XS-Stabilizer
Xiaotong Ni joint work with Buerschaper, Van den Nest
http://arxiv.org/abs/1404.5327 QEC14
XS-Stabilizer Xiaotong Ni joint work with Buerschaper, Van den Nest - - PowerPoint PPT Presentation
XS-Stabilizer Xiaotong Ni joint work with Buerschaper, Van den Nest QEC14 http://arxiv.org/abs/1404.5327 Definition Pauli-S group: n = h , X, S i n P s S 1 XS = iXZ S = diag(1 , i ) i = Given G = h g 1 , . . . ,
Xiaotong Ni joint work with Buerschaper, Van den Nest
http://arxiv.org/abs/1404.5327 QEC14
We call a state XS-stabilizer state if (uniquely) When not unique, we call it XS-stabilizer code
S = diag(1, i)
Ps
n = hα, X, Si⊗n
α = √ i
G = hg1, . . . , gmi ⇢ Ps
n
|ψi
gj|ψi = |ψi
S−1XS = −iXZ
+1 +i
X ⊗ S
| + 0i
| + 1i | 0i | 1i
(X ⊗ S)2 = I ⊗ Z
+1
P
P(X ⊗ S) is Hermitian
+1 +i
i3S ⊗ S ⊗ S
|001i |010i |100i
Only one of x1, x2, x3 is equal to 1 (Positive) 1-in-3 SAT problem NP-Complete
g1 = X ⊗ S ⊗ I g2 = I ⊗ S ⊗ X
g2
1 = I ⊗ Z ⊗ I = g2 2
g1 = X ⊗ X ⊗ S ⊗ S g2 = S ⊗ S ⊗ X ⊗ X
g1g2g−1
1 g−1 2
= Z ⊗ Z ⊗ Z ⊗ Z
P12 = 1 2(1 + Z ⊗ Z ⊗ Z ⊗ Z)
P1 P2
g1|ψi = g2|ψi = |ψi
P1P12|ψi = P2P12|ψi = P12|ψi = |ψi
subgroup as .
find a computational basis stabilized by .
procedure is efficient.
G = hg1, . . . , gmi ⇢ Ps
n
GD
|ψxi |xi
GD GD
Van den Nest 2011
Each element of G has the form: Zgx1
1 . . . gxm m ,
where Z is generated by {g2
j } ∪ {gjgkg−1 j g−1 k }
So we can write down a set of generators of GD by using linear algebra
g1 = X ⊗ S3 ⊗ S3 ⊗ S ⊗ X ⊗ X, g2 = S3 ⊗ X ⊗ S3 ⊗ X ⊗ S ⊗ X, g3 = S3 ⊗ S3 ⊗ X ⊗ X ⊗ X ⊗ S.
1
X
xj=0
(1)x1x2x3|x1, x2, x3, x2 x3, x1 x3, x1 x2i
Z ⊗ Z ⊗ Z
X
x1,x2
|x1, x2, x1 x2i
|0, x2i $ (1)x2|1, x2i X (1)x1x2|x1, x2i
|0, x2 x3, · · ·i $ ix2+x3(1)x2x3|1, x2 x3, · · ·i X ix1(x2+x3)(1)x1x2x3|x1, x2 x3, · · ·i X ⊗ S ⊗ · · ·
Bravyi, Haah 2012
S S S S S S X X X X X X Z Z Z
X
x is close loops
(1)number of loops|xi
2
Flip a (plaquette) loop, add a quadratic phase
Hu, Wan, Wu 2012
|0, x2 x3, · · ·i $ ix2+x3(1)x2x3|1, x2 x3, · · ·i X ix1(x2+x3)(1)x1x2x3|x1, x2 x3, · · ·i X ⊗ S ⊗ · · ·
X
x1,x2
|x1, x2, x1 x2i
S−1 ⊗ S−1 ⊗ S = CZ12 ⊗ I
If we have X ⊗ √ S ⊗ · · ·
(X ⊗ √ S ⊗ · · · )2 = I ⊗ S ⊗ · · ·
Hard to make it compatible with the string intuition
convenient for you
properties compared to Pauli states (~Flammia, Hamma, Hughes, Wen)
model) have transversal logical-X gate
Some error deserves not to be corrected