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 α , X, S i ⊗ n P s √ S − 1 XS = − iXZ S = diag(1 , i ) i α = • Given G = h g 1 , . . . , g m i ⇢ P s n We call a state XS-stabilizer state if (uniquely) | ψ i g j | ψ i = | ψ i When not unique, we call it XS-stabilizer code

Outline • Operator picture • State picture

Operator picture

Starting point of Pauli stabilizer • Either commute or anti-commute • Each generator evenly split the Hilbert space • Commutativity allows consecutively splitting

Example | + 1 i +i +1 X ⊗ S P ( X ⊗ S ) is Hermitian | � 0 i | + 0 i | � 1 i ( X ⊗ S ) 2 = I ⊗ Z -1 +1 P

Example +i i 3 S ⊗ S ⊗ S +1 | 001 i | 010 i | 100 i Only one of x 1 , x 2 , x 3 is equal to 1 (Positive) 1-in-3 SAT problem NP-Complete

Two operators 1. Commute and independent 2. Commute but not fully independent g 1 = X ⊗ S ⊗ I g 2 = I ⊗ S ⊗ X g 2 1 = I ⊗ Z ⊗ I = g 2 2

Two operators 3. Partially commute g 1 = X ⊗ X ⊗ S ⊗ S P 1 g 2 = S ⊗ S ⊗ X ⊗ X P 2 g 1 g 2 g − 1 1 g − 1 = Z ⊗ Z ⊗ Z ⊗ Z 2 P 12 = 1 2(1 + Z ⊗ Z ⊗ Z ⊗ Z )

Commuting projectors g 1 | ψ i = g 2 | ψ i = | ψ i P 1 P 12 | ψ i = P 2 P 12 | ψ i = P 12 | ψ i = | ψ i

Find codeword state • Given , define diagonal G = h g 1 , . . . , g m i ⇢ P s n subgroup as . G D • We can construct a codeword state , if we can | ψ x i find a computational basis stabilized by . | x i G D • When is generated by Z-type operators, this G D procedure is efficient. Van den Nest 2011

Diagonal subgroup Each element of G has the form: Z g x 1 1 . . . g x m m , where Z is generated by { g 2 j } ∪ { g j g k g − 1 j g − 1 k } So we can write down a set of generators of G D by using linear algebra

Operator picture • Properties of operators • Computational complexity • Equivalent commuting projectors • Find code states

The state picture

The state picture • Concrete • Easiest way to utilize the uniqueness condition • (Innsbruck-Munich influence)

Example g 1 = X ⊗ S 3 ⊗ S 3 ⊗ S ⊗ X ⊗ X, g 2 = S 3 ⊗ X ⊗ S 3 ⊗ X ⊗ S ⊗ X, g 3 = S 3 ⊗ S 3 ⊗ X ⊗ X ⊗ X ⊗ S. 1 X ( � 1) x 1 x 2 x 3 | x 1 , x 2 , x 3 , x 2 � x 3 , x 1 � x 3 , x 1 � x 2 i x j =0

Mechanism Z ⊗ Z ⊗ Z X | x 1 , x 2 , x 1 � x 2 i x 1 ,x 2

Mechanism X ⊗ Z | 0 , x 2 i $ ( � 1) x 2 | 1 , x 2 i X ( � 1) x 1 x 2 | x 1 , x 2 i

Mechanism X ⊗ S ⊗ · · · | 0 , x 2 � x 3 , · · · i $ i x 2 + x 3 ( � 1) x 2 x 3 | 1 , x 2 � x 3 , · · · i X i x 1 ( x 2 + x 3 ) ( � 1) x 1 x 2 x 3 | x 1 , x 2 � x 3 , · · · i Bravyi, Haah 2012

Twisted quantum double • Double semion: X ( � 1) number of loops | x i x is close loops S S X Z X X S S Z X X Z X S S

Twisted quantum double • twisted double on Z n 2 Flip a (plaquette) loop, add a quadratic phase Hu, Wan, Wu 2012

Mechanism X ⊗ S ⊗ · · · | 0 , x 2 � x 3 , · · · i $ i x 2 + x 3 ( � 1) x 2 x 3 | 1 , x 2 � x 3 , · · · i X i x 1 ( x 2 + x 3 ) ( � 1) x 1 x 2 x 3 | x 1 , x 2 � x 3 , · · · i

S-CZ gadget X | x 1 , x 2 , x 1 � x 2 i x 1 ,x 2 S − 1 ⊗ S − 1 ⊗ S = CZ 12 ⊗ I

Why quadratic? √ If we have X ⊗ S ⊗ · · · √ S ⊗ · · · ) 2 = I ⊗ S ⊗ · · · ( X ⊗ Hard to make it compatible with the string intuition

Discussion • Should we add CZ to the Pauli- S group? • There’s some tradeoff. Choose what is the most convenient for you

Other funny facts • XS states have very similar entanglement properties compared to Pauli states (~Flammia, Hamma, Hughes, Wen) • Double semion (and probably other twisted double model) have transversal logical- X gate

Some error deserves not to be corrected Thanks