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Let V and V be the set of qubits in the red |
1 2 1. If p 1 ǫ and p <1 ǫ, then |
Gpass ψpass |
dotted box and the blue dotted box in Fig. 1, ≥ − − |
respectively. Alice stores qubits in V 2 in her px∈/L < q+ 1 −q + 1 −q (1 ǫ) β . |
memory, and measures each qubit in V in Z acc 2 2 − ≡ 1 |
1 |
4 |
2. If p <1 ǫ and p 1 ǫ, then which means |
Gpass ψpass |
− ≥ − |
px ac∈/ cL < q+ 1 − q (1 −ǫ)+ 1 − q =β 1. (cid:12)Tr(Πρ) −Tr(ΠG Ψ′) (cid:12)≤δ |
2 2 |
(cid:12) (cid:12) |
for the POVM element Π corresponding to the accep- |
3. If p Gpass <1 ǫ and p ψpass <1 ǫ, then tance of the measurement-based quantum computing. |
− − |
Therefore, the total acceptance probability, px∈/L, is |
1 q 1 q acc |
px∈/L < q+ − (1 ǫ)+ − (1 ǫ) |
acc 2 − 2 − 1 q 1 q |
px∈/L q(b+δ)+ − + − |
= q+(1 q)(1 ǫ) β 2. acc ≤ 2 2 |
− − ≡ |
= q(b+δ)+(1 q) β . |
3 |
− ≡ |
Letus considerthe remainingcase, p 1 ǫ and |
Gpass |
≥ − |
p ψpass 1 ǫ. From the triangle inequality and the Let us define |
≥ − |
invariance of the trace norm under a unitary operation, |
we obtain ǫ(1 q) |
∆ (q) α β =q(a 1)+ − , |
1 1 |
≡ − − 2 |
1 1 |
2 ρ −G Ψ′ 1 = 2 ρ −G σ+G σ −G Ψ′ 1 ∆ 2(q) ≡ α −β 2 =q(a −1)+ǫ(1 −q), |
(cid:13) (cid:13) (cid:13) (cid:13) ∆ (q) α β =q(a b δ). |
(cid:13) (cid:13) 1(cid:13) 1 (cid:13) 3 ≡ − 3 − − |
≤ 2 ρ −G σ 1+ 2 G σ −G Ψ′ 1 The optimal value |
(cid:13) (cid:13) (cid:13) (cid:13) |
= 1(cid:13) ρ G (cid:13) + 1(cid:13) σ Ψ′ Ψ(cid:13) ′ , ǫ |
2 − σ 1 2 −| ih | 1 q∗ 2 |
(cid:13) (cid:13) (cid:13) (cid:13) ≡ 1+ ǫ b δ |
(cid:13) (cid:13) (cid:13) (cid:13) 2 − − |
where G Ψ′ G Ψ′ G Ψ′ , |
≡| ih | |
of q is that satisfies ∆ (q) = ∆ (q). Then, if we take |
1 3 |
G CZ (σ G G) CZ , a=1 2−r and b=2−r for a polynomial r, |
σ ≡ e ⊗| ih | e − |
(cid:0)e∈EO (cid:1) (cid:0)e∈EO (cid:1) |
connect connect ǫ(a b δ) |
∆ (q∗) = 2 − − |
and σ is any 3m-qubit state on V 2. Since p Gpass 1 ǫ, 3 1+ ǫ b δ |
the first term is upperbounded as ≥ − 2 − − |
ǫ 2 |
1 2−r+1 2√2ǫ +ǫ |
1 ρ G √2ǫ, ≥ 4 (cid:16) − − −r3 (cid:17) |
2 − σ 1 ≤ 1 |
(cid:13) (cid:13) . |
(cid:13) (cid:13) ≥ poly(x) |
from Eq. (1). We can show that if p ψpass 1 ǫ, the | | |
≥ − |
second term is upperbounded as |
As usual, the inverse polynomial gap can be amplified |
with a polynomial overhead. |
1 2 |
σ Ψ′ Ψ′ √2ǫ+ +ǫ. (3) |
2 −| ih | 1 ≤ r3 |
(cid:13) (cid:13) |
(cid:13) (cid:13) |
The proof is given in Appendix. Acknowledgments |
Therefore, we obtain |
TM is supported by the Grant-in-Aid for Scientific |
1 ρ −G Ψ′ 2√2ǫ+ r2 +ǫ R Jae ps aea nr ,ch ano dn thIn en Gov ra at niv t-e inA -Are idas foN ro. Y1 o5 uH n0 g08 S5 c0 ieo nf tisM tsE (X BT |
2 1 ≤ 3 ) |
(cid:13) (cid:13) |
(cid:13) (cid:13) δ, No.26730003of JSPS. |
≡ |
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