Day 6 — Estimator & Error Mitigation

Math Prep

Solve this BEFORE lesson.md. EstimatorV2 returns one number per (circuit, observable): $\langle O \rangle$ . This prep makes that number concrete, then gives the minimal noise math behind "error mitigation."

Requires: Day 3 math prep (Paulis, expectation values) + Day 4 §4–5. Video back-up: Watrous Lesson 09 Density Matrices, Lesson 10 Quantum Channels; notes in resource-artifactory/notes/video-notes/.

1. Observables = Hermitian operators = weighted Pauli sums

Any measurable quantity is a Hermitian operator $O$ ; any $n$ -qubit Hermitian operator expands uniquely in the Pauli basis:

O = \sum_{P \in \{I,X,Y,Z\}^{\otimes n}} c_P\, P, \qquad c_P = \tfrac{1}{2^n}\mathrm{Tr}(PO),\; c_P \in \mathbb{R}

That's precisely what SparsePauliOp([...], coeffs) encodes. Examples:

$|1\rangle\langle1| = \tfrac{1}{2}(I - Z)$ — "probability of 1" as an observable.
$|0\rangle\langle0| = \tfrac{1}{2}(I + Z)$ .
Ising-type Hamiltonian: $H = ZZ + 0.5\,XI + 0.5\,IX$ .

Linearity does the work: $\langle O\rangle = \sum_P c_P \langle P\rangle$ — the Estimator measures each Pauli term (grouping commuting ones into shared measurement bases) and sums with coefficients. $\langle I\cdots I\rangle = 1$ always, so identity terms shift the answer by their coefficient with zero variance.

2. Values and bounds

$\langle O\rangle$ lies between $\lambda_{\min}(O)$ and $\lambda_{\max}(O)$ . For a single Pauli: $[-1, +1]$ . Reference values (compute these until instant):

State	$\langle Z\rangle$	$\langle X\rangle$	$\langle Y\rangle$
$\\|0\rangle$	$+1$	0	0
$\\|1\rangle$	$-1$	0	0
$\\|+\rangle$	0	$+1$	0
$\\|-\rangle$	0	$-1$	0
$\\|{+i}\rangle$	0	0	$+1$
$R_y(\theta)\\|0\rangle$	$\cos\theta$	$\sin\theta$	0

Bell $|\Phi^+\rangle$ : $\langle ZZ\rangle = \langle XX\rangle = +1$ , $\langle YY\rangle = -1$ , all single-qubit Paulis $= 0$ .

Variance of a Pauli estimate: outcomes are ±1, so $\mathrm{Var} = 1 - \langle P\rangle^2$ , and the shot-noise on $\widehat{\langle P\rangle}$ is $\sqrt{(1-\langle P\rangle^2)/N}$ . Estimator's precision parameter is exactly this target standard error.

3. How noise moves expectation values

Model noise as a channel (Day 3's density matrices earn their keep). The two canonical ones:

Depolarizing noise — with probability $p$ , replace the state by $I/2^n$ :

\rho \mapsto (1-p)\rho + p\,\tfrac{I}{2^n} \;\;\Rightarrow\;\; \langle P\rangle_{\text{noisy}} = (1-p)\langle P\rangle_{\text{ideal}}

Noise shrinks expectation values toward 0 (for traceless observables). Under $k$ noisy layers: $(1-p)^k$ — exponential signal decay with depth.

Readout (measurement) error — classical bit-flip on the recorded outcome with probabilities $\epsilon_{0\to1}, \epsilon_{1\to0}$ . Effect on a single-qubit $\langle Z\rangle$ with symmetric $\epsilon$ : $\langle Z\rangle_{\text{meas}} = (1-2\epsilon)\langle Z\rangle_{\text{true}}$ — again a shrink factor.

4. Why mitigation methods work (one line of math each)

TREX / readout twirling: measure with random X-flips folded into readout; converts asymmetric readout error into a known symmetric shrink factor, then divide it out.
ZNE (zero-noise extrapolation): deliberately amplify noise by factor $\lambda$ (e.g. gate folding $G \to GG^\dagger G$ ), measure $\langle O\rangle(\lambda)$ at $\lambda = 1, 3, 5$ , fit, extrapolate to $\lambda = 0$ . Works because signal $\approx (1-p)^{\lambda k}$ is smooth in λ. Costs extra circuits; extrapolation can overshoot (estimate, not guarantee).
Pauli twirling: conjugate each 2q gate by random Paulis that leave the ideal gate invariant; averages coherent errors into stochastic (depolarizing-like) ones, which ZNE/PEC model better.
PEC (probabilistic error cancellation): write the ideal gate as a quasi-probability combination of noisy implementable ones and sample it — unbiased but with sampling overhead $\gamma^{2k}$ , exponential in depth.
DD (dynamical decoupling): insert X–X (or XY4) pulse pairs into idle windows; the pair is identity ( $XX = I$ ) but refocuses slow dephasing — a spin echo.

You don't need implementation depth for the exam — you need: what each acronym targets (readout vs gate noise vs idling), whether it's biased, and its cost.

5. Estimator vs Sampler in one probability statement

Sampler: returns samples from $P(x)$ — you post-process.
Estimator: returns $\widehat{\langle O\rangle} \pm$ std, having internally chosen measurement bases (X/Y terms need basis rotations — Day 4 §5), grouped commuting terms, and applied mitigation.

If the observable is diagonal (all I/Z), you could get the same number from Sampler counts: $\langle Z_i\rangle = P(\text{bit}_i{=}0) - P(\text{bit}_i{=}1)$ . For X/Y terms you cannot, without adding basis-change gates yourself.

PROBLEMS

P1. Expand $|11\rangle\langle11|$ (2 qubits) in the Pauli basis. (Hint: $|1\rangle\langle1| = \tfrac12(I-Z)$ per qubit; tensor and multiply out.)

P2. For $|\psi\rangle = R_y(\pi/3)|0\rangle$ , compute $\langle Z\rangle$ and $\langle X\rangle$ , then $\langle 2Z + X\rangle$ .

P3. For the Bell state, compute $\langle O \rangle$ with $O = \tfrac12(ZZ + XX)$ . Must an expectation value be an eigenvalue of $O$ ?

P4. Global depolarizing noise with $p = 0.2$ acts on the Bell state. Give $\langle ZZ\rangle_{\text{noisy}}$ . Now two such layers. Generalize to $k$ layers.

P5. Symmetric readout error $\epsilon = 0.05$ on ONE qubit of a Bell pair (other qubit read perfectly): what's the measured $\langle ZZ\rangle$ ?

P6. ZNE gives $\langle O\rangle = 0.60$ at noise factor 1 and $0.36$ at factor 3; assume exponential decay $\langle O\rangle(\lambda) = A e^{-b\lambda}$ . Extrapolate to $\lambda = 0$ .

P7. An estimator must resolve $\langle P \rangle \approx 0$ to precision 0.01. Using $\mathrm{Var} = 1 - \langle P\rangle^2$ , how many effective shots does the term need?

P8. True/false with reason: "Increasing shots reduces the bias caused by depolarizing noise."

ANSWERS

A1. $|11\rangle\langle11| = \tfrac12(I{-}Z)\otimes\tfrac12(I{-}Z) = \tfrac14(II - IZ - ZI + ZZ)$ .

A2. $\langle Z\rangle = \cos\frac{\pi}{3} = \tfrac12$ ; $\langle X\rangle = \sin\frac{\pi}{3} = \tfrac{\sqrt3}{2}$ . Linearity: $2\cdot\tfrac12 + \tfrac{\sqrt3}{2} = 1 + \tfrac{\sqrt3}{2} \approx 1.87$ .

A3. $\langle O\rangle = \tfrac12(1 + 1) = 1$ . Here 1 happens to be an eigenvalue ( $|\Phi^+\rangle$ is an eigenvector), but in general no — $\langle O\rangle$ is a weighted average of eigenvalues (e.g. $\langle Z\rangle_{|+\rangle} = 0$ , which is not ±1).

A4. One layer: $(1-p)\cdot1 = 0.8$ . Two: $0.8^2 = 0.64$ . $k$ : $0.8^k$ — exponential signal decay with depth, the fundamental depth-limit of noisy hardware.

A5. The recorded parity flips exactly when the noisy bit flips: probability $\epsilon$ . $\langle ZZ\rangle_{\text{meas}} = (1-2\epsilon)\cdot1 = 0.9$ .

A6. $e^{-2b} = 0.36/0.60 = 0.6$ ⇒ $b = -\tfrac12\ln 0.6 \approx 0.255$ . $A = 0.60\,e^{b} \approx 0.775$ . Extrapolated ideal value ≈ 0.775 — larger than any measured value; mitigation "un-shrinks" the signal.

A7. Var ≈ 1, need std 0.01: $N = 1/0.01^2 = 10\,000$ effective shots for that term.

A8. False. Shots shrink the statistical error bar around the noisy mean $(1-p)\langle O\rangle$ ; the bias (gap to the ideal value) is untouched. Removing bias is mitigation's job (ZNE/PEC/TREX), not the shot budget's.

Self-check: you can Pauli-decompose small projectors, compute ⟨Pauli sums⟩ by linearity, predict the $(1-p)^k$ shrink, and say in one sentence each what DD, TREX, ZNE, PEC target.