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Exact state evolution from a four-level codeword

PDF p. 8: Eq. (S1.1)

Statement

Assume the four-level model with real ω0,Δ, decay rate γ≥0, and t≥0. For a normalized initial code state ∣ψ⟩=a∣0⟩+b∣1⟩, its evolved density operator is

ρ(t)=e−γt∣ψ⟩⟨ψ∣+(1−e−γt)(∣a∣2∣2⟩⟨2∣+∣b∣2∣3⟩⟨3∣)+[ab∗c(t)∣2⟩⟨3∣+h.c.],

where c(t)=γ(eiΔt−e−γt)/(γ+iΔ) when (γ,Δ)≠(0,0) and c(t)=0 at (γ,Δ)=(0,0). Here h.c. denotes the Hermitian conjugate of the preceding off-diagonal term. The value at the origin is the continuous extension.

Facts & Assumptions

[F1]

In the postulated model, H=ω0(∣0⟩⟨0∣+∣1⟩⟨1∣)+Δ∣3⟩⟨3∣, L=γ(∣2⟩⟨0∣+∣3⟩⟨1∣), and the state evolves by the corresponding finite GKSL channel. Four-level spontaneous-decay model

[F2]

The ordered basis is orthonormal; bras, adjoints and outer products act according to the finite operator conventions. Finite operator and ket conventions

[F3]

The finite GKSL interface defines Nt=etL for the stated Hermitian Hamiltonian and jump. Finite GKSL evolution and the QEC channel interface

[F4]

For finite GKSL data, the channel is the sum of the no-jump term and ordered integrals of jump amplitudes, with G=−iH−L†L/2 and E(u)=euG. Finite no-jump and ordered-jump expansion

[F5]

A density operator on a nonzero finite-dimensional complex inner-product space is a non-negative endomorphism with trace one. Density operator

[F6]

A pure state is a ray, and a unit representative u has operator representative ∣u⟩⟨u∣. Use this operator representative for the initial code state. Pure state and ray

[F7]

On a nonzero finite-dimensional complex inner-product space with the root pairing linear in its first argument, the outer product ∣u⟩⟨u∣ of a unit vector is non-negative and has trace one, hence is a density operator. A unit-vector outer product is a density operator

Proof

Given: The model, normalized ∣ψ⟩, and nonnegative time in the statement.

Goal: Calculate Nt(∣ψ⟩⟨ψ∣) and establish the stated extension.

1.1givenF1F2F3F5F6F7algebra

By [F6], the initial pure code state has operator representative ρ(0)=∣ψ⟩⟨ψ∣. Its underlying space is C4 with the root pairing linear in its first argument, and the normalized vector ∣ψ⟩ is unit. The outer-product action in [F2] is ∣ψ⟩⟨ψ∣x=ψ(ψ†x)=⟨x,ψ⟩rootψ, which agrees with the convention in [F7]. Thus [F7] and [F5] make ρ(0) a density operator. By [F1] and [F3], its physical model evolution is ρ(t)=Nt(ρ(0)).

2.1givenF1F2F4step 1.1algebra

By [F1] and [F2], L†L=γ(∣0⟩⟨0∣+∣1⟩⟨1∣). Thus E(u) acts by e(−iω0−γ/2)u on both code basis vectors, by 1 on ∣2⟩, and by e−iΔu on ∣3⟩. In particular, the no-jump contribution in [F4] is e−γt∣ψ⟩⟨ψ∣.

3.1givenF1F2F4step 2.1algebra

For 0≤s≤t, the single-jump amplitude on the initial state is E(t−s)LE(s)∣ψ⟩=γe(−iω0−γ/2)s(a∣2⟩+be−iΔ(t−s)∣3⟩). Its image lies in the span of ∣2⟩,∣3⟩, which is preserved by E(u) and annihilated by L. Hence every term with two or more jumps vanishes on the initial density operator. The single-jump diagonal coefficients are ∣a∣2(1−e−γt) and ∣b∣2(1−e−γt).

4.1givenF3F4step 2.1step 3.1algebra∎

The single-jump ∣2⟩⟨3∣ coefficient is ab∗γ∫0te−γseiΔ(t−s) ds. For (γ,Δ)≠(0,0), direct integration gives ab∗γ(eiΔt−e−γt)/(γ+iΔ); the opposite coefficient is its complex conjugate. At (0,0) the integral is zero, and its absolute value is at most γt, so this value is continuous as (γ,Δ)→(0,0) within γ≥0. Combining the no-jump and single-jump terms gives the displayed density operator.

Depends on

Used by · 3

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