XS-Stabilizer Xiaotong Ni joint work with Buerschaper, Van den Nest - - PowerPoint PPT Presentation

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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 , . . . ,


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XS-Stabilizer

Xiaotong Ni joint work with Buerschaper, Van den Nest

http://arxiv.org/abs/1404.5327 QEC14

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  • Pauli-S group:
  • Given

We call a state XS-stabilizer state if (uniquely) When not unique, we call it XS-stabilizer code

Definition

S = diag(1, i)

Ps

n = hα, X, Si⊗n

α = √ i

G = hg1, . . . , gmi ⇢ Ps

n

|ψi

gj|ψi = |ψi

S−1XS = −iXZ

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Outline

  • Operator picture
  • State picture
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Operator picture

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Starting point of Pauli stabilizer

  • Either commute or anti-commute
  • Each generator evenly split the Hilbert space
  • Commutativity allows consecutively splitting
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Example

+1 +i

X ⊗ S

| + 0i

| + 1i | 0i | 1i

(X ⊗ S)2 = I ⊗ Z

+1

P

P(X ⊗ S) is Hermitian

  • 1
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Example

+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

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Two operators

  • 1. Commute and independent
  • 2. Commute but not fully independent

g1 = X ⊗ S ⊗ I g2 = I ⊗ S ⊗ X

g2

1 = I ⊗ Z ⊗ I = g2 2

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Two operators

  • 3. Partially commute

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

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Commuting projectors

g1|ψi = g2|ψi = |ψi

P1P12|ψi = P2P12|ψi = P12|ψi = |ψi

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  • Given , define diagonal

subgroup as .

  • We can construct a codeword state , if we can

find a computational basis stabilized by .

  • When is generated by Z-type operators, this

procedure is efficient.

Find codeword state

G = hg1, . . . , gmi ⇢ Ps

n

GD

|ψxi |xi

GD GD

Van den Nest 2011

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Diagonal subgroup

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

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Operator picture

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

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The state picture

  • Concrete
  • Easiest way to utilize the uniqueness condition
  • (Innsbruck-Munich influence)
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Example

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

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Mechanism

Z ⊗ Z ⊗ Z

X

x1,x2

|x1, x2, x1 x2i

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Mechanism

X ⊗ Z

|0, x2i $ (1)x2|1, x2i X (1)x1x2|x1, x2i

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Mechanism

|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

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Twisted quantum double

  • Double semion:

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

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Twisted quantum double

  • twisted double on Zn

2

Flip a (plaquette) loop, add a quadratic phase

Hu, Wan, Wu 2012

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Mechanism

|0, x2 x3, · · ·i $ ix2+x3(1)x2x3|1, x2 x3, · · ·i X ix1(x2+x3)(1)x1x2x3|x1, x2 x3, · · ·i X ⊗ S ⊗ · · ·

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S-CZ gadget

X

x1,x2

|x1, x2, x1 x2i

S−1 ⊗ S−1 ⊗ S = CZ12 ⊗ I

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Why quadratic?

If we have X ⊗ √ S ⊗ · · ·

(X ⊗ √ S ⊗ · · · )2 = I ⊗ S ⊗ · · ·

Hard to make it compatible with the string intuition

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Discussion

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

convenient for you

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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

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Thanks

Some error deserves not to be corrected