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Appendix: Proof of Eq. (3) |
In this appendix, we show Eq. (3). Due to the triangle inequality and the invariance of the trace norm under a |
unitary operation, |
3m 3m |
1 σ Ψ′ Ψ′ = 1 σ XxjZzj P ψ 0 ⊗m + ⊗m P† XxjZzj |
2 (cid:13) (cid:13) −| ih | (cid:13) (cid:13)1 2(cid:13) (cid:13) (cid:13) − (cid:0)Oj=1 j j (cid:1) (cid:0) ⊗ ⊗ (cid:1) (cid:0)Oj=1 j j (cid:1)(cid:13) (cid:13) (cid:13)1 |
3m 3m |
= 1 P† XxjZzj σ XxjZzj P ψ 0 ⊗m + ⊗m |
2(cid:13) (cid:13) (cid:0)Oj=1 j j (cid:1) (cid:0)Oj=1 j j (cid:1) − ⊗ ⊗ (cid:13) (cid:13)1 |
(cid:13) (cid:13) |
3m 3m |
1 |
P† XxjZzj σ XxjZzj P ρ |
≤ 2(cid:13) (cid:13) (cid:0)Oj=1 j j (cid:1) (cid:0)Oj=1 j j (cid:1) − before (cid:13) (cid:13)1 |
(cid:13)1 (cid:13) |
⊗m ⊗m |
+ ρ ψ 0 + , |
before |
2(cid:13) − ⊗ ⊗ (cid:13)1 |
(cid:13) (cid:13) |
(cid:13) (cid:13) |
where ρ is the state before measuring Λ ,Λ . |
before 0 1 |
{ } |
From the monotonicity of the trace distance under a CPTP map, the first term is upperbounded as |
3m 3m |
1 1 |
P† XxjZzj σ XxjZzj P ρ G ρ √2ǫ. |
2(cid:13) (cid:13) (cid:0)Oj=1 j j (cid:1) (cid:0)Oj=1 j j (cid:1) − before (cid:13) (cid:13)1 ≤ 2k σ − k1 ≤ |
(cid:13) (cid:13) |
As is shown below, the second term is upperbounded as |
1 ⊗m ⊗m 2 |
ψ 0 + ρ +ǫ. (A.1) |
2(cid:13) ⊗ ⊗ − before (cid:13)1 ≤ r3 |
(cid:13) (cid:13) |
(cid:13) (cid:13) |
6 |
Therefore, we have shown Eq. (3). |
Let us show Eq. (A.1). Note that |
ρ = 1 1 P†σ E σ P ψ 0 ⊗m + ⊗m P†σ E†σ P, |
before (3m)!43m α k α ⊗ ⊗ α k α |
PX,α,k (cid:0) (cid:1) |
where σ is a 3m-qubit Pauli operator,and E is a Kraus operator. Let us decompose each Kraus operator in terms |
α k |
of Pauli operators as E = Ckσ . Since |
k β β β |
P |
I = E†E |
k k |
Xk |
= Ck∗Ckσ σ |
β γ β γ |
kX,β,γ |
= Ck 2I + Ck∗Ckσ σ , |
| β| β γ β γ |
Xk,β kX,β6=γ |
we obtain |
Ck 2 =1. |
| β| |
Xk,β |
Then, |
ρ = 1 1 CkCk∗P†σ σ σ P ψ 0 ⊗m + ⊗m P†σ σ σ P |
before (3m)!43m β γ α β α ⊗ ⊗ α γ α |
P,αX,k,β,γ (cid:0) (cid:1) |
= 1 Ck 2P†σ P ψ 0 ⊗m + ⊗m P†σ P |
(3m)! | β| β ⊗ ⊗ β |
PX,k,β (cid:0) (cid:1) |
= 1 D P†σ P ψ 0 ⊗m + ⊗m P†σ P |
β β β |
(3m)! ⊗ ⊗ |
XP,β (cid:0) (cid:1) |
= D ψ 0 ⊗m + ⊗m + 1 D P†σ P ψ 0 ⊗m + ⊗m P†σ P |
0 β β β |
⊗ ⊗ (3m)! ⊗ ⊗ |
(cid:0) (cid:1) PX,β6=0 (cid:0) (cid:1) |
ρ +ρ . |
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