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Stabilizer measurement on $∣ 0 ⟩$ in the $X$ basis

Example

Start in the one-qubit basis state $∣ 0 ⟩$ , whose stabilizer is $S = ⟨ Z ⟩$ . Measure $X$ , then measure $X$ again. The first measurement has two equiprobable outcomes and updates the signed stabilizer generator to the corresponding $X$ eigenvalue. The second measurement is deterministic; its opposite branch has probability zero and no normalized conditional state.

Facts & Assumptions

Given: The ordered one-qubit computational basis, $∣ 0 ⟩$ , and the Pauli matrices $X, Z$ .

[F1]

The Pauli matrices: $X ∣ 0 ⟩ = ∣ 1 ⟩$ , $Z ∣ 0 ⟩ = ∣ 0 ⟩$ , $Z X = - X Z$ , and $X^{2} = Z^{2} = I$ .

[F2]

Pauli measurement branch semantics: for $M = X$ , the projectors are $(I + X) / 2$ and $(I - X) / 2$ .

[F3]

An anticommuting stabilizer measurement is fair: if the measured Pauli anticommutes with a stabilizer generator, each bit occurs with probability $1 / 2$ .

[F4]

Random Pauli measurement updates stabilizer generators: an allowed bit $b$ replaces the anticommuting pivot by $(- 1)^{b} X$ .

[F5]

Zero-probability Pauli measurement branch: a zero branch map is zero and does not define a normalized conditional state.

Verification

Given: The data in the Example.

1.1F1F3given

The initial generator $Z$ fixes $∣ 0 ⟩$ , and it anticommutes with $X$ by [F1]. Hence [F3] gives $p_{0} = p_{1} = 1 / 2$ for the first measurement.

2.1F1F2F4step 1.1algebra

Directly, $P_{0} ∣ 0 ⟩ = (∣ 0 ⟩ + ∣ 1 ⟩) / 2$ and $P_{1} ∣ 0 ⟩ = (∣ 0 ⟩ - ∣ 1 ⟩) / 2$ . Each vector has squared norm $1 / 2$ , so normalization gives $∣ + ⟩ = (∣ 0 ⟩ + ∣ 1 ⟩) / \sqrt{2}$ and $∣ - ⟩ = (∣ 0 ⟩ - ∣ 1 ⟩) / \sqrt{2}$ . By [F4], their stabilizers are $+ X$ and $- X$ , respectively.

3.1F2step 2.1F5∎

Measuring $X$ again on either state is deterministic: $∣ + ⟩$ lies in $P_{0} H_{1}$ and $∣ - ⟩$ lies in $P_{1} H_{1}$ . The opposite projector annihilates the state, so its branch has probability zero and no normalized conditional state by [F5].

Sources

The $∣ 0 ⟩$ stabilizer and $X$ -eigenstate examples appear in de Wolf §20.1 and Watrous Lesson 14 §14.1, printed pp. 405–407 (PDF pp. 411–413). The branch probabilities and update follow the exact local calculations above: https://arxiv.org/pdf/1907.09415; https://jhwatrous.github.io/UQIC.pdf.

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Used by

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Dependency tree · two levels

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Sources

Ronald de Wolf, Quantum Computing: Lecture Notes, §20.1 and §20.3 (standard reference, not scraped)
John Watrous, Understanding Quantum Information and Computation, Lesson 14 (standard reference, not scraped)

Stabilizer measurement on ∣0⟩∣0⟩ in the XX basis

Example

Facts & Assumptions

Verification

Sources

Depends on

Used by

Dependency tree · two levels

Sources

Stabilizer measurement on $∣ 0 ⟩$ in the $X$ basis