The LU-LC Conjecture is FALSE∗ Zhengfeng Ji† December 18th, 2007 @ QIP 2008

∗ †

Joint work with Jianxin Chen, Zhaohui Wei and Mingsheng Ying State Key Laboratory of Computer Science, Institute of Software, Chinese Academy of Sciences

Introduction

2 / 23

The Problem The LU-LC conjecture considers the local equivalence of stabilizer states (graph states).

3 / 23

The Problem The LU-LC conjecture considers the local equivalence of stabilizer states (graph states). ■ ■

Stabilizer states Graph states

3 / 23

The Problem The LU-LC conjecture considers the local equivalence of stabilizer states (graph states). ■ ■

Stabilizer states Graph states 1

X I X Z Z

Z X I X Z

Z Z X I Z

X Z Z X Z

I X Z Z Z

X

2 Z

Z 5

3

3 / 23

4

Z

The Problem The LU-LC conjecture considers the local equivalence of stabilizer states (graph states). ■ ■

Stabilizer states Graph states 1

X I X Z Z ■

Z X I X Z

Z Z X I Z

X Z Z X Z

I X Z Z Z

X

2 Z

Z 5

3

Local equivalence

3 / 23

4

Z

The Problem The LU-LC conjecture considers the local equivalence of stabilizer states (graph states). ■ ■

Stabilizer states Graph states 1

X I X Z Z ■

Z X I X Z

Z Z X I Z

X Z Z X Z

I X Z Z Z

X

2 Z

Z 5

3

4

Z

Local equivalence •

Given two entangled states, decide whether they are locally equivalent or not.

3 / 23

The Problem The LU-LC conjecture considers the local equivalence of stabilizer states (graph states). ■ ■

Stabilizer states Graph states 1

X I X Z Z ■

Z X I X Z

Z Z X I Z

X Z Z X Z

I X Z Z Z

X

2 Z

Z 5

3

4

Z

Local equivalence Given two entangled states, decide whether they are locally equivalent or not. • SLOCC, LOCC, LU, LC •

3 / 23

The Problem The LU-LC conjecture considers the local equivalence of stabilizer states (graph states). ■ ■

Stabilizer states Graph states 1

X I X Z Z ■

Z X I X Z

Z Z X I Z

X Z Z X Z

I X Z Z Z

X

2 Z

Z 5

3

4

Z

Local equivalence Given two entangled states, decide whether they are locally equivalent or not. • SLOCC, LOCC, LU, LC (Local Clifford Operation) •

3 / 23

The Problem The LU-LC conjecture considers the local equivalence of stabilizer states (graph states). ■ ■

Stabilizer states Graph states 1

X I X Z Z ■

Z X I X Z

Z Z X I Z

X Z Z X Z

I X Z Z Z

X

2 Z

Z 5

3

4

Z

Local equivalence Given two entangled states, decide whether they are locally equivalent or not. • SLOCC = LOCC = LU, LC •

3 / 23

The Problem The LU-LC conjecture considers the local equivalence of stabilizer states (graph states). ■ ■

Stabilizer states Graph states 1

X I X Z Z ■

Z X I X Z

Z Z X I Z

X Z Z X Z

I X Z Z Z

X

2 Z

Z 5

3

4

Z

Local equivalence Given two entangled states, decide whether they are locally equivalent or not. • SLOCC = LOCC = LU = ? LC •

3 / 23

The Problem The LU-LC conjecture considers the local equivalence of stabilizer states (graph states). ■ ■

Stabilizer states Graph states 1

X I X Z Z ■

Z X I X Z

Z Z X I Z

X Z Z X Z

I X Z Z Z

X

2 Z

Z 5

3

4

Z

Local equivalence Given two entangled states, decide whether they are locally equivalent or not. • SLOCC = LOCC = LU = ? LC •

■

The conjecture 3 / 23

Previous Progresses

1. LU/LC invariants for stabilizer states M. van den Nest, J. Dehaene, and B. Moor

2. Proved for subsets of stabilizer states M. van den Nest, J. Dehaene, and B. Moor B. Zeng, H. Chung, A. W. Cross, and I. L. Chuang

3. Proved for states with up to 7 qubits M. Hein, J. Eisert, and R. Raussendorf et. al.

4. ......

4 / 23

Previous Progresses

5 / 23

Previous Progresses 1. DLU-LC ⇔ LU-LC

5 / 23

Previous Progresses 1. DLU-LC ⇔ LU-LC D. Gross and M. van den Nest B. Zeng, A. Cross, and I. L. Chuang

5 / 23

Previous Progresses 1. DLU-LC ⇔ LU-LC

3→

1

D. Gross and M. van den Nest B. Zeng, A. Cross, and I. L. Chuang

5 / 23

Previous Progresses 1. DLU-LC ⇔ LU-LC

3→

1

D. Gross and M. van den Nest B. Zeng, A. Cross, and I. L. Chuang

2. An explicit formula of stabilizer states

5 / 23

Previous Progresses 1. DLU-LC ⇔ LU-LC

3→

1

D. Gross and M. van den Nest B. Zeng, A. Cross, and I. L. Chuang

2. An explicit formula of stabilizer states J. Dehaene and B. Moor

5 / 23

Previous Progresses 1. DLU-LC ⇔ LU-LC

3→

1

D. Gross and M. van den Nest B. Zeng, A. Cross, and I. L. Chuang

2. An explicit formula of stabilizer states J. Dehaene and B. Moor

1 X l(x) p i (−1)q(x) |xi |T | x∈T

5 / 23

Previous Progresses 1. DLU-LC ⇔ LU-LC

3→

1

D. Gross and M. van den Nest B. Zeng, A. Cross, and I. L. Chuang

2. An explicit formula of stabilizer states J. Dehaene and B. Moor

5 / 23

Previous Progresses 1. DLU-LC ⇔ LU-LC

3→

1

D. Gross and M. van den Nest B. Zeng, A. Cross, and I. L. Chuang

2. An explicit formula of stabilizer states J. Dehaene and B. Moor

3. Quadratic Form Phase Problem (QFP) D. Gross and M. van den Nest

5 / 23

Previous Progresses 1. DLU-LC ⇔ LU-LC

3→

1

D. Gross and M. van den Nest B. Zeng, A. Cross, and I. L. Chuang

2. An explicit formula of stabilizer states J. Dehaene and B. Moor

3. Quadratic Form Phase Problem (QFP) D. Gross and M. van den Nest

QFP ⇒ LU-LC

5 / 23

Previous Progresses 1. DLU-LC ⇔ LU-LC

3→

1

D. Gross and M. van den Nest B. Zeng, A. Cross, and I. L. Chuang

2. An explicit formula of stabilizer states J. Dehaene and B. Moor

3. Quadratic Form Phase Problem (QFP) D. Gross and M. van den Nest

QFP ⇒ LU-LC In this work: The QFP problem is our starting point.

5 / 23

Previous Progresses 1. DLU-LC ⇔ LU-LC

3→

1

D. Gross and M. van den Nest B. Zeng, A. Cross, and I. L. Chuang

2. An explicit formula of stabilizer states J. Dehaene and B. Moor

3. Quadratic Form Phase Problem (QFP) D. Gross and M. van den Nest

QFP ⇒ LU-LC In this work: The QFP problem is our starting point. QFP is FALSE

5 / 23

Previous Progresses 1. DLU-LC ⇔ LU-LC

3→

1

D. Gross and M. van den Nest B. Zeng, A. Cross, and I. L. Chuang

2. An explicit formula of stabilizer states J. Dehaene and B. Moor

3. Quadratic Form Phase Problem (QFP) D. Gross and M. van den Nest

QFP ⇒ LU-LC In this work: The QFP problem is our starting point. QFP is FALSE LU-LC is FALSE

5 / 23

Previous Progresses 1. DLU-LC ⇔ LU-LC

3→

1

D. Gross and M. van den Nest B. Zeng, A. Cross, and I. L. Chuang

2. An explicit formula of stabilizer states J. Dehaene and B. Moor

3. Quadratic Form Phase Problem (QFP) D. Gross and M. van den Nest

QFP ⇒ LU-LC 4. Efficient algorithm of deciding LC equivalence M. van den Nest, J. Dehaene, and B. Moor

5 / 23

Finding Counterexamples

6 / 23

Starting Point The QFP Problem For all quadratic function Q(x) and subspace S of Fn2 , if there exists a set of phases {cj } such that for all x ∈ S (−1)Q(x) =

n Y

x

cj j ,

j=1

then all cj ’s can be chosen from {±1, ±i}.

7 / 23

Starting Point The QFP Problem For all quadratic function Q(x) and subspace S of Fn2 , if there exists a set of phases {cj } such that for all x ∈ S (−1)Q(x) =

n Y

x

cj j ,

j=1

then all cj ’s can be chosen from {±1, ±i}. ■

Q(x) = x1 x2 + x1 x4 + x1 x5 + x3 x4 + x4 x5 + x4 .

7 / 23

Starting Point The QFP Problem For all quadratic function Q(x) and subspace S of Fn2 , if there exists a set of phases {cj } such that for all x ∈ S (−1)Q(x) =

n Y

x

cj j ,

j=1

then all cj ’s can be chosen from {±1, ±i}. ■ ■

Q(x) = x1 x2 + x1 x4 + x1 x5 + x3 x4 + x4 x5 + x4 . Fix Q(x), S: A system of equations in cj .

7 / 23

Starting Point The QFP Problem For all quadratic function Q(x) and subspace S of Fn2 , if there exists a set of phases {cj } such that for all x ∈ S (−1)Q(x) =

n Y

x

cj j ,

j=1

then all cj ’s can be chosen from {±1, ±i}. ■ ■ ■ ■

Q(x) = x1 x2 + x1 x4 + x1 x5 + x3 x4 + x4 x5 + x4 . Fix Q(x), S: A system of equations in cj . QFP ⇒ LU-LC. We will give a “proof” of QFP! 7 / 23

A “PROOF”: The Representation of Q(x) ■

Linear terms in Q(x).

■

Represent Q(x) by a graph G: Vertex set V = {1, 2, . . . , n}, Edge set E = { (i, j) | xi xj occurs in Q(x) }. For subset A of V , define a cor1 responding subgraph G|A 2 Q(x) = E(G|Ix ) , where Ix is the set of positions in which x is 1.

• • ■

■

3

5 A

4

Suggraph G|A

8 / 23

A “PROOF”: The Structure of S 1 2 ··· ··· ··· n

■

S ⊆ Fn2

ξ1 ξ2 .. . ξd

■

Choose a basis of S: {ξ 1 , · · · , ξ d }. For all x ∈ S, x =

d P

hk ξ k .

k=1 ■

Define h =

d P

hk 2k−1 , label the

k=1 h

above x with x . x

■

Let D = 2d , subspace S consists of D elements x0 , x1 , . . . , xD−1 .

9 / 23

A “PROOF”: Pattern of a position in S

S ⊆ Fn2

1 2 ··· j ··· n

■

0 1 .. .

ξ1 ξ2 .. .

1

ξd

The pattern of position j consists of the j’s bit of all the basis ξ k .

x

10 / 23

A “PROOF”: Pattern of a position in S

S ⊆ Fn2

1 2 ··· j ··· n

■

0 1 .. .

0 1 .. .

ξ1 ξ2 .. .

1

1

ξd

?

?

x

■

The pattern of position j consists of the j’s bit of all the basis ξ k . The reason to consider patterns.

10 / 23

A “PROOF”: Pattern of a position in S

S ⊆ Fn2

1 2 ··· j ··· n

■

0 1 .. .

0 1 .. .

ξ1 ξ2 .. .

■

1

1

ξd

■

?

?

x

■

The pattern of position j consists of the j’s bit of all the basis ξ k . The reason to consider patterns. Delete all redundant variables, “One Pattern, One Variable”

10 / 23

A “PROOF”: Pattern of a position in S

S ⊆ Fn2

1 2 ··· j ··· n

■

0 1 .. .

0 1 .. .

ξ1 ξ2 .. .

■

1

1

ξd

■

?

?

x

■

The pattern of position j consists of the j’s bit of all the basis ξ k . The reason to consider patterns. Delete all redundant variables, “One Pattern, One Variable”

10 / 23

A “PROOF”: Pattern of a position in S

S ⊆ Fn2

1 2 ··· j ··· n

■

0 1 .. .

0 1 .. .

ξ1 ξ2 .. .

■

1

1

ξd

■

?

?

■

The pattern of position j consists of the j’s bit of all the basis ξ k . The reason to consider patterns. Delete all redundant variables, “One Pattern, One Variable”

(−1)

x

Q(x)

=

D−1 Y

m=1

10 / 23

x[m]

C[m]

A “PROOF”: Pattern of a position in S

S ⊆ Fn2

1 2 ··· j ··· n

■

0 1 .. .

0 h 1 1 h .. 2 .

ξ1 ξ2 .. .

■

1

1 h d

ξd

■

?

?

■

The pattern of position j consists of the j’s bit of all the basis ξ k . The reason to consider patterns. Delete all redundant variables, “One Pattern, One Variable”

(−1)

x

Q(x)

=

D−1 Y

x[m]

C[m]

m=1

xh[m] = hh, mi (xh =

d X k=1

10 / 23

hk ξ k ).

A “PROOF”: Pattern of a position in S

S ⊆ Fn2

1 2 ··· j ··· n

■

0 1 .. .

0 h 1 1 h .. 2 .

ξ1 ξ2 .. .

■

1

1 h d

ξd

■

?

?

■

The pattern of position j consists of the j’s bit of all the basis ξ k . The reason to consider patterns. Delete all redundant variables, “One Pattern, One Variable”

(−1)

x

Q(x)

=

D−1 Y

x[m]

C[m]

m=1

xh[m] = hh, mi (xh =

d X

hk ξ k ).

k=1

X hk mk ( mo d 2) k 10 / 23

A “PROOF”: Simplification 1 2

5

3

11 / 23

4

A “PROOF”: Simplification (−1)

Q(x)

=

D−1 Y

1

x[m] C[m]

m=1

2

5

3

11 / 23

4

A “PROOF”: Simplification (−1)

Q(x)

=

D−1 Y m=1

(−1)Q(x

h

)

=

D−1 Y

1

x[m] C[m]

2

5

hm,hi C[m]

m=1

3

11 / 23

4

A “PROOF”: Simplification (−1)

Q(x)

=

D−1 Y m=1

(−1)Q(x

h

)

=

D−1 Y

1

x[m] C[m]

2

5

hm,hi C[m]

m=1

˛ ˛ ˛ ˛ Q(x ) = ˛E(G|I h )˛ x h

11 / 23

3

4

A “PROOF”: Simplification (−1)

Q(x)

=

D−1 Y

1

x[m] C[m]

m=1

(−1)Q(x

h

)

=

D−1 Y

2

5

hm,hi C[m]

m=1

˛ ˛ ˛ ˛ Q(x ) = ˛E(G|I h )˛ x [ Ixh = Am h

4

3

m:hm,hi=1

h

Q(x )

=

X m:hm,hi=1

˛ ˛ ˛ ˛ ˛E(G|Am )˛ +

11 / 23

X m
Emm′

A “PROOF”: Simplification (−1)

Q(x)

=

D−1 Y

1

x[m] C[m]

m=1

(−1)Q(x

h

)

=

D−1 Y

2

5

hm,hi C[m]

m=1

˛ ˛ ˛ ˛ Q(x ) = ˛E(G|I h )˛ x [ Ixh = Am h

4

3

m:hm,hi=1

h

Q(x )

X

=

m:hm,hi=1

=

X m

˛ ˛ ˛ ˛ ˛E(G|Am )˛ +

X

Emm′

m
˛ ˛ X ˛ ˛ hm, hihm′ , hiEmm′ hm, hi ˛E(G|Am )˛ + m
11 / 23

A “PROOF”: Simplification (−1)

Q(x)

=

D−1 Y

1

x[m] C[m]

m=1

(−1)Q(x

h

)

=

D−1 Y

2

5

hm,hi C[m]

m=1

˛ ˛ ˛ ˛ Q(x ) = ˛E(G|I h )˛ x [ Ixh = Am h

4

3

m:hm,hi=1

h

Q(x )

X

=

m:hm,hi=1

=

X m

X

Emm′

m
˛ ˛ X ˛ ˛ hm, hihm′ , hiEmm′ hm, hi ˛E(G|Am )˛ + m
X (−1)

˛ ˛ ˛ ˛ ˛E(G|Am )˛ +

hm, hihm′ , hiEmm′

m
=

D−1 Y m=1

11 / 23

hm,hi

C[m]

.

A “PROOF”: Take Logarithm D−1 Y

hm,hi

C[m]

= (−1)

X

m
m=1

12 / 23

hm, hihm′ , hiEmm′

A “PROOF”: Take Logarithm D−1 Y

hm,hi

C[m]

= (−1)

X

hm, hihm′ , hiEmm′

m
m=1

C[m] = ir[m]

12 / 23

A “PROOF”: Take Logarithm D−1 Y

hm,hi

C[m]

= (−1)

X

hm, hihm′ , hiEmm′

m
m=1

C[m] = ir[m]

i

D−1 X

hm, hir[m]

m=1

= (−1)

12 / 23

X

m
hm, hihm′ , hiEmm′

A “PROOF”: Take Logarithm D−1 Y

hm,hi

C[m]

= (−1)

X

hm, hihm′ , hiEmm′

m
m=1

C[m] = ir[m]

i

D−1 X

hm, hir[m]

m=1

= (−1)

X

hm, hihm′ , hiEmm′

m
G~r ≡ 2T~e (mod 4)

12 / 23

A “PROOF”: Take Logarithm D−1 Y

hm,hi

C[m]

= (−1)

X

hm, hihm′ , hiEmm′

m
m=1

C[m] = ir[m]

) Q(x

(− 1 )

D−1 X

hm, hir[m]

i

m=1

=

n xj Y cj j =1

= (−1)

X

hm, hihm′ , hiEmm′

m
G~r ≡ 2T~e (mod 4)

Where ~r and ~e encode cj and Q(x) respectively.

12 / 23

A “PROOF”: Take Logarithm D−1 Y

hm,hi

C[m]

= (−1)

X

hm, hihm′ , hiEmm′

m
m=1

C[m] = ir[m]

) Q(x

(− 1 )

D−1 X

hm, hir[m]

i

m=1

=

n xj Y cj j =1

= (−1)

X

hm, hihm′ , hiEmm′

m
G~r ≡ 2T~e (mod 4)

Where ~r and ~e encode cj and Q(x) respectively. G is the matrix G = (hi, ji), T is a matrix with Ti,(j,k) = hi, jihi, ki. 12 / 23

A “PROOF”: The Inverse of G G~r ≡ 2T~e (mod 4) 01

■



1 G = 10  0 11 1 01

G = (hi, ji),

13 / 23

10

0 1 1

11



1 1 0

A “PROOF”: The Inverse of G G~r ≡ 2T~e (mod 4) 01

■

G = (hi, ji),

■

G has an inverse: G−1

10



0 1 1

1 = 2



1 G = 10  0 11 1 01

11



1 1 0 01

2 = (2hi, ji − 1), D

13 / 23

G−1

10

11



1 −1 1 10  −1 1 1  11 1 1 −1 01

A “PROOF”: The Inverse of G G~r ≡ 2T~e (mod 4) 01

■

G = (hi, ji),

■

G has an inverse: G−1

■

10

11



0 1 1

1 = 2



1 G = 10  0 11 1 01



1 1 0 01

2 = (2hi, ji − 1), D

2G−1 T is a matrix consists of 0, ±1 entries only.

13 / 23

G−1

0 T = 10  0 11 1 01

11



1 −1 1 10  −1 1 1  11 1 1 −1 01

01,10



10

01,11

1 0 0

10,11



0 1  0

A “PROOF”: The Inverse of G G~r ≡ 2T~e (mod 4) 01

■

G = (hi, ji),

■

G has an inverse: G−1

■

■

10

11



0 1 1

1 = 2



1 G = 10  0 11 1 01



1 1 0 01

2 = (2hi, ji − 1), D

2G−1 T is a matrix consists of 0, ±1 entries only. QED 13 / 23

G−1

0 T = 10  0 11 1 01

11



1 −1 1 10  −1 1 1  11 1 1 −1 01

01,10



10

01,11

1 0 0

10,11



0 1  0

A “PROOF”: The Inverse of G G~r ≡ 2T~e (mod 4) 01

■

G = (hi, ji),

■

G has an inverse: G−1

■

■

11

0 1 1

1 = 2



1 G = 10  0 11 1 01



1 1 0 01

2 = (2hi, ji − 1), D

2G−1 T is a matrix consists of 0, ±1 entries only. QED

10



? 13 / 23

G−1

0 T = 10  0 11 1 01

11



1 −1 1 10  −1 1 1  11 1 1 −1 01

01,10



10

01,11

1 0 0

10,11



0 1  0

Where is the problem?

14 / 23

Where is the problem? One Pattern, One Variable? 1 2 ··· j ··· n

S ⊆ Fn2

■

■

0 1 .. .

0 1 .. .

ξ1 ξ2 .. .

1

1

ξd

■ ■

The reason to consider patterns. Delete all redundant variables, “One Pattern, One Variable”

Q(x)

(−1) ?

14 / 23

?

x

=

D−1 Y

m=1

x[m] C[m]

Where is the problem? ■

One Pattern, One Variable?

S ⊆ Fn2

1 2 ··· j ··· n

■

0 1 .. .

0 1 .. .

ξ1 ξ2 .. .

1

1

ξd

■ ■

The reason to consider patterns. Delete all redundant variables, “One Pattern, One Variable”

Q(x)

(−1) ?

■

?

x

When pattern m never occurs in S,

14 / 23

=

D−1 Y

m=1

x[m] C[m]

Where is the problem? ■

One Pattern, One Variable?

S ⊆ Fn2

1 2 ··· j ··· n

■

0 1 .. .

0 1 .. .

ξ1 ξ2 .. .

1

1

ξd

■ ■

The reason to consider patterns. Delete all redundant variables, “One Pattern, One Variable”

Q(x)

(−1) ?

■ ■

?

x

When pattern m never occurs in S, Let C[m] = 1.

14 / 23

=

D−1 Y

m=1

x[m] C[m]

Where is the problem? ■

One Pattern, One Variable?

S ⊆ Fn2

1 2 ··· j ··· n

■

0 1 .. .

0 1 .. .

ξ1 ξ2 .. .

1

1

ξd

■ ■

The reason to consider patterns. Delete all redundant variables, “One Pattern, One Variable”

Q(x)

(−1) ?

■ ■ ■

?

x

When pattern m never occurs in S, Let C[m] = 1. Fix r[m] to be 0 in G~r ≡ 2T~e (mod 4)

14 / 23

=

D−1 Y

m=1

x[m] C[m]

QFP “Linearized” ■

Where are we now?

15 / 23

QFP “Linearized” ■

Where are we now? If the following equation of ~r has a real solution, G~r ≡ 2T~e (mod 4) then it also has an integral solution that preserves all zero entries in the real solution.

15 / 23

QFP “Linearized” ■

Where are we now? If the following equation of ~r has a real solution, G~r ≡ 2T~e (mod 4) then it also has an integral solution that preserves all zero entries in the real solution.

■

The above statement is true without the zero-preserving requirement.

15 / 23

QFP “Linearized” ■

Where are we now? If the following equation of ~r has a real solution, G~r ≡ 2T~e (mod 4) then it also has an integral solution that preserves all zero entries in the real solution.

■

■

The above statement is true without the zero-preserving requirement. Goal: Find an ~r that has as many zero entries as possible.

15 / 23

Random Generation of Counterexamples The solutions of the equation is

~r = 2G−1 T~e + 4G−1~s

16 / 23

Random Generation of Counterexamples The solutions of the equation is

~r = 2G−1 T~e + 4G−1~s

The Method: 1. 2. 3. 4. 5.

Randomly generate ~s0 ; Shifting, to ~s1 ; (more integers) Find ~e to cancel integers in 4G−1~s1 ; Calculate Q(x), S, ci ; Verify.

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~r ←

~e ←

~s

Modified Gaussian Elimination Verify:

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Modified Gaussian Elimination Verify: Show that the following equation does not have an integral solution A~x ≡ b (mod 4).

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Modified Gaussian Elimination Verify: Show that the following equation does not have an integral solution A~x ≡ b (mod 4).

Gaussian Elimination

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Modified Gaussian Elimination Verify: Show that the following equation does not have an integral solution A~x ≡ b (mod 4).

Gaussian Elimination

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Modified Gaussian Elimination Verify: Show that the following equation does not have an integral solution A~x ≡ b (mod 4).

Gaussian Elimination ■ ■

pivot = 1 pivot = 3

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Modified Gaussian Elimination Verify: Show that the following equation does not have an integral solution A~x ≡ b (mod 4).

Gaussian Elimination ■ ■ ■

pivot = 1 pivot = 3 pivot = 2

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Modified Gaussian Elimination Verify: Show that the following equation does not have an integral solution A~x ≡ b (mod 4).

Gaussian Elimination ■ ■ ■

pivot = 1 pivot = 3 pivot = 2 0≡2

(mod 4)

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Counterexamples of QFP Counterexamples of QFP with n = 27 or 35.

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Counterexamples of QFP Counterexamples of QFP with n = 27 or 35. Q(x) equals to x1 x3 + x1 x4 + x1 x5 + x1 x12 + x1 x13 + x1 x14 + x1 x18 + x1 x19 + x1 x20 + x1 x21 + x1 x24 + x1 x27 + x2 x3 + x2 x4 + x2 x9 + x2 x10 + x2 x11 + x2 x15 + x2 x16 + x2 x18 + x2 x19 + x2 x20 + x3 x4 + x3 x5 + x3 x8 + x3 x21 + x3 x24 + x3 x26 + x4 x5 + x4 x16 + x4 x20 + x4 x22 + x4 x25 + x5 x18 + x5 x25 + x6 x10 + x6 x12 + x6 x13 + x6 x17 + x8 x9 + x8 x27 .

S is spanned by ξ 1 = 110111010010100010100010100 ξ 2 = 010010110110011000011000011 ξ 3 = 101110001110001010001010001 ξ 4 = 011110000001111001111001111 ξ 5 = 000001111111111000000111111 ξ 6 = 000000000000000111111111111. 18 / 23

Counterexamples: From QFP To LU-LC

1 X (−1)Q(x) |xi |Q, Si = p |S| x∈S

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Counterexamples: From QFP To LU-LC

1 X (−1)Q(x) |xi |Q, Si = p |S| x∈S 1 X |Si = p |xi |S| x∈S

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Counterexamples: From QFP To LU-LC

←→

1 X (−1)Q(x) |xi |Q, Si = p |S| x∈S LU

1 X |Si = p |xi |S| x∈S

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Counterexamples: From QFP To LU-LC

←→

1 X (−1)Q(x) |xi |Q, Si = p |S| x∈S

|GQ,S i

LC

|GS i

LU

1 X |Si = p |xi |S| x∈S

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LC

←→

←→

Counterexamples: From QFP To LU-LC

LU

1 X |Si = p |xi |S| x∈S

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LC

←→

|GQ,S i ←→

←→

1 X (−1)Q(x) |xi |Q, Si = p |S| x∈S

LC

←→

NOT LC

|GS i

Counterexamples

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Counterexamples

GS

——

GQ,S — + —

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Conclusions

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Conclusions

■

Local equivalences of stabilizer states are more complex than we have imagined.

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Conclusions

■

■

Local equivalences of stabilizer states are more complex than we have imagined. Find smaller scale counterexamples and a more direct proof (without resorting to the LC algorithm).

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Conclusions

■

■

■

Local equivalences of stabilizer states are more complex than we have imagined. Find smaller scale counterexamples and a more direct proof (without resorting to the LC algorithm). Design efficient algorithms for deciding the LU equivalence.

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Conclusions

■

■

■

■

Local equivalences of stabilizer states are more complex than we have imagined. Find smaller scale counterexamples and a more direct proof (without resorting to the LC algorithm). Design efficient algorithms for deciding the LU equivalence. Applications?

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b

b b

b b b

Thank You! b

b b

b

b

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b