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Noise in Quantum Information

Noise is the difference between an intended quantum experiment and its physical implementation. It includes environmental decoherence, but also an isolated qubit rotated through the wrong angle, population in unwanted levels, and a detector that reports the wrong result. These mechanisms need different models and different remedies. This page explains how to distinguish them and why a single error percentage seldom predicts an entire circuit.

Helpful background. The circuit model defines the intended operations; density operators describe mixtures and reduced states. Quantum channels provide the completely positive maps used below.

Suppose input label xx selects a preparation ρx\rho_x, the device applies a channel E\mathcal E, and a measurement has effects MyM_y. Its predictions are

p(y∣x,t,c)=Tr⁡ ⁣[My,t,c Et,c(ρx,t,c)].p(y\mid x,t,c)=\operatorname{Tr}\!\left[ M_{y,t,c}\,\mathcal E_{t,c}(\rho_{x,t,c})\right].

Here tt is the acquisition time and cc describes the circuit context: neighboring drives, simultaneous measurements, idle intervals and earlier pulses. A fixed channel is useful only when those dependencies can be ignored for the experiment being predicted. We assume a common preparation domain supports the channel description; arbitrary initial system–environment correlations need not admit one channel for all preparations.

An observed discrepancy can come from preparation, evolution, measurement, or several together. For example, if preparation produces Bloch polarization ss, the gate shrinks it by λ\lambda, and readout has contrast mm, the measured polarization is sλms\lambda m. One number does not determine all three factors. The physical boundary also matters: populating a third atomic level is an error for a qubit gate even if the full three-level motion is perfectly unitary.

The following categories overlap. Crosstalk can cause a coherent phase; drift can turn that phase into an apparent mixture when many runs are averaged.

Mechanism What changes A useful distinction
Coherent control error amplitudes evolve under the wrong unitary aligned rotations can accumulate before probabilities are formed
Relaxation and dephasing populations or reduced-state coherences decay a lifetime is specific to a state, basis and experiment
Leakage or loss population leaves the physical computational space a detected erasure differs from hidden loss
Crosstalk a process depends on activity elsewhere isolated-gate performance may not transfer to parallel gates
Preparation and readout error the input state or measurement differs from the intended one an output bias need not be a gate error
Drift and memory parameters change, or successive errors are dependent one-step data need not predict a long sequence

Coherent errors and stochastic faults accumulate differently

Section titled “Coherent errors and stochastic faults accumulate differently”

A systematic overrotation is a unitary error. For example, a nominal identity cycle might instead apply

Vϵ=e−iϵX/2.V_\epsilon=e^{-i\epsilon X/2}.

Starting in ∣0⟩|0\rangle and repeating the same error LL times gives

VϵL=e−iLϵX/2,P(1)=sin⁡2(Lϵ/2)≃L2ϵ2/4.\begin{aligned} V_\epsilon^L&=e^{-iL\epsilon X/2},\\ P(1)&=\sin^2(L\epsilon/2) \simeq L^2\epsilon^2/4. \end{aligned}

The approximation requires L∣ϵ∣≪1L|\epsilon|\ll1. Angles add first, so this probability grows quadratically at short times. It eventually oscillates; there is no unlimited quadratic growth. Other ideal gates can rotate the error axis. In particular, ZVϵZ=Vϵ†ZV_\epsilon Z=V_\epsilon^\dagger, so an ideal refocusing sequence can cancel this error rather than amplify it.

Compare this with independently applying XX with probability p=sin⁡2(ϵ/2)p=\sin^2(\epsilon/2) on each cycle. Both models have the same one-cycle ZZ-basis flip probability on ∣0⟩|0\rangle, but the stochastic model gives

P(1)=1−(1−2p)L2.P(1)=\frac{1-(1-2p)^L}{2}.

This is the probability of an odd number of flips: two XX errors cancel. By contrast, the probability of at least one fault is 1−(1−p)L≃Lp1-(1-p)^L\simeq Lp for Lp≪1Lp\ll1. Counting faults and predicting an output error are different calculations. The chapter’s accumulation plot shows the coherent and stochastic results on common axes.

An incoherent reduced description can arise from entanglement with an environment or from averaging over unobserved classical variables. The same detuning may be coherent within one shot and produce dephasing across shots. Decoherence is loss of phase coherence in a specified reduced description; a wrong unitary alone is not decoherence. Dephasing and dissipation also need not change populations in the same way.

Average infidelity and worst-case distance

Section titled “Average infidelity and worst-case distance”

For a CPTP map on a fixed dd-dimensional space and an ideal unitary UU, define

Favg(E,U)=∫dψ ⟨ψ∣U†E(∣ψ⟩⟨ψ∣)U∣ψ⟩,r=1−Favg,D⋄=12∥E−U∥⋄.\begin{aligned} F_{\rm avg}(\mathcal E,U) &=\int d\psi\,\langle\psi|U^\dagger \mathcal E(|\psi\rangle\langle\psi|)U|\psi\rangle,\\ r&=1-F_{\rm avg},\\ D_\diamond&=\tfrac12\|\mathcal E-\mathcal U\|_\diamond. \end{aligned}

The integral averages uniformly over pure inputs. The diamond distance instead maximizes output distinguishability, allowing an input entangled with an unchanged reference. With equal prior probabilities, the best one-use success probability for distinguishing the two channels is (1+D⋄)/2(1+D_\diamond)/2.

For the qubit overrotation above, and for the independent Pauli channel (1−p)ρ+pXρX(1-p)\rho+pX\rho X matched by p=sin⁡2(ϵ/2)p=\sin^2(\epsilon/2),

Error channel Average infidelity Diamond distance from the identity
coherent rotation r=2p/3r=2p/3 D⋄=pD_\diamond=\sqrt p
stochastic Pauli flip r=2p/3r=2p/3 D⋄=pD_\diamond=p

For a pure qubit with Bloch component xx, the rotation’s survival probability is 1−p+px21-p+px^2. Averaging x2x^2 over the sphere gives 1/31/3, explaining r=2p/3r=2p/3. An input perpendicular to the rotation axis attains output trace distance p\sqrt p. For the Pauli channel, convexity bounds that distance by pp, and ∣0⟩|0\rangle attains it. An ancillary system cannot improve either maximum. Thus the same average infidelity can conceal quite different worst-case behavior (Kueng et al.).

Average infidelity does give bounds; it is not devoid of worst-case information:

d+1dr ≤ D⋄ ≤ min⁡ ⁣{1,d(d+1)r}.\frac{d+1}{d}r\ \leq\ D_\diamond \ \leq\ \min\!\left\{1,\sqrt{d(d+1)r}\right\}.

These bounds assume the stated fixed-space CPTP channel and unitary target. They do not make a calibration valid in a different circuit context, turn a trace-decreasing postselected map into a channel, or account for omitted leakage levels. Benchmark decay parameters also require a model before they can be interpreted as rr; gate-dependent noise complicates that interpretation (Proctor et al.).

For a two-level system, T1T_1 describes longitudinal population relaxation; T2T_2 describes transverse coherence decay. In the Markovian model with independent energy relaxation and pure dephasing,

1T2=12T1+1Tϕ,T2≤2T1.\frac1{T_2}=\frac1{2T_1}+\frac1{T_\phi},\qquad T_2\leq2T_1.

The factor 1/21/2 reflects that an amplitude decays at half the corresponding population-decay rate. This relation assumes compatible exponential rates for the same operating conditions. It is not a definition for arbitrary driven, multilevel or non-Markovian dynamics. Ramsey measurements often quote T2∗T_2^*, which includes slowly varying frequency offsets; echo can remove some of that broadening. Decoherence timescales explains the experiments and assumptions.

A gate much shorter than T1T_1 can still have a substantial overrotation, unwanted coupling or leakage. A lifetime alone therefore cannot predict gate fidelity.

Let PP project onto the physical computational space, with input ρ=PρP\rho=P\rho P. For trace-preserving evolution on the full space,

ℓ(ρ)=Tr⁡[(I−P)E(ρ)]=1−Tr⁡[PE(ρ)]\ell(\rho)=\operatorname{Tr}[(I-P)\mathcal E(\rho)] =1-\operatorname{Tr}[P\mathcal E(\rho)]

is the output leakage probability. Higher transmon levels, unwanted atomic states or a photonic vacuum outside a fixed-photon encoding are examples. The retained map PE(ρ)PP\mathcal E(\rho)P is generally trace decreasing; its trace is the acceptance probability, not a normalization to discard silently.

Distinguish this physical space from a smaller error-correcting code. For the repetition code spanned by ∣000⟩,∣111⟩|000\rangle,|111\rangle, an XX error on one qubit leaves the code while remaining inside the three-qubit Hilbert space. That is an ordinary correctable error, not physical leakage. The carrier and encoding hierarchy makes this distinction explicit.

Return from a leaked level, often called seepage, need not restore the unknown state or its entanglement. Likewise, detecting a lost dual-rail photon reveals an erasure location, but reloading a photon alone cannot reconstruct the lost logical amplitudes. See erasure and loss channels and leakage and crosstalk for channel models and leakage measurements (Wood and Gambetta).

Crosstalk means that an operation depends on neighboring controls, states or measurements beyond the intended local model. Drive spillover, residual ZZZZ coupling and measurement-induced dephasing are examples. A parallel layer intended as GA⊗GB\mathcal G_A\otimes\mathcal G_B may not be describable by fixed independent local maps. Comparing isolated with simultaneous operation tests that assumption (Sarovar et al.; Rudinger et al.).

SPAM means state-preparation and measurement error. A useful restricted readout model is a classical assignment matrix:

p~(y)=∑zAyzp(z),Ayz≥0,∑yAyz=1.\begin{aligned} \widetilde p(y)&=\sum_z A_{yz}p(z),\\ A_{yz}&\geq0,\qquad \sum_y A_{yz}=1. \end{aligned}

It describes a computational-basis measurement followed by a classical misreporting process. Equivalently, its effects are My=∑zAyz∣z⟩⟨z∣M_y=\sum_z A_{yz}|z\rangle\langle z|. The indices may be whole bit strings: a joint assignment matrix can describe correlated readout. Such correlations invalidate a product approximation AA⊗ABA_A\otimes A_B, not necessarily the joint assignment model.

What fails beyond that model? If M0=(I+X)/2M_0=(I+X)/2, then ∣+⟩|+\rangle and ∣−⟩|-\rangle give probabilities one and zero even though both have the same computational-basis populations. No fixed classical assignment matrix acting on those populations can reproduce both. Leakage outside the calibrated basis, changing settings and detector memory can also require a larger model. Inverting a fitted matrix can amplify noise, and preparation errors can bias its calibration. SPAM errors develops these identifiability and measurement assumptions.

Memory can hide behind the same one-step channel

Section titled “Memory can hide behind the same one-step channel”

Consider idle steps with an unwanted ZZ rotation Us=exp⁡(−isδZ/2)U_s=\exp(-is\delta Z/2), where s=±1s=\pm1 with equal probabilities. Compare two experiments: choose ss once per complete sequence, or redraw it independently at every step. Their averaged one-step channels are identical on every input. Yet on the initial state ∣+⟩|+\rangle,

⟨X⟩fixed within sequence=cos⁡(Lδ),⟨X⟩independently redrawn=(cos⁡δ)L.\begin{aligned} \langle X\rangle_{\rm fixed\ within\ sequence}&=\cos(L\delta),\\ \langle X\rangle_{\rm independently\ redrawn}&=(\cos\delta)^L. \end{aligned}

In the first case the accumulated phase is ±Lδ\pm L\delta; in the second, averaging independent phase factors multiplies their means. The oscillation and revival of the first model cannot be predicted by composing its averaged one-step map LL times.

X expectation after repeated random-sign Z rotations, with oscillating fixed-sign noise and slowly decaying independent signs

Both models apply +δ+\delta or −δ-\delta with equal probabilities in one step. They differ only in whether that sign persists. Here δ=0.1\delta=0.1 rad and the initial state is ∣+⟩|+\rangle. Negative ⟨X⟩\langle X\rangle means that the probability of the X=−1X=-1 outcome exceeds one half; it is not a negative probability. An ideal echo cancels the fixed sign, as explored below.

Slow drift can produce a similar discrepancy between calibration and later use. Record acquisition times, interleave reference circuits and compare short with long sequences. Averaging epochs may hide excess uncertainty: if NN conditionally independent shots share a random success probability qq, the number KK of successes obeys

Var⁡K=Nqˉ(1−qˉ)+N(N−1)Var⁡q.\operatorname{Var}K =N\bar q(1-\bar q)+N(N-1)\operatorname{Var}q.

The second term vanishes only when this shared fluctuation vanishes. Ordinary binomial error bars can therefore be too small. Explicit time dependence, memory and failure of a chosen exponential fit are distinct statements; Markovian and non-Markovian noise and the open-system discussion explain the formal distinctions (Breuer et al.).

Experiment What it helps determine
Rabi, Ramsey, echo and relaxation measurements rotation calibration and selected decay mechanisms
Process tomography with trusted references, or gate-set tomography a channel or a self-consistent preparation–gate–measurement model; stationarity and reference assumptions matter
Randomized benchmarking an ensemble decay with reduced sensitivity to constant SPAM offsets, not automatic workload accuracy
Simultaneous operations and cycle benchmarks errors in relevant parallel layers and repeated operations
Held-out application circuits whether the fitted model predicts outputs, leakage and uncertainty at the depths and times actually used

Use circuits excluded from fitting to test predictions, including different depths and simultaneous operations. Device characterization connects these tests; noise simulation explains how to compare models. A benchmark score such as cross entropy needs its own interpretation and does not alone certify closeness of complete distributions.

A deterministic detuning suggests recalibration or refocusing. Leakage may require pulse changes and explicit detection. Time-varying readout may need fresh calibration. Error mitigation can reduce bias in selected estimated observables, while increasing variance or relying on a model; it generally does not restore a protected state. Quantum error correction instead uses redundancy and syndrome measurements to preserve encoded information. Both approaches must be tested with the correlations and failed attempts present in the intended experiment.

For ϵ=10−3\epsilon=10^{-3} and L=100L=100, compare the unwanted XX rotation’s flip probability with 100100 times its one-cycle value. How can ideal ZZ pulses cancel two equal rotations?

Solution

sin⁡2(0.05)≃0.002498\sin^2(0.05)\simeq0.002498, whereas 100sin⁡2(0.0005)≃0.000025100\sin^2(0.0005)\simeq0.000025. The angles add before squaring. For the cancellation sequence, ZVϵZVϵ=Vϵ†Vϵ=IZV_\epsilon ZV_\epsilon=V_\epsilon^\dagger V_\epsilon=I. The conclusion assumes ideal instantaneous ZZ pulses and equal errors.

For L=200L=200 independent XX faults with p=2×10−4p=2\times10^{-4}, compare the probability of at least one fault with the final flip probability on ∣0⟩|0\rangle.

Solution

1−(1−p)200≃0.039211-(1-p)^{200}\simeq0.03921, while [1−(1−2p)200]/2≃0.03845[1-(1-2p)^{200}]/2\simeq0.03845. The difference includes nonzero even numbers of faults that cancel. Both approach Lp=0.04Lp=0.04 at leading order, but they are not the same event.

In the exponential two-level model, T1=80 μsT_1=80\,\mu\mathrm{s} and T2=60 μsT_2=60\,\mu\mathrm{s}. Find TϕT_\phi. Could T2=180 μsT_2=180\,\mu\mathrm{s} describe the same model at that T1T_1?

Solution

Tϕ−1=1/60−1/160=1/96T_\phi^{-1}=1/60-1/160=1/96 in inverse microseconds, so Tϕ=96 μsT_\phi=96\,\mu\mathrm{s}. The second value exceeds 2T1=160 μs2T_1=160\,\mu\mathrm{s} and would require a negative pure-dephasing rate. Check measurement uncertainties, pulse sequences and the validity of the matched exponential model before assigning such a rate.

The output of a three-level device is ρ′=0.70∣0⟩⟨0∣+0.25∣1⟩⟨1∣+0.05∣2⟩⟨2∣\rho'=0.70|0\rangle\langle0|+0.25|1\rangle\langle1|+0.05|2\rangle\langle2|. If the computational space is spanned by ∣0⟩,∣1⟩|0\rangle,|1\rangle, find the leakage and the reported bit-one probability when ∣2⟩|2\rangle is assigned to bit one. What happens if leaked outcomes are discarded instead?

Solution

The leakage is 0.050.05. Misassignment gives bit-one probability 0.300.30. Discarding the leaked outcomes accepts with probability 0.950.95 and gives conditional bit-one probability 0.25/0.95=5/190.25/0.95=5/19. These are three different quantities; the accepted probability alone omits a physical failure channel.

A detector flips both reported bits with probability qq and otherwise reports them correctly. Write its assignment matrix in the order 00,01,10,1100,01,10,11. Why do independent one-bit calibration matrices fail?

Solution A=(1−q00q01−qq00q1−q0q001−q).A=\begin{pmatrix} 1-q&0&0&q\\0&1-q&q&0\\0&q&1-q&0\\q&0&0&1-q \end{pmatrix}.

This is a valid joint classical model. Each marginal detector flips with probability qq, but their product predicts single-bit errors with probability q(1−q)q(1-q) each. Those events never occur in the actual model. At q=0.1q=0.1, input 0000 produces only 0000 and 1111 with probabilities 0.9,0.10.9,0.1, whereas the product approximation predicts 0.81,0.09,0.09,0.010.81,0.09,0.09,0.01.

A qubit channel has average infidelity r=10−4r=10^{-4}. Compare the diamond distance if it is a coherent Pauli overrotation or a stochastic Pauli flip.

Solution

For the overrotation, ∣sin⁡(ϵ/2)∣=3r/2|\sin(\epsilon/2)|=\sqrt{3r/2}, so ∣ϵ∣≃0.024495|\epsilon|\simeq0.024495 for the small-angle branch and D⋄≃0.012247D_\diamond\simeq0.012247. For the stochastic flip, D⋄=p=3r/2=0.00015D_\diamond=p=3r/2=0.00015. Knowing rr without the mechanism does not choose between these values.

Insert an ideal instantaneous XX pulse halfway through an even number LL of the random-sign ZZ rotations above. Start in ∣+⟩|+\rangle and measure XX. Compare the fixed-sign and independently redrawn models. Separately, if 100100 shots share q=0.4q=0.4 or 0.60.6 with equal probabilities, find the mean and variance of their success count.

Solution

Because XZX=−ZXZX=-Z, the fixed-sign propagator is UsL/2XUsL/2=XU_s^{L/2}XU_s^{L/2}=X. The initial state is an XX eigenstate, so the echo restores ⟨X⟩=1\langle X\rangle=1. Independent symmetric signs are unaffected statistically by reversing the signs in the second half; their expectation remains (cos⁡δ)L(\cos\delta)^L.

For the counts, qˉ=0.5\bar q=0.5 and Var⁡q=0.01\operatorname{Var}q=0.01. Thus EK=50\mathbb E K=50 and Var⁡K=25+100⋅99⋅0.01=124\operatorname{Var}K=25+100\cdot99\cdot0.01=124, not the binomial value 2525. Redrawing qq independently before every shot would instead give that binomial variance.

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