycle per sampling interval agree at all integer sampling times but differ at an added quarter-interval measurement.

With V=1V=1, γ=ϕ=0\gamma=\phi=0, frequencies fτ=0.2f\tau=0.2 and 1.21.2 give identical XX probabilities at every integer t/τt/\tau. The same alias survives YY measurements. An added time t/τ=1/4t/\tau=1/4 separates these two candidates, although a broader frequency range can contain other aliases.

The fitted visibility can also be ambiguous. If preparation has transverse Bloch length cc and symmetric readout has contrast vv, these experiments see only V=cvV=cv. Pairs (c,v)=(0.8,1)(c,v)=(0.8,1) and (1,0.8)(1,0.8) predict the same record. An independent preparation or detector measurement is needed to attribute the contrast loss. Degen, Reinhard and Cappellaro discuss how experimental design converts such phase information into parameter estimates.

For independent shots at fixed settings, the number of positive outcomes is binomial:

nx∼Binomial⁡(Nx,pθ(x)).n_x\sim\operatorname{Binomial}(N_x,p_\theta(x)).

The log likelihood, apart from constants, is

ℓ(θ)=∑xnxlog⁡pθ(x)+∑x(Nx−nx)log⁡(1−pθ(x)).\ell(\theta)=\sum_x n_x\log p_\theta(x) +\sum_x(N_x-n_x)\log(1-p_\theta(x)).

A likelihood fit respects the probability-dependent sampling variance. Ordinary least squares with constant error bars can be inappropriate near probabilities zero or one. Correlated shots, drifting parameters or uncertain calibration need an expanded likelihood or a valid analysis of correlated data.

The Jacobian Jxa=∂pθ(x)/∂θaJ_{xa}=\partial p_\theta(x)/\partial\theta_a describes local sensitivity. In a regular model its rank identifies the parameter combinations accessible to first order. A singular Jacobian need not prove global nonidentifiability in every nonlinear model; it can also mark a special point with vanishing first-order sensitivity. Conversely, full rank near one fit does not exclude distant aliases. Confidence regions should retain multiple plausible solutions when the experiment cannot distinguish them.

Predictions reveal different failures depending on what is held out:

New data What it tests What it does not establish
More shots of the same circuits Sampling predictions in those settings New-circuit behavior
New lengths or pulse orders Composition and history dependence Stability at a later date
Different spectator states Transfer between contexts Every many-qubit configuration
A later acquisition period Stability over that period Permanent stationarity

Compare predictions with the actual counts. For example, a model predicting p=0.50p=0.50 with N=1000N=1000 gives a shot-noise scale 0.25/1000≈0.0158\sqrt{0.25/1000}\approx0.0158 for the observed frequency. A residual of 0.100.10 is about 6.326.32 such scales. It suggests a failed prediction if the data were held out and independent; it is not automatically a universal significance calculation after fitting many parameters or searching many residuals.

A constant unwanted rotation Uϵ=e−iϵZ/2U_\epsilon=e^{-i\epsilon Z/2} has average infidelity

r=23sin⁡2(ϵ/2)≈ϵ26.r=\frac23\sin^2(\epsilon/2)\approx\frac{\epsilon^2}{6}.

The same commuting error repeated mm times rotates by mϵm\epsilon. Phase-sensitive sequences can therefore reveal a signed error hidden by a small one-gate infidelity. Target/inverse sequences must be chosen carefully: some error terms cancel rather than amplify.

A trace-preserving qubit channel acts affinely on Bloch vectors:

r′=Tr+t.\boldsymbol r'=T\boldsymbol r+\boldsymbol t.

It is unital precisely when t=0\boldsymbol t=0. Unitary evolution rotates the Bloch ball; nonunitary evolution can contract and translate it. Amplitude damping can even increase the purity of some inputs by moving them toward the ground state. Purity loss is therefore not a universal definition of irreversible noise.

Wallman and colleagues define unitarity using the traceless transfer block. For a qubit,

u=13Tr⁡(TTT).u=\frac13\operatorname{Tr}(T^{\mathsf T}T).

A unitary channel has u=1u=1; a depolarizing channel T=pIT=pI has u=p2u=p^2. A complete reset to ∣0⟩|0\rangle has pure outputs but T=0T=0, hence u=0u=0. This removes the misleading contribution of nonunital translation. Unitarity adds information to average fidelity, but does not specify a unique microscopic mechanism or a literal fraction of coherent errors.

A fitted contraction must still be physical

Section titled “A fitted contraction must still be physical”

For a Pauli-diagonal unital qubit map with diagonal entries (λx,λy,λz)(\lambda_x,\lambda_y,\lambda_z), the four Pauli probabilities are

pI=(1+λx+λy+λz)/4,pX=(1+λx−λy−λz)/4,pY=(1−λx+λy−λz)/4,pZ=(1−λx−λy+λz)/4.\begin{aligned} p_I&=(1+\lambda_x+\lambda_y+\lambda_z)/4,\\ p_X&=(1+\lambda_x-\lambda_y-\lambda_z)/4,\\ p_Y&=(1-\lambda_x+\lambda_y-\lambda_z)/4,\\ p_Z&=(1-\lambda_x-\lambda_y+\lambda_z)/4. \end{aligned}

Complete positivity requires all four to be nonnegative. The tempting fit (0.98,0.98,0.90)(0.98,0.98,0.90) fails: it gives pZ=−0.015p_Z=-0.015. Contracting every Bloch coordinate is insufficient for a physical channel acting on part of an entangled system. A constrained fit must accommodate statistical and model uncertainty; one cannot interpret an unphysical fitted matrix as a mechanism.

For stationary two-level Markovian relaxation and pure dephasing,

1T2=12T1+1Tϕ.\frac1{T_2}=\frac1{2T_1}+\frac1{T_\phi}.

Here T1T_1 describes population relaxation and TϕT_\phi a nonnegative pure-dephasing rate. Ramsey, echo and driven decay constants generally probe different filters and need not satisfy this model when combined indiscriminately. An inferred negative dephasing rate calls for uncertainty and model checks.

To fix normalization in noise spectroscopy, take

Hβ(t)=ℏ2β(t)ZH_\beta(t)=\frac\hbar2\beta(t)Z

with stationary zero-mean classical Gaussian noise. Define its two-sided spectrum by

Cβ(t)=E[β(t)β(0)],Sβ(ω)=∫−∞∞Cβ(t)e−iωt dt.\begin{aligned} C_\beta(t)&=\mathbb E[\beta(t)\beta(0)],\\ S_\beta(\omega)&=\int_{-\infty}^{\infty}C_\beta(t)e^{-i\omega t}\,dt. \end{aligned}

Ideal control gives a modulation y(t)=±1y(t)=\pm1 and accumulated phase φ=∫0Ty(t)β(t)dt\varphi=\int_0^T y(t)\beta(t)dt. Gaussian averaging yields

⟨e−iφ⟩=e−χ(T),χ(T)=Var⁡φ2=14π∫−∞∞Sβ(ω)∣Y(ω)∣2dω,Y(ω)=∫0Ty(t)eiωtdt.\begin{aligned} \langle e^{-i\varphi}\rangle&=e^{-\chi(T)},\\ \chi(T)&=\frac{\operatorname{Var}\varphi}{2}\\ &=\frac1{4\pi}\int_{-\infty}^{\infty}S_\beta(\omega)|Y(\omega)|^2d\omega,\\ Y(\omega)&=\int_0^T y(t)e^{i\omega t}dt. \end{aligned}

The factor 1/(4π)1/(4\pi) belongs to these definitions; other spectrum and Hamiltonian conventions change prefactors. For Cβ(t)=2Dδ(t)C_\beta(t)=2D\delta(t), the result is χ=DT\chi=DT. Ideal sign flips do not suppress this white-noise dephasing. By contrast, a constant random detuning with variance σ2\sigma^2 gives Ramsey χ=σ2T2/2\chi=\sigma^2T^2/2 and is canceled by an ideal midpoint echo. Szańkowski and colleagues explain the assumptions and limitations of filter-based spectroscopy. Finite pulses, non-Gaussian noise and quantum environments require additional treatment.

Ordinary process tomography assumes known preparation and measurement. Gate-set tomography estimates them jointly with the gates, using short sequences to span preparation/readout directions and repeated blocks to amplify gate errors. It still assumes a system dimension and a sufficiently stable, composable model.

In a vector representation, write a sequence probability as p=eGL⋯G1rp=eG_L\cdots G_1r. For any invertible BB, the changes

r′=Br,e′=eB−1,Gk′=BGkB−1\begin{aligned} r'&=Br,\qquad e'=eB^{-1},\\ G_k'&=BG_kB^{-1} \end{aligned}

leave every probability unchanged. This gauge freedom prevents unique recovery of all matrix entries. A general coordinate transformation need not preserve the usual matrix expression of positivity, so it should not be interpreted as a physical operation. Nielsen and colleagues explain physical estimation and gauge choices in detail. Sequence probabilities remain the direct operational predictions.

Context can carry additional physics. For a target AA and spectator BB, take

H/ℏ=ΔZA/2+JZAZB/2.H/\hbar=\Delta Z_A/2+JZ_AZ_B/2.

Spectator states with ZB=±1Z_B=\pm1 shift the target detuning to Δ±J\Delta\pm J. An equal incoherent spectator mixture produces target coherence proportional to e−iΔtcos⁡(Jt)e^{-i\Delta t}\cos(Jt). A simple exponential dephasing fit can miss its revivals. Interleaved spectator preparations can distinguish this model from slow temporal drift; the observation alone does not identify every possible coupling mechanism.

Leakage makes the choice of modeled system equally consequential. The unnormalized retained output is a linear trace-decreasing map; dividing by its state-dependent survival probability generally makes the conditional transformation nonlinear. Report survival as well as conditional performance, and test return from leaked levels. Leakage and crosstalk and memory-dependent noise develop these distinctions.

A useful final test of a diagnosis is its predicted response to a controlled change. If retuning removes a Ramsey phase slope, verify that prediction with fresh data and check relevant neighboring operations. Calibration loops explains how repeated measurements turn such models into control updates.

Let τ=10 μs\tau=10\,\mu\mathrm{s} and compare f=20f=20 kHz with 120120 kHz, where Δ=2πf\Delta=2\pi f. For ideal visibility and zero initial phase, calculate pXp_X at t=τ/4t=\tau/4.

Solution

The phases are 0.1π0.1\pi and 0.6π0.6\pi, giving probabilities about 0.975530.97553 and 0.345490.34549. At all integer multiples of τ\tau the phase difference is an integer multiple of 2π2\pi, so both quadratures coincide. This added time distinguishes these candidates, not all unbounded frequencies.

At one time t0t_0, suppose both quadratures determine θ0\theta_0. Describe every (Δ,ϕ)(\Delta,\phi) giving that phase. Does scanning the analysis phase at that same time help?

Solution

For every chosen Δ\Delta, take ϕ=θ0−Δt0\phi=\theta_0-\Delta t_0 modulo 2π2\pi. An analysis-phase scan measures the same combination and does not remove the degeneracy. Additional evolution times or independent knowledge of ϕ\phi are required; multiple times still need alias control.

For ϵ=0.01\epsilon=0.01 rad, estimate the one-gate average infidelity and the error after 100 commuting repetitions.

Solution

The infidelity is approximately ϵ2/6=1.67×10−5\epsilon^2/6=1.67\times10^{-5}. The error angle becomes 11 rad. Its average infidelity is then (2/3)sin⁡2(1/2)≈0.1532(2/3)\sin^2(1/2)\approx0.1532, rather than 100 times the one-gate value. The coherent approximation and commuting-error assumption are essential.

4. Check complete positivity before unitarity

Section titled “4. Check complete positivity before unitarity”

Find the Pauli probabilities for (0.98,0.98,0.90)(0.98,0.98,0.90). Repeat for the physical alternative (0.94,0.94,0.90)(0.94,0.94,0.90) and calculate its unitarity.

Solution

The first gives (0.965,0.025,0.025,−0.015)(0.965,0.025,0.025,-0.015) and is not CP. The alternative gives (0.945,0.025,0.025,0.005)(0.945,0.025,0.025,0.005), all nonnegative, with

u=(0.942+0.942+0.902)/3≈0.85907.\begin{aligned} u&=(0.94^2+0.94^2+0.90^2)/3\\ &\approx0.85907. \end{aligned}

The first matrix has a formal squared-norm value too, but calling that the unitarity of a quantum channel would assume a channel that does not exist.

An amplitude-damping channel with decay probability γ\gamma acts as

T=diag⁡(1−γ,1−γ,1−γ),t=(0,0,γ).\begin{aligned} T&=\operatorname{diag}(\sqrt{1-\gamma},\sqrt{1-\gamma},1-\gamma),\\ \boldsymbol t&=(0,0,\gamma). \end{aligned}

For γ=1/2\gamma=1/2, find the output purity of I/2I/2 and the unitarity.

Solution

The output has Bloch vector (0,0,1/2)(0,0,1/2) and purity (1+1/4)/2=0.625(1+1/4)/2=0.625, larger than 0.50.5. Yet

u=2(1−γ)+(1−γ)23=512.u=\frac{2(1-\gamma)+(1-\gamma)^2}{3}=\frac5{12}.

The increase comes from relaxation toward a preferred state, not preservation of unknown input coherence.

For T1=40 μsT_1=40\,\mu\mathrm{s} and a compatible exponential T2=30 μsT_2=30\,\mu\mathrm{s}, find TϕT_\phi. What would T2=100 μsT_2=100\,\mu\mathrm{s} imply under the same model?

Solution

In inverse microseconds, 1/Tϕ=1/30−1/80=1/481/T_\phi=1/30-1/80=1/48, so Tϕ=48 μsT_\phi=48\,\mu\mathrm{s}. The second choice gives 1/100−1/80=−0.00251/100-1/80=-0.0025, incompatible with nonnegative Markovian pure dephasing at those exact parameters. Revisit the uncertainties, filters and two-level assumptions instead of assigning a negative microscopic rate.

For a static Gaussian detuning β\beta of variance σ2\sigma^2, derive the Ramsey and ideal midpoint-echo coherence directly, without a spectral integral.

Solution

Ramsey accumulates φ=βT\varphi=\beta T, with variance σ2T2\sigma^2T^2, so Gaussian averaging gives e−σ2T2/2e^{-\sigma^2T^2/2}. Echo accumulates β(T/2−T/2)=0\beta(T/2-T/2)=0, giving unit coherence in this idealized model. These results agree with Sβ(ω)=2πσ2δ(ω)S_\beta(\omega)=2\pi\sigma^2\delta(\omega) and the stated 1/(4π)1/(4\pi) prefactor.

Use the transformations above to show that p′=pp'=p for a two-gate sequence. Which matrices can more shots uniquely determine?

Solution p′=(eB−1)(BG2B−1)(BG1B−1)(Br)=eG2G1r=p.\begin{aligned} p'&=(eB^{-1})(BG_2B^{-1})(BG_1B^{-1})(Br)\\ &=eG_2G_1r=p. \end{aligned}

The adjacent inverse factors cancel. More shots improve operational probability estimates but cannot select a unique representative among equivalent matrices without extra conventions or trusted information.

Take Δ=0\Delta=0 and an equal mixture of spectator ZZ states. What happens to a target initially in ∣+⟩|+\rangle at Jt=π/2Jt=\pi/2 and Jt=πJt=\pi?

Solution

The target’s transverse component is cos⁡(Jt)\cos(Jt). At π/2\pi/2 it is zero and the target is maximally mixed; at π\pi it is −1-1 and the target is the pure state ∣−⟩|-\rangle. A time-homogeneous exponential dephasing channel cannot reproduce this loss and revival. The larger target-plus-spectator system still evolves unitarily.

10. Conditional filtering is generally nonlinear

Section titled “10. Conditional filtering is generally nonlinear”

Let a successful filter have Kraus operator K=diag⁡(1,1/2)K=\operatorname{diag}(1,1/2). Compare filtering I/2I/2 and renormalizing with mixing the separately normalized outputs for ∣0⟩|0\rangle and ∣1⟩|1\rangle equally.

Solution

The unnormalized mixed-state output is diag⁡(1/2,1/8)\operatorname{diag}(1/2,1/8), with survival 5/85/8, so the conditional state is diag⁡(4/5,1/5)\operatorname{diag}(4/5,1/5). Each basis-state output separately normalizes back to itself; their equal mixture is I/2I/2. Conditioning reweights the two inputs according to their different survival probabilities.

  • Degen, C. L., Reinhard, F., and Cappellaro, P. “Quantum sensing.” Reviews of Modern Physics 89, 035002 (2017). doi:10.1103/RevModPhys.89.035002.
  • Nielsen, E., Gamble, J. K., Rudinger, K., Scholten, T., Young, K., and Blume-Kohout, R. “Gate set tomography.” Quantum 5, 557 (2021). doi:10.22331/q-2021-10-05-557.
  • Szańkowski, P., Ramon, G., Krzywda, J., Kwiatkowski, D., and Cywiński, Ł. “Environmental noise spectroscopy with qubits subjected to dynamical decoupling.” Journal of Physics: Condensed Matter 29, 333001 (2017). doi:10.1088/1361-648X/aa7648.
  • Wallman, J. J., Granade, C., Harper, R., and Flammia, S. T. “Estimating the coherence of noise.” New Journal of Physics 17, 113020 (2015). doi:10.1088/1367-2630/17/11/113020.