Pauli Noise and Depolarizing Channels

Pauli noise replaces a quantum error by a probability distribution over Pauli conjugations. This makes two useful calculations possible: channel composition becomes a convolution of probabilities, and ideal Clifford circuits transport each fault to another Pauli. In a stabilizer code, the resulting distribution predicts both measured syndromes and the logical error after a specified recovery. These are distinct predictions; a small syndrome rate need not mean a small logical error rate.

Helpful background. Quantum channels supplies transfer coordinates and composition; stabilizer formalism supplies commuting checks and their syndromes. Common noise models compares Pauli errors with relaxation, leakage and coherent rotations.

Pauli probabilities and observable decay

Fix the one-qubit order $(I,X,Y,Z)$ and omit global phases from Pauli labels. A stochastic Pauli channel is

$$\begin{aligned} \Phi(\rho)&=\sum_Pp_P P\rho P,\\ p_P&\geq0,\qquad\sum_Pp_P=1. \end{aligned}$$

A trajectory simulator samples a label and applies its conjugation. Averaging trajectories gives the channel above. This is a possible realization of the channel, not proof that a device secretly selected a definite Pauli in each run.

Write $\rho=(I+r_xX+r_yY+r_zZ)/2$. Conjugation preserves its own axis and reverses the other two, so $r_j\mapsto\lambda_jr_j$, where

$$\begin{aligned} \lambda_x&=p_I+p_X-p_Y-p_Z,\\ \lambda_y&=p_I-p_X+p_Y-p_Z,\\ \lambda_z&=p_I-p_X-p_Y+p_Z. \end{aligned}$$

Conversely,

$$\begin{aligned} p_I&=(1+\lambda_x+\lambda_y+\lambda_z)/4,\\ p_X&=(1+\lambda_x-\lambda_y-\lambda_z)/4,\\ p_Y&=(1-\lambda_x+\lambda_y-\lambda_z)/4,\\ p_Z&=(1-\lambda_x-\lambda_y+\lambda_z)/4. \end{aligned}$$

For a trace-preserving, unital map diagonal in this Pauli basis, complete positivity is equivalent to all four inverse probabilities being nonnegative. Checking only $|\lambda_j|\leq1$ is insufficient. For example, $(\lambda_x,\lambda_y,\lambda_z)=(1,1,-1)$ preserves the Bloch ball but gives $p_Z=-1/2$. It is positive on isolated states but fails on a suitable entangled input. The Pauli-channel treatment derives the resulting tetrahedral geometry and Choi condition.

For $n$ qubits, use phase-free strings $P,Q\in\{I,X,Y,Z\}^{\otimes n}$. If $[P,Q]$ is zero for commuting strings and one for anticommuting strings, then

$$\begin{aligned} \lambda_Q&=\sum_P(-1)^{[P,Q]}p_P,\\ p_P&=4^{-n}\sum_Q(-1)^{[P,Q]}\lambda_Q. \end{aligned}$$

This finite character transform is the multiqubit version of the formulas above. A general joint law still has $4^n-1$ independent parameters; simple algebra does not make an arbitrary correlated distribution small.

Composition, cancellations and depolarization

If successive labels are independently sampled from laws $p$ and $q$, their product has distribution

$$(p*q)_R=\sum_{PQ=R\ {\rm mod\ phase}}p_Pq_Q.$$

Equal labels cancel to identity. For example,

$$(p*q)_I=p_Iq_I+p_Xq_X+p_Yq_Y+p_Zq_Z.$$

The sign characters multiply under label products, so equivalently

$\lambda_j(p*q)=\lambda_j(p)\lambda_j(q).$

As a concrete check, take $p=(0.70,0.10,0.05,0.15)$ and $q=(0.80,0.05,0.10,0.05)$. Their decay triples are $(0.60,0.50,0.70)$ and $(0.70,0.80,0.70)$, whose product is $(0.42,0.40,0.49)$. Inverting gives

$p*q=(0.5775,0.1325,0.1225,0.1675).$

The probability convolution gives the same answer. In contrast, using independent marginals when the successive labels are correlated gives the wrong product law. If exactly the same random Pauli is applied twice, it always cancels even though its one-step averaged channel may be noisy.

A qubit depolarizing channel has equal nonidentity weights $p_X=p_Y=p_Z=p/3$ and common shrink $\eta$. The parameter conversions are

Quantity	Relation
Total nonidentity Pauli probability	$p=1-p_I$
Replacement coordinate	$q=4p/3$
Bloch shrink	$\eta=1-4p/3=1-q$
Entanglement infidelity	$1-F_{\mathrm e}=p$
Average infidelity	$r=2p/3=q/2$

Both fidelities here compare with the identity channel; $F_{\mathrm e}$ uses a maximally entangled purification of the maximally mixed input, and $F_{\mathrm{avg}}$ averages pure-input state fidelity. For a nonidentity target, first form its consistently defined error channel. The depolarizing-channel page derives these conventions in general dimension.

The Pauli mixture is physical for $0\leq p\leq1$, hence $-1/3\leq\eta\leq1$. Only $q\leq1$ has the literal interpretation “replace the state by $I/2$ with probability $q$.” Negative shrink is physical; it is not a negative probability.

For $m$ independent identical steps,

$$\begin{aligned} \eta_m&=\eta^m,\qquad q_m=1-\eta^m,\\ p_m&=\tfrac34(1-\eta^m),\qquad r_m=\tfrac12(1-\eta^m). \end{aligned}$$

Adding $mp$ is only a first-order small-error approximation. It neither gives the probability $1-(1-p)^m$ of at least one sampled fault nor accounts for cancellations in the final channel. A negative eigenvalue also produces alternating signs under perfectly memoryless repetition: oscillation by itself does not establish coherent or non-Markovian noise.

Moving faults through Clifford gates

To move a fault before an ideal unitary $U$ to its output, replace $P$ by $UPU^\dagger$. Clifford gates permute Pauli labels up to irrelevant global phases:

$$\begin{aligned} H:&\quad X\leftrightarrow Z,\\ S:&\quad X\mapsto Y,\quad Z\mapsto Z,\\ \mathrm{CNOT}_{c\to t}:&\quad X_c\mapsto X_cX_t,\\ &\quad Z_t\mapsto Z_cZ_t. \end{aligned}$$

Control-$Z$ and target-$X$ stay on their original qubits. Thus a fault’s circuit location matters: an input $Z_1$ before $H_1$ and then $\mathrm{CNOT}_{1\to2}$ becomes $X_1X_2$, while a $Z_1$ inserted between those gates remains $Z_1$.

An anisotropic Pauli law must therefore be relabeled when moved through a Clifford. A generic non-Clifford can send a Pauli to a non-Pauli operator, so this finite-label description is no longer closed. For Clifford circuits, sampled Pauli faults can be combined with efficient tableau simulation (Aaronson and Gottesman, 2004); stabilizer simulation describes the algorithm. Exact distribution propagation can still require exponentially many strings, and efficient sampling assumes the input law can itself be sampled efficiently.

From physical faults to syndrome and logical error

Assume the input lies in a stabilizer code, the commuting check measurements are ideal, and the noise is the specified stochastic Pauli mixture. Let $s_a(P)=1$ when $P$ anticommutes with generator $S_a$. Then

$\Pr(s)=\sum_{P:s(P)=s}p_P.$

Choose a Pauli recovery $R(s)$ with the same syndrome $s$. The product $R(s(P))P$ then returns to the code and has a logical Pauli class $L(P)$ modulo its stabilizers. Its probability is

$\Pr(L)=\sum_{P:L(P)=L}p_P.$

The recovery condition is essential: an arbitrary operation labeled “recovery” need not return to the code. Noisy checks and recovery gates require modeling the complete correction circuit. The decoder determines the recovery rule; the sums above assume that choice has been made.

For an explicit example, use the three-qubit repetition code with checks $(Z_1Z_2,Z_2Z_3)$. Take $00\mapsto I$, $10\mapsto X_1$, $11\mapsto X_2$, $01\mapsto X_3$. Independent bit flips with probability $t=0.1$ give the following paired errors. Each table entry shows a physical error and its probability.

Syndrome	Recovery leaves logical $I$	Recovery leaves logical $X$
$00$	$III:\ 0.729$	$XXX:\ 0.001$
$10$	$XII:\ 0.081$	$IXX:\ 0.009$
$11$	$IXI:\ 0.081$	$XIX:\ 0.009$
$01$	$IIX:\ 0.081$	$XXI:\ 0.009$

Adding a row gives its syndrome probability; adding the last column gives logical failure. Thus

$$\begin{aligned} \Pr(P\ne I)&=1-0.9^3=0.271,\\ \Pr(s\ne00)&=0.27,\\ \Pr(L=X)&=3(0.009)+0.001=0.028. \end{aligned}$$

The undetected $XXX$ error has logical effect, whereas single flips are detected and corrected. Detected double flips are miscorrected. These differences are physical, not merely different names for one error rate.

Now replace independence by $p_{III}=0.9$, $p_{XXX}=0.1$. Each qubit still has marginal flip probability $0.1$, but the syndrome is always $00$ and logical failure is $0.1$. Marginals therefore do not fix logical performance.

Overlapping checks also correlate syndromes under independent faults. In the independent example, each syndrome bit equals one with probability $0.18$, while both equal one with probability $0.09$. Their covariance is $0.09-0.18^2=0.0576$. Nonzero syndrome correlation alone is not a test for correlated physical noise: compare with the correlation predicted by the independent model and the actual check circuit.

What twirling removes

The ideal Pauli twirl averages an error channel over conjugations:

$$\mathcal T_P(\mathcal E)(\rho) =4^{-n}\sum_R R^\dagger\mathcal E(R\rho R^\dagger)R.$$

It removes off-diagonal Pauli-transfer entries, preserving the diagonal. The result need not be depolarizing. For example, twirling $e^{-i\theta Z/2}$ gives a mixture of $I$ and $Z$ with $p_Z=\sin^2(\theta/2)$; the $Z$ axis remains fixed. A Haar twirl, or exact unitary 2-design, additionally makes the contraction isotropic (Dankert et al., 2009). Both ideal twirls preserve average fidelity relative to the identity, but need not preserve the observable or logical behavior of a particular circuit.

Syndrome measurement likewise removes coherence between different syndromes, not necessarily between different errors within the same syndrome. For the repetition code, let every physical qubit undergo $U=cI-iaX$, where $c=\cos(\theta/2)$ and $a=\sin(\theta/2)$. Expanding $U^{\otimes3}$ and applying the ideal recoveries gives logical branch operators

$$\begin{aligned} K_{00}&=c^3I+ia^3X_L,\\ K_{10}=K_{11}=K_{01}&=-iac^2I-a^2cX_L. \end{aligned}$$

Each branch contains two errors from one row of the table. Forgetting the syndrome gives

$$\begin{aligned} \mathcal E_L(\rho) &=(1-p_L)\rho+p_LX_L\rho X_L -2ic^3a^3[X_L,\rho],\\ p_L&=3a^4c^2+a^6. \end{aligned}$$

The commutator term is absent if each physical rotation is replaced first by its Pauli twirl. Error correction suppresses this coherent contribution at small angle, but does not make it exactly zero. This finite example complements the more general analysis of Beale et al. (2018).

Randomized compiling is an operational protocol that can tailor an ensemble of dressed circuits toward Pauli noise (Wallman and Emerson, 2016). Its assumptions about gates, randomizers and correlations must be checked; an abstract twirl of a fixed channel is not automatically what a device implements.

Learning only what the experiment identifies

Axis-resolved decay measurements probe the $\lambda_Q$, but unknown preparation and measurement errors can obscure their relation to individual gate parameters. Chen et al. (2023) identify combinations of Pauli-noise parameters that are learnable under specified Clifford-gate and SPAM assumptions. One should not infer all $4^n-1$ probabilities from a few marginal decay constants.

Test proposed models on new axes, circuit depths and simultaneous-gate contexts. For example, local depolarization with single-qubit shrink $\eta$ gives $\eta^w$ on a weight-$w$ Pauli observable, whereas global depolarization gives the same shrink on every nonidentity string. A weight-two measurement can distinguish models matched at weight one. When fitted inverse probabilities are negative beyond their uncertainty, investigate the calibration and model; clipping each $\lambda_Q$ separately does not enforce complete positivity.

Exercises

1. Invert Pauli-transfer eigenvalues

In order $(I,X,Y,Z)$, recover the Pauli probabilities from $(\lambda_x,\lambda_y,\lambda_z)=(0.60,0.50,0.70)$. Decide whether the resulting trace-preserving, unital, Pauli-diagonal map is completely positive.

Solution

Apply the four inverse-transform formulas:

$$\begin{aligned} p_I&=0.70,\\ p_X&=0.10,\\ p_Y&=0.05,\\ p_Z&=0.15. \end{aligned}$$

They sum to one and are all nonnegative. Under the stated trace-preserving, unital, Pauli-diagonal assumptions, that is necessary and sufficient for complete positivity. The conclusion would not follow from the three eigenvalue bounds alone for a generic channel representation.

2. Compose two biased Pauli channels

Compose $p=(0.70,0.10,0.05,0.15)$ with $q=(0.80,0.05,0.10,0.05)$ in the fixed phase-free order. Use both Klein-four convolution and transfer-eigenvalue multiplication.

Solution

The identity output collects equal labels:

$$\begin{aligned} (p*q)_I&=0.70(0.80)+0.10(0.05) +0.05(0.10)+0.15(0.05)\\ &=0.5775. \end{aligned}$$

Collecting the four ordered pairs for each remaining phase-free product gives

$$p*q=(0.5775,0.1325,0.1225,0.1675).$$

Separately, $\lambda(p)=(0.60,0.50,0.70)$ and $\lambda(q)=(0.70,0.80,0.70)$, so componentwise multiplication gives $(0.42,0.40,0.49)$. Inverting this triple reproduces the same four probabilities. The agreement checks both the label order and the discarded-phase convention.

3. Convert and repeat depolarizing conventions

A qubit depolarizing step has total nonidentity Pauli probability $p=0.12$. Convert it to $q$, $\eta$, $1-F_{\mathrm e}$, and $r$, then compose five identical memoryless steps.

Solution

For one step,

$$\begin{aligned} q&=0.16,&\eta&=0.84,\\ 1-F_{\mathrm e}&=0.12,&r&=0.08. \end{aligned}$$

Five steps multiply the shrink:

$$\eta_5=0.84^5\simeq0.418212.$$

Therefore $q_5=1-\eta_5\simeq0.581788$, $p_5=3q_5/4\simeq0.436341$, and $r_5=q_5/2\simeq0.290894$ (rounded). The value $5p=0.60$ is wrong because label products can cancel or become other Paulis. Probability addition is only first-order bookkeeping at small error.

4. Propagate a Pauli fault through a Clifford circuit

Qubit $1$ is the control and qubit $2$ the target. An input-frame fault $Z_1$ occurs immediately before $H_1$, followed by controlled-NOT from $1$ to $2$. Transport the fault to the output frame. Then state what changes if the same printed fault occurs between the two gates.

Solution

Conjugation by the Hadamard gives

$$H_1Z_1H_1=X_1.$$

Controlled-NOT sends a control-$X$ to $X_1X_2$, so the input fault becomes $X_1X_2$ in the final frame, up to an irrelevant phase. If $Z_1$ occurs after the Hadamard but before controlled-NOT, it remains $Z_1$ because a control-$Z$ does not spread under that conjugation. The example shows why a Pauli label, qubit order, and circuit location must be recorded together.

5. Push faults to syndrome and logical classes

For the three-qubit repetition code and recovery declared above, reconstruct the four paired rows of the table at $t=0.1$. Compute the at-least-one physical-$X$, nonzero-syndrome, and logical-$X$ probabilities.

Solution

A weight-$w$ pattern has probability $0.1^w0.9^{3-w}$. Anticommutation with $(Z_1Z_2,Z_2Z_3)$ gives $III:00$, $XII:10$, $IXI:11$, $IIX:01$, $XXI:01$, $XIX:11$, $IXX:10$, and $XXX:00$. The chosen recovery corrects all weight-zero and weight-one patterns. Each weight-two pattern and $XXX$ becomes logical $X$ modulo the stabilizer.

Thus

$$\Pr(P\neq I)=1-0.9^3=0.271,$$ $$\begin{aligned} \Pr(s\neq0)&=3(0.081)+3(0.009)\\ &=0.27, \end{aligned}$$

and

$$\begin{aligned} \Pr(L=X)&=3(0.009)+0.001\\ &=0.028. \end{aligned}$$

The undetected physical event is $XXX$, whereas every weight-two event is detected but miscorrected into the same logical class.

6. Compare equal marginals with unequal correlations

Compare independent bit flips at $t=0.1$ with the all-or-none law $\Pr(III)=0.9$, $\Pr(XXX)=0.1$. Which listed observations distinguish them?

Solution

Both laws give marginal $X$ probability $0.1$ on each qubit, so any collection of those three marginals fails to distinguish them. Independence gives nonzero-syndrome probability $0.27$ and logical-$X$ probability $0.028$. The all-or-none law always has syndrome $00$, because both $III$ and $XXX$ commute with the stabilizers, and it has logical-$X$ probability $0.1$.

Syndrome frequencies distinguish the laws immediately in this idealized code, as do three-body correlations or the final logical class. One-qubit calibration data do not. This conclusion is specific to the stated code, measurement, and recovery; it is not a general decoder-performance claim.

7. Check the residual coherent logical error

For the coherent repetition-code example, take $a=\sin(\theta/2)=1/5$ and $c=\sqrt{24}/5$. Start from logical $|0\rangle$. Compare the logical $Y$ expectation with the model obtained by independently Pauli-twirling each physical rotation. Why can the logical-$Z$ population test miss the difference?

Solution

The flip coefficient is $p_L=3(1/25)^2(24/25)+(1/25)^3=73/15625=0.004672$. Since $[X,|0\rangle\langle0|]=-iY$, the coherent channel gives

$\langle Y_L\rangle=-4c^3a^3\simeq-0.030099.$

The independently twirled physical model gives zero. Both give $\langle Z_L\rangle=1-2p_L=0.990656$, so a population measurement alone misses the remaining logical coherence. These statements assume ideal encoding, syndrome measurement and the specified recovery.

References

• S. Aaronson and D. Gottesman, “Improved simulation of stabilizer circuits,” Physical Review A 70, 052328 (2004), doi:10.1103/PhysRevA.70.052328.
• S. J. Beale, J. J. Wallman, M. Gutiérrez, K. R. Brown, and R. Laflamme, “Quantum error correction decoheres noise,” Physical Review Letters 121, 190501 (2018), doi:10.1103/PhysRevLett.121.190501.
• S. Chen, Y. Liu, M. Otten, A. Seif, B. Fefferman, and L. Jiang, “The learnability of Pauli noise,” Nature Communications 14, 52 (2023), doi:10.1038/s41467-022-35759-4.
• C. Dankert, R. Cleve, J. Emerson, and E. Livine, “Exact and approximate unitary 2-designs and their application to fidelity estimation,” Physical Review A 80, 012304 (2009), doi:10.1103/PhysRevA.80.012304.
• D. Gottesman, Stabilizer Codes and Quantum Error Correction, Ph.D. thesis, California Institute of Technology, 1997, arXiv:quant-ph/9705052.
• M. A. Nielsen and I. L. Chuang, Quantum Computation and Quantum Information, 10th anniversary edition, Cambridge University Press, 2010, doi:10.1017/CBO9780511976667.
• J. J. Wallman and J. Emerson, “Noise tailoring for scalable quantum computation via randomized compiling,” Physical Review A 94, 052325 (2016), doi:10.1103/PhysRevA.94.052325.