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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 of reproducing the effect of 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 obtained for three qubits called Mermin's inequality. • The Mermin operator is defined as                         B a . a . a . a . b . b . M 1 2 3 1 2 3                         b . a . b . b . b . a . 1 2 3 1 2 3         where 3 a and b ( j 1, 2,3) are unit vectors in R , and =( , , ). j j x y z  • For any three qubit state , the Mermin inequality is   B 2 M

      1 [                 I I I l . I I I m . I I I n . 8                             u . v . I u . I w . I v . w .       t ijk i j k  i j k , , x y z , ,           with l Tr ( ( I I )), m Tr ( ( I I )), i i i i       n Tr ( ( I I )), ( i x y z , , ) i i             u v Tr ( ( I )), u w Tr ( ( I )), i j i j i j i j        v w Tr ( ( I )), ( , i j x y z , , ) i j i j         t Tr ( ( )), ( , i j k , x y z , , ) ijk i j k

Mermin inequality • Violation of 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.

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 optimization 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 operator in terms of eigenvalues • The expectation value of the Mermin operator with  respect to the state is given by                       T T T T B Tr B ( ) ( a , a Ta ) ( a , b Tb ) ( b b , Ta ) ( b , a Tb )  M M 1 3 2 1 3 2 1 3 2 1 3 2      where a ( a , a , a ), b ( b , b , b ), ( s 1, 2, 3) s sx sy sz s sx sy sz     ( , x y ) denote the inner product of two vectors and and is defined as x y  ,         ( x y ) x y cos , being the angle between and . x y   T ( T , T , T ) x y z       t t t t t t t t t xxx xyx xzx yxx yyx yzx zxx zyx zzx          , , , where T t t t T t t t T t t t       x xxy xyy xzy y yxy yyy yzy z zxy zyy zzy       t t t t t t t t t       xxz xyz xzz yxz yyz yzz zxz zyz zzz

Result-1 T T T If the symmetric matrices T T , T T , and T T has unique x x y y z z    max max max largest eigenvalue , , respectively then M ermin's x y z inequality is violated if        max max max max B max 2 , 2 , 2 2 M x y z  Pr oof :In the expression for B , we first simplify the vectors  M                 T T T T a Ta , b Tb , b Ta , a Tb . W e choose the vectors a , a , b , b 3 2 3 2 3 2 3 2 2 3 2 3 in such a way that they maximize the quantity B over all  M ope rators B . M

   (1) max Case I :In this case we will show B 2 . M x     T ( ) The vector i a Ta can be simplified as 3 2              T max a Ta (( a , T a ), ( a , T a ), ( a , T a )) ( a T a , 0, 0) 3 2 3 x 2 3 y 2 3 z 2 3 x 2   max where a is the unit vector along T a and perpendicular 3 x 2      max max to T a and T a .Since a is the unit vector so a 1. y 2 z 2 3 3      2    max T Again T a ( T a , T a ) ( a , T T a ). If is the x 2 x 2 x 2 2 x x 2 x  T max largest eigenvalue of the symmetric matrix T T and a x x 2    2  T is the c orresponding unit vector then T a ( a , T T a ) x 2 2 x x 2       max max max ( a , a ). If a is the unit vector along a then 2 x 2 2 2     2     max T max T a .Thus a Ta ( , 0, 0). x 2 x 3 2 x

   T ( ii ) The vector a Tb can be simplified as 3 2               T min a Tb (( a , T b ), ( a , T b ), ( a , T b )) (0, a T b , 0) 3 2 3 x 2 3 y 2 3 z 2 3 y 2   min where a is the unit vector antiparallel to T b and 3 y 2    min perpendicular to T b and T b .Si nce a is the unit vector, x 2 z 2 3       2     min T max a 1. Again T b ( T b , T b ) ( b , T T b ). If 3 y 2 y 2 y 2 2 y y 2 y T is the largest eigenvalue of the symmetric matrix T T and y y   2  max b is the corresponding unit vector then T b 2 y 2         T max max max ( b , T T b ) ( b , b ). If b is the unit vector along b 2 y y 2 2 y 2 2 2     2      max T max then T b .Thus a Tb (0, , 0). y 2 y 3 2 y

   T ( iii ) The vector b Tb can be simplified as 3 2               T min b Tb (( b , T b ), ( b , T b ), ( b , T b )) (0, b T b , 0) 3 2 3 x 2 3 y 2 3 z 2 3 y 2    max (0, , 0) y   min where b is the unit vector antiparallel to T b and 3 y 2    min perpendicular to T b and T b .Since b is the unit vector, x 2 z 2 3   min b 1. 3    T ( iv ) The vector b Ta can be simplified as 3 2              T max b Ta (( b , T a ), ( b , T a ), ( b , T a )) ( b T a , 0, 0) 3 2 3 x 2 3 y 2 3 z 2 3 x 2   max ( , 0, 0) x   max where b is the unit vector along T a and perpendicular 3 x 2      max max to T a and T a .Since b is the unit vector, b 1. y 2 z 2 3 3

W e have       (1) max max B max [( a , ( , 0, 0)) ( a , (0, , 0))   M 1 x 1 y  a , b 1 1       max max ( b , ( , 0, 0)) ( b , (0, , 0))] 1 x 1 y       max max max max a ( , 0, 0) b ( , 0, 0) 1 x 1 x   max 2 x   max max where a is the unit vector parallel to ( , 0, 0) and 1 x   max max per pendicular to (0, , 0); is the unit vector b y 1   max max antiparallel to ( , 0, 0) and perpendicular to (0, , 0) x y