Error-Aware Compilation

Error-aware compilation chooses how to run a circuit using what is known about the processor’s noise. The shortest circuit need not give the most accurate answer: it might use a poor connection, leave a fragile state idle, or measure an important output on an unreliable detector. The aim is to choose among implementations of the same intended computation.

The guiding distinction is between a predicted score and measured task performance. Calibration data help compare candidates, but drift, correlated errors and an imperfect noise model can reverse their ranking. We develop simple scores first, then examine when they fail.

Required background. Circuit representations specify outputs and timing; circuit optimization defines allowable rewrites; mapping and routing tracks physical sites; quantum noise distinguishes coherent errors, decoherence and leakage.

Choosing among equivalent circuits

Let $\mathcal X(C,\Theta)$ be the set of implementations that satisfy the intended input–output behavior and hardware constraints:

$$\mathcal X(C,\Theta) = \bigl\{ x: \ \mathsf{Semantics}(x) \simeq \mathsf{Semantics}(C), \mathsf{Legal}(x,\Theta)=1 \bigr\}.$$

Here $\Theta$ specifies available gates, connections, timings and resources. The equivalence relation $\simeq$ may allow global phase, an approximation tolerance, ancillas returned in a specified state, a final qubit permutation, or classical relabeling. Those permissions must already be explicit. Noise awareness does not license a compiler to silently change the algorithm.

In practice the compiler searches only a bounded candidate subset $\mathcal X_{\mathcal B}\subseteq\mathcal X$. It selects

$$x^\star \in \operatorname*{arg\,min}_{x\in\mathcal X_{\mathcal B}} \widehat L \bigl( x;\mathcal D,t_{\mathrm{exec}},\mathcal O \bigr).$$

Here $\mathcal D$ is the calibration data, $t_{\mathrm{exec}}$ the expected execution time, and $\mathcal O$ the quantity to optimize. The prediction $\widehat L$ estimates loss on that task. It is not a guarantee about the future run. The distinction matters: the candidate generator, score model, calibration sample, and queue delay can all change the winner.

Useful objectives include the probability of a correct answer, the error in an expectation value, or the time needed to reach a chosen statistical precision. They are different goals: maximizing the number of accepted shots, for example, need not minimize bias from state-dependent rejection.

What calibration data describe

A number called a gate error is incomplete without its measurement method, uncertainty, time and operating conditions. Hardware metrics explains the underlying quantities; here we ask how they affect a choice.

Quantity	Possible use	Limitation
Gate benchmark	Compare gate families and edges	A fitted decay is not a per-use failure probability
Duration and coherence time	Estimate idle exposure	State, drive and noise spectrum matter
Readout confusion matrix	Choose measurement sites	Errors can depend on the ideal outcome
Simultaneous-gate tests	Avoid harmful overlap	Untested spectators can change the result
Leakage estimates	Compare pulses and reset choices	A qubit-only noise model misses escaped population

Use data from a relevant context. An isolated gate measurement need not predict the same gate next to a driven neighbor. Acquisition time also matters, but age alone does not determine usefulness: a stable parameter measured yesterday may be more informative than a noisy estimate obtained minutes ago. Predicting future performance requires a drift model or repeated measurements, not an arbitrary expiry factor. Proctor and colleagues discuss both benchmark interpretation and drift detection in the references below.

A product score and its assumptions

A common first score uses reported error numbers $0\leq\widehat r<1$ as though their complements were independent probabilities of no fault:

$$\widehat P_{\mathrm{ind}}(x) = \prod_{g\in x} \bigl(1-\widehat r_g\bigr) \prod_{m\in x} \bigl(1-\widehat r_m\bigr).$$

Taking a negative logarithm converts the product into an additive path cost:

$$\begin{aligned} J_{\mathrm{ind}}(x) &= -\log\widehat P_{\mathrm{ind}}(x)\\ &= \sum_{g\in x} -\log\bigl(1-\widehat r_g\bigr) + \sum_{m\in x} -\log\bigl(1-\widehat r_m\bigr). \end{aligned}$$

For small $\widehat r$,

$$-\log(1-\widehat r) = \widehat r + O(\widehat r^2),$$

so minimizing the log score resembles minimizing a weighted error count. This is useful for shortest paths, assignment, integer programming, and fast candidate ranking.

It is still only a surrogate. Its literal probabilistic interpretation requires independent, context-invariant events. Even then it estimates the probability of no modeled fault, which need not equal task success: some faults have no effect on the requested output, and some can cancel. A randomized-benchmarking decay parameter is not literally the probability that one occurrence of a named gate fails. Coherent overrotations may add or cancel. Leakage survives beyond one gate. Readout errors depend on the prepared state. Crosstalk depends on the schedule. Temporally correlated noise violates the factorization itself.

Call $\widehat P_{\mathrm{ind}}$ a calibration-derived product score unless its probabilistic interpretation has been independently justified. Do not rename it “circuit fidelity” merely because its value lies between zero and one.

Choosing what to minimize

A broader loss can separate terms:

$$\widehat L(x) =\; \alpha J_{\mathrm{gate}}(x) + \beta J_{\mathrm{idle}}(x) + \gamma J_{\mathrm{read}}(x) + \delta J_{\mathrm{xtalk}}(x) + \epsilon J_{\mathrm{leak}}(x) + \zeta J_{\mathrm{resource}}(x).$$

Normalize terms or give their weights the units needed to combine them. Avoid counting the same mechanism twice: a measured gate error may already include relaxation during that gate. The coefficients express the relative importance of the terms for this task. A circuit returning a full bit string may care strongly about readout assignment. An expectation-estimation workload may tolerate some outcomes differently. A dynamic circuit can be dominated by measurement, reset, and classical-feedback latency. A fault-tolerant subroutine should ultimately be judged by logical failure and resource cost, not a product of physical gate benchmarks.

Whenever possible, predict the operational loss directly. Examples include:

• probability of a declared correct output for a checkable circuit;
• total-variation or Hellinger distance between output distributions;
• bias, variance, or mean-squared error of a target observable;
• logical failure probability per round or operation;
• accepted samples per unit wall time;
• an application utility with explicit classical postprocessing.

When direct prediction is infeasible, report which surrogate is used and validate whether it preserves the candidate ranking relevant to the workload. A model can be badly calibrated in absolute value yet useful for ranking; it can also predict plausible values while ranking candidates incorrectly.

Pareto decisions

Compressing everything into one scalar can hide meaningful tradeoffs. Define a cost vector

$$\mathbf J(x) = \bigl( J_{\mathrm{error}}, T_{\mathrm{exec}}, T_{\mathrm{compile}}, N_{\mathrm{cal}}, N_{\mathrm{shots}} \bigr).$$

A candidate $x$ dominates $y$ if every component is no worse and at least one is better. Nondominated candidates form a Pareto set. A scheduler can then choose according to a declared latency or calibration budget while keeping the tradeoff visible.

Uncertainty in the ranking

Finite characterization shots, fit uncertainty, drift, context mismatch, and queue delay make the device and noise parameters $\eta$ uncertain. Three defensible objectives are:

$$\begin{aligned} J_{\mathrm{mean}}(x) &= \mathbb E_{\eta\mid\mathcal D} \bigl[ L(x,\eta) \bigr], \\ J_{\mathrm{rob}}(x) &= \sup_{\eta\in\mathcal U(\mathcal D)} L(x,\eta), \\ J_{\mathrm{tail}}(x) &= \operatorname{CVaR}_{\alpha} \bigl( L(x,\eta) \bigr). \end{aligned}$$

For an integrable loss and $0<\alpha<1$, the conditional value at risk is the mean loss in the worst $1-\alpha$ fraction of the modeled distribution. For a discrete distribution, use the needed fraction of probability at the quantile boundary. Its convenient definition is

$$\operatorname{CVaR}_{\alpha}(L) = \min_z \Biggl[ z + \frac{1}{1-\alpha} {}\times \mathbb E \bigl[ (L-z)_+ \bigr] \Biggr].$$

These choices rely on different assumptions. A mean or tail average requires a predictive probability distribution; a worst-case bound requires an uncertainty set that actually contains the relevant possibilities. Increasing $\alpha$ emphasizes rarer adverse outcomes. A weighted mean plus CVaR is also possible, but is not generally an interpolation to the worst case over an independently chosen set. See Rockafellar and Uryasev for the tail-risk definition, including discrete losses.

The compiler should also estimate rank stability. For two candidates,

$$\Delta L(\eta) = L(x_2,\eta)-L(x_1,\eta).$$

Form the uncertainty interval for $\Delta L$ from the joint model of the two losses. Shared fluctuations can cancel in this difference even when the separate loss intervals overlap. If its interval crosses zero, that interval alone does not establish an ordering at its stated confidence or coverage level. The right action may be to choose the simpler candidate, run a targeted characterization, compile a small portfolio, or report a tie. Extra decimal places do not resolve epistemic uncertainty.

Historical calibration can help when immediate data are unavailable, but it should be conditioned on relevant modes and monitored for distribution shift. The acquisition cost also belongs in the decision: a fresh characterization that consumes the entire execution window can be worse than a robust choice from older data.

Placement and routing with unequal errors

Connectivity defines which routes are legal. Error awareness ranks those routes. On a fixed coupling graph, assign edge weights

$$w_e = -\log \bigl( 1-\widehat r_e \bigr),$$

and site or readout weights analogously. A minimum-hop path and a minimum-weight path need not agree. Nor must the lowest-weight path minimize scheduled loss, because its gates may serialize or overlap with hostile neighbors.

Consider two legal implementations of a measured output. Candidate $A$ uses three two-qubit operations with reported rate $0.006$ and a readout with rate $0.025$. Candidate $B$ uses two two-qubit operations at $0.014$ and a readout at $0.008$. Their independent scores are

$$\begin{aligned} \widehat P_A &= (0.994)^3(0.975) \simeq 0.9576, \\ \widehat P_B &= (0.986)^2(0.992) \simeq 0.9644. \end{aligned}$$

The proxy selects $B$: its gates look worse individually, but it uses fewer of them and ends on a better readout site. This is a ranking under a model, not proof that $B$ will have higher experimental fidelity.

Error-aware routing can influence:

• the initial placement of high-degree or long-lived program qubits;
• which physical edge carries repeated entangling operations;
• which of several shortest SWAP paths is used;
• whether a final permutation is accepted rather than restored;
• where measured outputs finish;
• whether an isomorphic embedding is substituted after routing;
• when a marginally longer route avoids a volatile or disabled component.

Track the final qubit assignment when comparing measured outputs. A route with a good gate score can still finish on a poor readout site.

Crosstalk versus extra waiting

Let $y_{gh}=1$ when operations $g$ and $h$ overlap in a context declared relevant. A pairwise scheduling penalty is

$$J_{\mathrm{xtalk}} = \sum_{g\lt h} \xi_{gh}y_{gh},$$

where $\xi_{gh}$ may be inferred from simultaneous benchmarking or another contextual experiment. Pairwise additivity is an approximation; triple interactions and global modes can invalidate it.

Serializing every gate eliminates concurrency but increases live time. A simple comparison illustrates the tradeoff. Suppose the task measures Ramsey contrast. Parallel execution reduces that contrast by a factor $1-\chi$, while serialization adds delay $\Delta$ with exponential coherence decay $e^{-\Delta/T_2}$. Holding other effects fixed, serialization preserves more contrast when

$$\begin{aligned} e^{-\Delta/T_2} &\gt 1-\chi, \\ \Delta &\lt -T_2\log(1-\chi). \end{aligned}$$

For $T_2=80\,\mu\mathrm{s}$ and $\chi=0.06$, the threshold is about $4.95\,\mu\mathrm{s}$. A $3\,\mu\mathrm{s}$ delay favors serialization in this model; an $8\,\mu\mathrm{s}$ delay favors overlap. This is a comparison of contrast, not a general formula for circuit success or state fidelity. For a dephasing qubit prepared in $|+\rangle$, for example, the probability of returning the $X=+1$ outcome is $[1+e^{-\Delta/T_2}]/2$. Driven-gate noise, spectators, leakage and echo pulses can change the schedule comparison. Murali and colleagues study this competition between crosstalk and additional decoherence.

The scheduler needs operation intervals, exclusion resources, conditional rates, state live ranges, measurement deadlines, and classical latencies. This is why error-aware compilation cannot be reduced to assigning static weights to a coupling graph.

Gate choices and approximation error

Two implementations can be exactly equivalent yet expose different physical error channels. The compiler may choose:

• one native entangler family over another;
• an interaction direction or locally equivalent orientation;
• a continuously parameterized entangler instead of repeated maximal gates;
• a mirrored or sign-reversed decomposition with lower total interaction angle;
• a decomposition that trades entangling operations for local rotations;
• an approximate synthesis with a declared algorithmic tolerance;
• a frame update instead of a driven physical operation.

Continuously parameterized gates make the boundary especially clear. The gate decomposer establishes which exact or approximate words implement the target. The error-aware selector ranks those words using angle-, pair-, and context-dependent measurements. The pulse-control layer realizes the selected native operation.

Algorithmic approximation error and physical implementation error must not be added unless they use compatible metrics. For channels $\mathcal U$, $\mathcal V$, and a physical implementation $\widetilde{\mathcal V}$, the diamond norm gives the rigorous triangle bound

$$\frac12 \left\| \widetilde{\mathcal V}-\mathcal U \right\|_{\diamond} \leq\; \frac12 \left\| \widetilde{\mathcal V}-\mathcal V \right\|_{\diamond} + \frac12 \left\| \mathcal V-\mathcal U \right\|_{\diamond}.$$

A calibration-derived product score is not a diamond-norm estimate, so it cannot simply be inserted into this bound.

Asymmetric readout changes the best site

Readout is generally asymmetric. Define

$$\begin{aligned} e_{1\mid0} &= \Pr(\widetilde m=1\mid m=0), \\ e_{0\mid1} &= \Pr(\widetilde m=0\mid m=1). \end{aligned}$$

With observed outcomes as rows and ideal outcomes as columns, the one-qubit confusion matrix is

$$M = \begin{pmatrix} 1-e_{1\mid0} & e_{0\mid1} \\ e_{1\mid0} & 1-e_{0\mid1} \end{pmatrix}.$$

If a trustworthy task model predicts probabilities $p_0$ and $p_1$, the expected bit-error proxy is

$$\widehat r_{\mathrm{read}} = p_0e_{1\mid0} + p_1e_{0\mid1}.$$

Using $(e_{1\mid0}+e_{0\mid1})/2$ instead assumes balanced ideal outcomes. That may be appropriate for a benchmark and wrong for a workload. For an unknown quantum output, using a guessed distribution can introduce a hidden task prior into placement.

Readout-aware compilation maps important measured values to suitable channels, preserves the final physical-to-program permutation, and respects simultaneous-readout conflicts. Measurement-error mitigation instead estimates or corrects noisy output statistics; it is a separate procedure.

Testing whether the choice helps

Three separate questions matter: does the circuit compute the intended quantity, can the hardware execute its schedule, and does the chosen implementation perform better? A noise score answers neither of the first two. Recompute the ideal behavior and check the timing and final qubit map before comparing noisy results.

Use fresh application shots to evaluate a candidate selected from noisy calibration measurements. Selecting the smallest measured error among many candidates tends to select a favorable statistical fluctuation as well as a good candidate. Reusing that minimum as an unbiased performance estimate overstates the improvement. Independent shots address this selection effect; they do not remove drift or a mismatch between benchmark and application.

For a fair comparison, keep the input circuits, approximation tolerance and available compilation time comparable. If studying the score alone, compare the same candidate set. If studying the complete compiler, differences in candidate generation are part of the method and should be described, not artificially suppressed. Report compilation and characterization time along with task accuracy, since obtaining a slightly better circuit may consume the entire useful execution window.

Interleave or randomize candidate execution order so that drift does not systematically favor one candidate. Paired measurements can reduce shared drift uncertainty, but their statistical analysis must retain correlations. Repeat across relevant device conditions and retain failures and ties. A fixed time limit for calibration validity is an operational choice, not a physical boundary at which the processor suddenly changes.

Keep the calibration time, measurement conditions, final assignment and settings needed to reproduce the comparison. Calibration loops explains how these quantities are refreshed; pulse-level control explains how the selected operations become physical drives.

Exercises

1. Compute an independent product score

Candidate $A$ uses four two-qubit gates with reported rate $0.008$. Candidate $B$ uses three with rate $0.013$. Ignore all other errors. Which candidate does the independent product proxy prefer?

Solution

The two scores are

$$\begin{aligned} \widehat P_A &= (1-0.008)^4 \simeq 0.9684, \\ \widehat P_B &= (1-0.013)^3 \simeq 0.9615. \end{aligned}$$

The proxy prefers $A$ even though it uses one more gate. Equivalently, $-4\log(0.992)\lt-3\log(0.987)$. This does not establish that $A$ has higher experimental fidelity; it is the ranking under the independent scalar model.

2. Add readout to a route decision

Candidate $A$ has the two-qubit score from Exercise 1 and ends on a readout site with error $0.06$. Candidate $B$ ends on a site with error $0.01$. Re-rank the candidates using the product proxy.

Solution

Including readout gives

$$\begin{aligned} \widehat P_A' &= 0.9684(0.94) \simeq 0.9103, \\ \widehat P_B' &= 0.9615(0.99) \simeq 0.9519. \end{aligned}$$

The ranking reverses, so $B$ is preferred. The example shows why a route cannot be ranked solely by the edges used when its final map determines measurement channels.

3. Balance crosstalk and delay

Two gates reduce Ramsey contrast by a factor $1-\chi$, where $\chi=0.04$, when run simultaneously. Serializing them delays the measured coherence by $\Delta=2\,\mu\mathrm{s}$, with $T_2=70\,\mu\mathrm{s}$. Which schedule is favored by the simple threshold model?

Solution

The largest delay favoring serialization is

$$\begin{aligned} \Delta_{\max} &= -T_2\log(1-\chi) \\ &= -70\,\mu\mathrm{s}\log(0.96) \simeq 2.86\,\mu\mathrm{s}. \end{aligned}$$

Because $2\,\mu\mathrm{s}\lt2.86\,\mu\mathrm{s}$, the model favors serialization. The conclusion is conditional on its exponential coherence decay and multiplicative contrast model.

4. Detect a robust rank reversal

Two candidates have loss intervals

$$\begin{aligned} L_A&\in[0.020,0.055], \\ L_B&\in[0.030,0.040]. \end{aligned}$$

Their nominal losses are $0.025$ and $0.035$. Which candidate is selected by nominal and worst-case rules?

Solution

Nominal minimization selects $A$ because $0.025\lt0.035$. Worst-case minimization compares the upper endpoints and selects $B$ because $0.040\lt0.055$. The separate intervals alone do not establish a uniform ordering. Correlations between the losses could supply additional information; the nominal and worst-case rules here use only the numbers supplied.

5. Use asymmetric readout data

Site $A$ has $e_{1\mid0}=0.02$ and $e_{0\mid1}=0.10$. Site $B$ has $e_{1\mid0}=0.04$ and $e_{0\mid1}=0.03$. Rank the sites when the ideal output has $(p_0,p_1)=(0.9,0.1)$, then when it is balanced.

Solution

For the skewed output,

$$\begin{aligned} \widehat r_A &= 0.9(0.02)+0.1(0.10) = 0.028, \\ \widehat r_B &= 0.9(0.04)+0.1(0.03) = 0.039. \end{aligned}$$

Site $A$ is preferred. For balanced outputs,

$$\widehat r_A=0.060, \qquad \widehat r_B=0.035,$$

so $B$ is preferred. The workload prior changes the ranking; it must therefore be part of the chosen objective.

6. Coherent errors can cancel

Two circuits implement the identity up to global phase: $A=R_z(\pi)R_z(\pi)$ and $B=R_z(\pi)X R_z(\pi)X$. Assume perfect $X$ gates and the same additive rotation error $\widetilde R_z(\pi)=R_z(\pi+\varepsilon)$ on both occurrences. Compare their physical action and their $X$-measurement results on $|+\rangle$. What does a product of identical per-rotation scores miss?

Solution

Using $XR_z(\phi)X=R_z(-\phi)$ gives

$$\widetilde A=-R_z(2\varepsilon), \qquad \widetilde B=I.$$

Thus $A$ returns $X=+1$ with probability $\cos^2\varepsilon$, whereas $B$ returns it with probability one. The two equal rotation costs alone cannot describe this cancellation. The example assumes a common, stable error and noiseless $X$ gates; imperfect echoes or rapidly changing offsets alter the comparison. Global phase is harmless here because these are standalone circuits, not controlled subroutines.

7. Identify a Pareto set

Three candidates have

$$\begin{aligned} \mathbf J(A)&=(0.030,\,12,\,4),\\ \mathbf J(B)&=(0.025,\,15,\,6),\\ \mathbf J(C)&=(0.040,\,18,\,7), \end{aligned}$$

where components are predicted loss, execution time, and compile time, all to be minimized. Which candidates are Pareto optimal?

Solution

$A$ dominates $C$: it has lower predicted loss, execution time, and compile time. Neither $A$ nor $B$ dominates the other. $B$ has lower predicted loss, while $A$ has lower execution and compile times. The Pareto set is therefore $\{A,B\}$.

8. Shared uncertainty need not obscure a ranking

Suppose $L_A=0.03+z$ and $L_B=0.04+z$, where the same drift variable obeys $z\in[-0.02,0.02]$. Their separate intervals overlap. Is the ranking ambiguous? Compare with independent $z_A,z_B$ in the same ranges.

Solution

With common drift, $L_B-L_A=0.01$ identically, so $A$ is better throughout the joint uncertainty set. With independently variable drift, $L_B-L_A=0.01+z_B-z_A\in[-0.03,0.05]$, and either candidate can be better. The separate intervals are identical in the two cases; the dependence structure changes the answer.

9. Design a fair hardware evaluation

Design an experiment to test whether an error-aware compiler improves a workload over a topology-only compiler.

Solution

Use the same source circuits, target, resource budget, semantic tolerance, final-map permissions, and execution allocation. Freeze or record all compiler versions and seeds. Acquire calibration data for selection, but reserve independent application shots for evaluation.

Run randomized or interleaved candidate order across several calibration epochs so drift does not favor one compiler. Report structural metrics, predicted scores, task-appropriate held-out metrics, confidence intervals, compiler time, characterization cost, queue delay, and failures. Include all prespecified circuits and seeds rather than only wins. This tests both average benefit and temporal robustness.

10. Tail risk with a discrete distribution

A candidate has loss zero with probability $0.95$ and loss $0.2$ with probability $0.05$. Find its mean loss and $\operatorname{CVaR}_{0.90}$. Compare it with a candidate whose loss is always $0.03$.

Solution

The mean is $0.05(0.2)=0.01$. The worst ten percent contains all the five-percent mass at $0.2$ and five percent at zero, so

$$\begin{aligned} \operatorname{CVaR}_{0.90} &=\frac{0.05(0.2)+0.05(0)}{0.10}\\ &=0.10. \end{aligned}$$

The constant-loss candidate has both mean and CVaR equal to $0.03$. Mean loss favors the first candidate; this tail-risk measure favors the second. Conditioning only on outcomes strictly above the quantile would incorrectly give $0.2$ for the first candidate’s CVaR.

References

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