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Conditional decoded state and decay coherence

PDF p. 8: Eqs. (S1.2)–(S1.3)

Statement

Assume the four-level model with real $ω_{0}, Δ$ , rate $γ \geq 0$ , and time $t \geq 0$ . Let $∣ ψ ⟩ = a ∣ 0 ⟩ + b ∣ 1 ⟩$ be a normalized code state, so $∣ a ∣^{2} + ∣ b ∣^{2} = 1$ . Apply the four-level syndrome recovery $R_{4}$ to its evolved density operator, retaining the null or decay outcome. For $γ t > 0$ , the outcome probabilities are $p_{n u l l} = e^{- γ t}$ and $p_{d e c a y} = 1 - e^{- γ t}$ . The normalized null output is $∣ ψ ⟩ ⟨ ψ ∣$ . The normalized decay output, expressed in the recovered code basis $∣ 0 ⟩, ∣ 1 ⟩$ , is

$ρ_{d} = ∣ a ∣^{2} ∣ 0 ⟩ ⟨ 0 ∣ + ∣ b ∣^{2} ∣ 1 ⟩ ⟨ 1 ∣ + [a b^{*} χ_{d} (t) ∣ 0 ⟩ ⟨ 1 ∣ + h . c .],$

where

$χ_{d} (t) = \frac{γ (e^{i Δ t} - e^{- γ t})}{(γ + i Δ) (1 - e^{- γ t})} .$

Here h.c. denotes the Hermitian conjugate of the preceding off-diagonal term. At $γ t = 0$ the decay outcome has zero probability, so no conditional decay state is assigned.

Facts & Assumptions

[F1]

The postulated model has the ordered orthonormal basis $∣ 0 ⟩, ∣ 1 ⟩, ∣ 2 ⟩, ∣ 3 ⟩$ , code span ${∣ 0 ⟩, ∣ 1 ⟩}$ , decay span ${∣ 2 ⟩, ∣ 3 ⟩}$ , real $ω_{0}, Δ$ , and $γ \geq 0$ . Its evolution is the specified finite GKSL channel. Four-level spontaneous-decay model

[F2]

For the normalized initial code state, the evolved operator is $e^{- γ t} ∣ ψ ⟩ ⟨ ψ ∣ + (1 - e^{- γ t}) (∣ a ∣^{2} ∣ 2 ⟩ ⟨ 2 ∣ + ∣ b ∣^{2} ∣ 3 ⟩ ⟨ 3 ∣) + [a b^{*} c (t) ∣ 2 ⟩ ⟨ 3 ∣ + h . c .]$ , where $c (t) = γ (e^{i Δ t} - e^{- γ t}) / (γ + i Δ)$ when $(γ, Δ) \neq (0, 0)$ and $c (t) = 0$ at $(0, 0)$ . Exact state evolution from a four-level codeword

[F3]

The recovery's null branch is $X \mapsto P X P$ and its decay branch is $X \mapsto U X U^{†}$ , where $P = ∣ 0 ⟩ ⟨ 0 ∣ + ∣ 1 ⟩ ⟨ 1 ∣$ and $U = ∣ 0 ⟩ ⟨ 2 ∣ + ∣ 1 ⟩ ⟨ 3 ∣$ . Four-level syndrome recovery

[F4]

Bras, adjoints and outer products act by the stated finite orthonormal-basis conventions. Finite operator and ket conventions

[F5]

An instrument outcome's probability is the trace of its unnormalized output; its conditional state is that output divided by its probability when positive, and no conditional state is assigned at zero probability. Finite instruments and conditional states

Proof

Given: The model, normalized initial code state, time, and recovery in the statement.

Goal: Compute both outcome probabilities and the positive-probability conditional outputs, including the zero-decay-probability boundary.

1.1givenF1F2F3F4F5algebra

The code and decay spans in [F1] are orthogonal. Applying the null operation in [F3] to [F2] therefore gives $e^{- γ t} ∣ ψ ⟩ ⟨ ψ ∣$ . Its trace is $e^{- γ t}$ because $∣ ψ ⟩$ is normalized. This number is positive for finite $γ, t$ , so [F5] gives the original state as the conditional null output.

1.2givenF1F2F3F4algebra

The operator $U$ annihilates the code basis vectors and maps $∣ 2 ⟩$ to $∣ 0 ⟩$ and $∣ 3 ⟩$ to $∣ 1 ⟩$ . Applying it on both sides of [F2] yields the unnormalized decay output $(1 - e^{- γ t}) (∣ a ∣^{2} ∣ 0 ⟩ ⟨ 0 ∣ + ∣ b ∣^{2} ∣ 1 ⟩ ⟨ 1 ∣) + [a b^{*} c (t) ∣ 0 ⟩ ⟨ 1 ∣ + h . c .]$ . Its trace is $1 - e^{- γ t}$ , since the off-diagonal terms have zero trace and $∣ a ∣^{2} + ∣ b ∣^{2} = 1$ .

2.1givenF2F5step 1.2algebra∎

If $γ t > 0$ , then $1 - e^{- γ t} > 0$ and $γ > 0$ , so $γ + i Δ \neq 0$ . Divide the output from step 1.2 by its trace and substitute the expression for $c (t)$ in [F2]. This gives the displayed $ρ_{d}$ and $χ_{d} (t)$ . If $γ t = 0$ , the decay probability in step 1.2 is zero, and [F5] assigns no conditional decay state. Thus all asserted branch probabilities and conditional outputs follow.

Depends on

Used by · 11

Dependency tree · 2 levels

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Sources

Monolithic Quantum Error Correction in the Presence of Distinguishability · PDF p. 8, Eqs. (S1.2)–(S1.3)