An analytical condition for the violation of Mermin's inequality by - - PowerPoint PPT Presentation

An analytical condition for the violation of Mermin's inequality by any three qubit state

Satyabrata Adhikari Birla Institute of Technology, Mesra

Collaborator: Prof. Archan S. Majumdar

Quantum non-locality

• The

impossiblity

• f

reproducing the effect

• f

quantum correlations between the outcomes of the distant measurements using local hidden variable theories is known as quantum non-locality.

• Quantum nonlocality finds applications in several

information theoretic protocols such as device independent quantum key generation, quantum state estimation and communication complexity, where the amount of violation of the Bell-CHSH inequality is important.

Mermin inequality

• Like two qubit non-locality, non-locality for three qubit

systems has also been studied using various approaches.

• A generalized form of the Bell-CHSH inequality was
• btained for three qubits called Mermin's inequality.
• The Mermin operator is defined as

where

• For any three qubit state , the Mermin inequality is

1 2 3 1 2 3 1 2 3 1 2 3

. . . . . . . . . . . .

M

B a a a a b b b a b b b a                                                

3

and ( 1, 2,3) are unit vectors in , and =( , , ).

j j x y z

a b j R         

2

M

B

 

, , , ,

1 [ . . . 8 . . . . . . ( ( )), ( ( )), ( ( )), ( , , ) ( ( )), (

ijk i j k i j k x y z i i i i i i i j i j i j

I I I l I I I m I I I n u v I u I w I v w t with l Tr I I m Tr I I n Tr I I i x y z u v Tr I u w Tr                       



                                     



                  ( )), ( ( )), ( , , , ) ( ( )), ( , , , , )

i j i j i j ijk i j k

I v w Tr I i j x y z t Tr i j k x y z                   

• Violation
• f

the Mermin inequality has been computed for several three qubit states such as GHZ and W-states earlier.[V. Scarani, and N. Gisin, J. Phys. A 34, 6043

(2001), C. Emary, and C. W. J. Beenakker, Phys. Rev. A 69, 032317 (2004),

• D. P. Chi, K. Jeong, T. Kim, K. Lee, and S. Lee, Phys. Rev. A 81, 044302

(2010)]

• In order to find the maximum violation of the

Mermin inequality for three qubit states one has to tackle the optimization problem numerically because there does not exist any analytical formula for even pure three qubit states.

Mermin inequality

Motivation

• Motivated by the work of Horodecki [R. Horodecki, P. Horodecki,

and M. Horodecki, Phys. Lett. A 200, 340 (1995).] in the context of two

qubit systems, in the present work we perform the

• ptimization problem involved in the Mermin inequality

analytically and obtain a formula for the maximal value of the expectation of the Mermin operator in terms of the eigenvalues of symmetric matrices, that gives the maximal violation of the Mermin inequality not only for pure states but also for mixed states.

Maximum expectation value of the Mermin

• perator in terms of eigenvalues
• The expectation value of the Mermin operator with

respect to the state is given by



1 3 2 1 3 2 1 3 2 1 3 2

( ) ( , ) ( , ) ( , ) ( , ) ( , , ), ( , , ), ( 1, 2, 3) ( , ) denote the inner product of two vectors and and is defined as (

T T T T M M s sx sy sz s sx sy sz

B Tr B a a Ta a b Tb b b Ta b a Tb where a a a a b b b b s x y x y x



                               , ) cos , being the angle between and . y x y x y        

( , , ) , , ,

x y z xxx xyx xzx yxx yyx yzx zxx zyx zzx x xxy xyy xzy y yxy yyy yzy z zxy zyy zzy xxz xyz xzz yxz yyz yzz zxz zyz zzz

T T T T t t t t t t t t t where T t t t T t t t T t t t t t t t t t t t t                                   

Result-1

 

max max max max max max max

If the symmetric matrices , , and has unique largest eigenvalue , , respectively then M ermin's inequality is violated if max 2 , 2 , 2 2

T T T x x y y z z x y z M x y z

T T T T T T B



       

3 2 3 2 3 2 3 2 2 3 2 3

Pr :In the expression for , we first simplify the vectors , , , . W e choose the vectors , , , in such a way that they maximize the quantity

• ver all
• pe

M T T T T M

• of

B a Ta b Tb b Ta a Tb a a b b B

 

                rators .

M

B

(1) max 3 2 max 3 2 3 2 3 2 3 2 3 2 max 3 2

:In this case we will show 2 . ( ) The vector can be simplified as (( , ), ( , ), ( , )) ( , 0, 0) where is the unit vector along and

M x T T x y z x x

Case I B i a Ta a Ta a T a a T a a T a a T a a T a



                    

max max 2 2 3 3 2 max 2 2 2 2 2 max 2

perpendicular to and .Since is the unit vector so 1. Again ( , ) ( , ). If is the largest eigenvalue of the symmetric matrix and is the c

y z T x x x x x x T x x

T a T a a a T a T a T a a T T a T T a              

2 2 2 2 max max max 2 2 2 2 2 max max 2 3 2

• rresponding unit vector then

( , ) ( , ). If is the unit vector along then .Thus ( , 0, 0).

T x x x x T x x x

T a a T T a a a a a T a a Ta                  

3 2 min 3 2 3 2 3 2 3 2 3 2 min 3 2 2 2

( ) The vector can be simplified as (( , ), ( , ), ( , )) (0, , 0) where is the unit vector antiparallel to and perpendicular to and .Si

T T x y z y y x z

ii a Tb a Tb a T b a T b a T b a T b a T b T b T b                     

min 3 2 min max 3 2 2 2 2 2 2 max 2 2

nce is the unit vector,

• 1. Again

( , ) ( , ). If is the largest eigenvalue of the symmetric matrix and is the corresponding unit vector then

T y y y y y y T y y y

a a T b T b T b b T T b T T b T b             

max max max 2 2 2 2 2 2 2 max max 2 3 2

( , ) ( , ). If is the unit vector along then .Thus (0, , 0).

T y y y T y y y

b T T b b b b b T b a Tb                  

3 2 min 3 2 3 2 3 2 3 2 3 2 max min 3 2

( ) The vector can be simplified as (( , ), ( , ), ( , )) (0, , 0) (0, , 0) where is the unit vector antiparallel to and perpendicular to

T T x y z y y y x

iii b Tb b Tb b T b b T b b T b b T b b T b T                      

min 2 2 3 min 3 3 2 max 3 2 3 2 3 2 3 2 3 2 max max 3

and .Since is the unit vector, 1. ( ) The vector can be simplified as (( , ), ( , ), ( , )) ( , 0, 0) ( , 0, 0) where is the unit

z T T x y z x x

b T b b b iv b Ta b Ta b T a b T a b T a b T a b                        

2 max max 2 2 3 3

vector along and perpendicular to and .Since is the unit vector, 1.

x y z

T a T a T a b b      

1 1

(1) max max 1 1 , max max 1 1 max max max max 1 1 max max max 1

W e have max [( , ( , 0, 0)) ( , (0, , 0)) ( , ( , 0, 0)) ( , (0, , 0))] ( , 0, 0) ( , 0, 0) 2 where is the unit vector parallel to ( , 0, 0) and per

M x y a b x y x x x x

B a a b b a b a



              

 

      

max max 1 max max

pendicular to (0, , 0); is the unit vector antiparallel to ( , 0, 0) and perpendicular to (0, , 0)

y x y

b    

( 2 ) max 3 2 3 2 3

:In this case w e choose the vectors in such a w ay that the m axim ized expectation value of the M erm in

• perator is given by

2 . ( ) The vector can be sim plified as (( ,

M y T T

Case II B i a Ta a Ta a



          

max 2 3 2 3 2 3 2 max 3 2 2 2 max 3 2

), ( , ), ( , )) (0, , 0) w here is the unit vector along and perpendicular to and . Repeating the steps of case-I, w e find (0, , 0). Sim ilarly w e

x y z y y x z T y

T a a T a a T a a T a a T a T a T a a Ta                 

max max 3 2 3 2 max 3 2

• btain

(ii) ( , 0, 0); (iii) ( , 0, 0); ( ) (0, , 0).

T T x x T y

a Tb b Tb iv b Ta                 

1 1

( 2 ) max max 1 1 , max max 1 1 max max max max 1 1 max max m 1

Therefore, we have max [( , (0, , 0)) ( , ( , 0, 0)) ( , (0, , 0)) ( , ( , 0, 0))] (0, , 0) (0, , 0) 2 where is the unit vector parallel to (0,

M y x a b y x y y y y

B a a b b a b a



              

 

      

ax max max 1 max max

, 0) and perpendicular to ( , 0, 0); is the unit vector antiparallel to (0, , 0) and perpendicular to ( , 0, 0).

x y x

b    

1 1

(3) max max 1 1 , max max 1 1 max max max ma 1 1

:Proceeding in a similar fashion as in C ase-I and Case-II, we obtain max [( , (0, 0, )) ( , ( , 0, 0)) ( , (0, 0, )) ( , ( , 0, 0))] (0, 0, ) (0, 0,

M z x a b z x z z

Case III B a a b b a b



            

 

     

x max max max 1 max max 1 max max

) 2 where is the unit vector parallel to (0, 0, ) and perpendicular to ( , 0, 0); is the unit vector antiparallel to (0, 0, ) and perpendicular to ( , 0, 0).

z z x z x

a b        

   

max (1) ( 2 ) (3) max max max max

Thus finally, the m axim um expectation value of the M erm in

• perator w ith respect to the state

is g iven by m ax , , m ax 2 , 2 , 2 The M erm in inequality is violated if

M M M M x y z M

B B B B B

   

     

 

max max max

2 m ax 2 , 2 , 2 2

x y z 

     

Result-2

 

max max max max max max max

If the symmetric matrices , , and have two equal largest eigenvalue , , respectively then M ermin's inequality is violated if max 4 , 4 , 4 2

T T T x x y y z z x y z M x y z

T T T T T T B



       

3 2 3 2 3 2 3 2 2 3 2 3

Pr :In the expression for , we first simplify the vectors , , , . W e choose the vectors , , , in such a way that they maximize the quantity

• ver all
• pe

M T T T T M

• of

B a Ta b Tb b Ta a Tb a a b b B

 

                rators .

M

B

max ( 4 ) max 3 2 3 2 3 2

: W e consider the symmetric matrix which has two equal largest eigenvalues . In this case, we will show 4 . ( ) The vector can be simplified as (( , ), (

T x x x M x T T x

Case I T T B i a Ta a Ta a T a



            

max 3 2 3 2 3 2 max 3 2 max max 2 2 3 3 2 2 2 2 2

, ), ( , )) ( , 0, 0) where is the unit vector along and perpendicular to and .Since is the unit vector so 1. Again ( , ) ( ,

y z x x y z T x x x x x

a T a a T a a T a a T a T a T a a a T a T a T a a T T                     

max 2 max 2 2 2 2 2 max max max 2 2 2 2

). If is the largest eigenvalue of the symmetric matrix and is the corresponding unit vector then ( , ) ( , ). If is the unit vector along then

x T x x T x x x x

a T T a T a a T T a a a a a            

2 max max 2 3 2

.Thus ( , 0, 0).

T x x x

T a a Ta        

3 2 min 3 2 3 2 3 2 3 2 3 2 min 3 2 2 2

( ) The vector can be sim plified as (( , ), ( , ), ( , )) ( , 0, 0) w here is the unit vector antiparallel to and perpendicular to and .Si

T T x y z x x y z

ii a Tb a Tb a T b a T b a T b a T b a T b T b T b                     

min 3 min 3 max max 2 2 max max 2 2 2 2 2

nce is the unit vector,

• 1. Since the m atrix

has tw o equal largest eigenvalue , so is another corresponding unit

• eigenvector. Then

( , ) ( , ).

T x x x T x x x x

a a T T b T b b T T b b b            

2 max max 2 2 2 max 3 2

If is the unit vector along then . Thus ( , 0, 0).

x x T x

b b T b a Tb           

3 2 min 3 2 3 2 3 2 3 2 3 2 max min 3 2

( ) The vector can be simplified as (( , ), ( , ), ( , )) ( , 0, 0) ( , 0, 0) where is the unit vector antiparallel to and perpendicular to

T T x y z y x x y

iii b Tb b Tb b T b b T b b T b b T b b T b T                      

2 2 3 2 min 3 2 3 2 3 2 3 2 3 2 max min 3 2

and . ( ) The vector can be simplified as (( , ), ( , ), ( , )) ( , 0, 0) ( , 0, 0) where is the unit vector antiparallel to and perpend

z T T x y z x x x

b T b iv b Ta b Ta b T a b T a b T a b T a b T a                        

2 2

icular to and .

y z

T a T a  

1 1

( 4 ) max max 1 1 , max max 1 1 max max max max 1 1 max max max max 1 1

W e have max [( , ( , 0, 0)) ( , ( , 0, 0)) ( , ( , 0, 0)) ( , ( , 0, 0))] 2 ( , 0, 0) 2 ( , 0, 0) 4 where and is the unit vector along ( , 0, 0

M x x a b x x x x x x

B a a b b a b a b



              

 

        ).

max (5 ) max 3 2 3 2 3

:Here we consider the symmetric matrix which has two equal largest eigenvalues . In this case, we will show 4 . ( ) The vector can be simplified as (( ,

T y y y M x T T x

Case II T T B i a Ta a Ta a T



           

max 2 3 2 3 2 3 2 max 3 2 2 2 2 2 2 2 2 2 max

), ( , ), ( , )) (0, , 0) where is the unit vector along and perpendicular to and . Again ( , ) ( , ). If is the largest eigenvalu

y z y y T x z y y y y y y

a a T a a T a a T a a T a T a T a T a T a T a a T T a                    

max 2 2 max max 2 2 2 2 2 2 2 max max ma 2 2 3 2

e of the symmetric matrix and is the corresponding unit vector then ( , ) ( , ). If is the unit vector along then .Thus (0,

T y y T x y y y T y y y

T T a T a a T T a a a a a T a a Ta                   

x , 0).

3 2 min 3 2 3 2 3 2 3 2 3 2 min 3 2 2 2

( ) The vector can be simplified as (( , ), ( , ), ( , )) (0, , 0) where is the unit vector antiparallel to and perpendicular to and . Si

T T x y z y y x z

ii a Tb a Tb a T b a T b a T b a T b a T b T b T b                     

max max 2 2 max max 2 2 2 2 2 2

nce the matrix has two equal largest eigenvalue , so let us consider be another corresponding unit eigenvector. Then ( , ) ( , ). If is the unit vector

T x x y T y y y y

T T b T b b T T b b b b          

2 max max 2 2 max 3 2

along then . Thus (0, , 0).

x y T x

b T b a Tb          

3 2 min 3 2 3 2 3 2 3 2 3 2 max min 3 2

( ) The vector can be simplified as (( , ), ( , ), ( , )) (0, , 0) (0, , 0) where is the unit vector antiparallel to and perpendicular to

T T x y z y y y x

iii b Tb b Tb b T b b T b b T b b T b b T a T                      

2 2 3 2 min 3 2 3 2 3 2 3 2 3 2 max min 3 2

and . ( ) The vector can be simplified as (( , ), ( , ), ( , )) (0, , 0) (0, , 0) where is the unit vector antiparallel to and perpend

z T T x y z y y y

b T b iv b Ta b Ta b T a b T a b T a b T a b T a                        

2 2

icular to and .

x z

T a T a  

1 1

(5 ) max max 1 1 , max max 1 1 max max max max 1 1 max max max max 1 1

W e have max [( , (0, , 0)) ( , (0, , 0)) ( , (0, , 0)) ( , (0, , 0))] 2 (0, , 0) 2 (0, , 0) 4 where and is the unit vector along (0, , 0

M y y a b y y y y y y

B a a b b a b a b



              

 

        ).

1 1

max ( 6 ) max max 1 1 , max 1 1

:H ere w e consider the sym m etric m atrix w hich has tw o equal largest eigenvalues . In this case, w e obtain m ax [( , (0, 0, )) ( , (0, 0, )) ( , (0, 0, )) ( ,

T z z z M z z a b z

Case III T T B a a b b



        

 

   

max max max max max 1 1 max max max 1 max max 1 max

(0, 0, ))] 2 (0, 0, ) 2 (0, 0, ) 4 w here is the unit vector parallel to (0, 0, ) and perpendicular to ( , 0, 0); is the unit vector antiparallel to (0, 0, ) and

z z z z z x z

a b a b              

max

perpendicular to ( , 0, 0).

x



   

max ( 4 ) (5 ) ( 6 ) max max max max

Thus, the maximum expectation value of the M ermin

• perator with respect to the state

is given by max , , max 4 , 4 , 4 The M ermin inequality is violated if 2 max 4

M M M M x y z M

B B B B B

    

       

 

max max max

, 4 , 4 2

x y z

   

 

max max max max max max max

Result-3: If has two equal largest eigenvalue and , and have unique largest eigenvalue , respectively then M ermin's inequality is violated if max 4 , 2 , 2

T x x x T T y y z z y z M x y z

T T T T T T B



        2

 

max max max max max max max

Result-4: If has two equal largest eigenvalue and , and have unique largest eigenvalues , respectively then M ermin's inequality is violated if max 2 , 4 , 2

T y y y T T x x z z x z M x y z

T T T T T T B



       2 

 

max max max max max max max

Result-5: If has two equal largest eigenvalue and , and have unique largest eigenvalue , respectively then M ermin's inequality is violated if max 2 , 2 , 4

T z z z T T x x y y x y M x y z

T T T T T T B



        2

 

max max max max max max max

Result-6: If and has two equal largest eigenvalues , respectively and has unique largest eigenvalue , then M ermin's inequality is violated if max 4 , 4 , 2

T T x x y y T x y z z z M x y z

T T T T T T B



        2

 

max max max max max max max

Result-7: If and has two equal largest eigenvalues , respectively and has unique largest eigenvalue , then M ermin's inequality is violated if max 4 , 2 , 4

T T x x z z T x z y y y M x y z

T T T T T T B



        2

 

max max max max max max max

Result-8: If and has two equal largest eigenvalues , respectively and has unique largest eigenvalue , then M ermin's inequality is violated if max 2 , 4 , 4

T T y y z z T y z x x x M x y z

T T T T T T B



        2

Example-1

2 2 2 2

G eneralised G H Z is given by 000 111 . T he m atrices , are given by 4 4 T he m atrix is a zero m atrix. T he largest eigenvalues of , are given b

G H Z T T x x y y T T x y y T z z T T x x y y

T T T T T T T T T T T T T T                     

m ax m ax 2 2 m ax

y 4 . T he m axim um expectation value of the M erm in operator w ith respect to the state is given by 8 . T herefore, the state violates M erm in's inequality is 1 2 . T he 2

G H Z G H Z

x y G H Z M G H Z

B

 

              sam e result has been found by Scarani and G isin.

[V. Scarani, and N. Gisin, J. Phys. A 34, 6043 (2001).]

, 3 2 2 3 2 2

Let us consider a pure state 1 000 . 1 w here ( 001 010 100 ). 3 T he m atrices , are given by 4 4 (1 ) (1 ) 9 3 3 , 4 4 (1 ) (1 )(1 2 ) 9 3 3 4 (1 ) 9 4 (1 9

W S T T x x y y T x x T y y

p W p W T T T T p p p T T p p p p T T p                               

2

) . p                 

Example-2

2 3 2 2

4 (1 )(1 2 ) 9 4 (1 ) 9 4 2 w here (1 ) 1 (2 1) 3 3 3 4 (1 ) (2 1) 3

T z z

p p f T T p f g f p p p p p g p p p                               

max 2 max 2 max 4 3 2 2

The largest eigenvalues of , and are given by 4 2 2 4 (1 )( 12 3 ), (1 ) 9 9 9 9 1 13 8 10 256 640 672 232 25 18 18 9 9 The maximum expectation value of the M ermin ope

T T T x x y y z z x y z

T T T T T T p p p p p p p p p p p                  

,

, max max max max

rator with respect to the state is given by 4 , 0 0.43 2 , 0.43 0.45 2 , 0.45 1

W S

W S M y x z

B p p p

 

            

,

The state violates Mermin inequality when 0 0.25. This result was obtained by Chi et.al.

W S

p   

• D. P. Chi, K. Jeong, T. Kim, K. Lee, and S. Lee, Phys. Rev. A 81, 044302 (2010).

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