C CS Phys C191 Entanglement can facilitate information processing Fall 2005 9 13 05 Lecture 5 1 Readings To date we have covered the following material in Benenti Casati and Strini Ch 2 Ch 3 1 3 4 2 EPR pairs and information transfer Nature is consistent with quantum mechanics and not with local realism confirming that for the wavefunction 0 1 nothing can be known about the coefficients until a measurement is made Entangled pairs of qubits such as 1 ab 2 0a 1b 1a 0b can be used to facilitate sharing or transmission of information but not to transmit information from A to B directly I e there is no superluminal transfer of information happening in an entangled state Why not Because Alice has no control over the result of her measurement and consequently she cannot control what Bob measures either Many names have been given to describe the effects of entanglement quantum non locality spooky action at a distance Einstein passion at a distance A Shimony I can add belonging at a distance to these descriptors of the relation between the measurements made on qubits a and b 2 1 Tensor product of operators Suppose v and w are unentangled states on C m and C n respectively The state of the combined system is v w on C mn If the unitary operator A is applied to the first subsystem and B to the second subsystem the combined state becomes A v B w In general the two subsystems will be entangled with each other so the combined state is not a tensorproduct state We can still apply A to the first subsystem and B to the second subsystem This gives the operator A B on the combined system defined on entangled states by linearly extending its action on unentangled states C CS Phys C191 Fall 2005 Lecture 5 1 For example A B 0 0 A 0 B 0 A B 1 1 A 1 B 1 Therefore we define A B 12 00 12 11 to be 12 A B 00 12 A B 11 12 A 0 B 0 A 1 B 1 Let e1 em be a basis for the first subsystem and write A m i j 1 ai j ei e j the i jth element of A is ai j Let f1 fn be a basis for the second subsystem and write B nk l 1 bkl fk fl Then a basis for the combined system is ei f j for i 1 m and j 1 n The operator A B is A B ai j ei ai j bkl i jkl ej ij bkl kl ei e j fk ai j bkl ei i jkl fk fl fl fk e j fl Therefore the i k j l th element of A B is ai j bkl If we order the basis ei f j lexicographically i e first according to the index i then according to the index j then the matrix for A B is a11 B a12 B a B a B 22 21 i e in the i jth subblock we multiply ai j by the matrix for B For practice with these tensor operators see the examples worked out in the previous lecture specifically in the section calculating quantum analogs of the classical correlation functions of different measurements for the Bell inequality For example work through the evaluation of A B0 A B0 1 0a 1b 1a 0b ab 2 with A za and B0 cos zb sin xb 3 Example of more efficient information processing by use of shared entanglement Consider the following communication protocol in the classical world Alice A and Bob B are two parties who share a common string S They receive independent random bits XA XB and try to output bits a b respectively such that XA XB a b The notation x y takes the AND of two binary variables x and y i e is one if x y 1 and zero otherwise x y x y mod 2 the XOR In the quantum mechanical analogue of this protocol A and B share the EPR pair As before they receive bits XA XB and try to output bits a b respectively such that XA XB a b However Alice and Bob s best protocol for the classical game as you will prove in the homework is to output a 0 and b 0 respectively Then a b 0 so as long as the inputs XA XB 6 1 1 they are successful a b 0 XA XB If XA XB 1 then they fail Therefore they are successful with probability exactly 3 4 C CS Phys C191 Fall 2005 Lecture 5 2 We will show that the quantum mechanical system can do better Specifically if Alice and Bob share an EPR pair we will describe a protocol for which the probability Pr XA XB a b is greater than 3 4 We can setup the following protocol if XA 0 then Alice measures in the standard basis and outputs the result if XA 1 then Alice rotates by 8 then measures and outputs the result if XB 0 then Bob measures in the standard basis and outputs the complement of the result if XB 1 then Bob rotates by 8 then measures and outputs the complement of the result So we need to calculate Pr a b 6 XA XB for each of the four possible cases Pr a b 6 XA XB XA XB 1 4 Pr a b 6 XA XB XA XB First we note that if measurement in the standard basis yields 0 with probability 1 then if a state is rotated by measurement will yield 0 with probability cos2 Recall that in general rotation of a state 0 1 by angle in the two dimensional state space gives the rotated state 0 0 0 0 1 where 0 cos sin 1 0 sin cos Hence the probability of measuring a 0 for the rotated state is given by 2 cos2 etc Now we claim Pr a b 6 XA XB Pr a b 6 XA XB Pr a b 6 XA XB Pr a b 6 XA XB XA 0 XB 0 0 XA 0 XB 1 sin2 8 XA 1 XB 0 sin2 8 XA 1 XB 1 cos2 4 1 2 Indeed for the first case XA XB 0 so XA XB 0 Alice and Bob each measure in the computational basis without any rotation If Alice measures a 0 then Bob s measurement is the opposite and Bob outputs the complement b 0 Therefore a b 0 XA XB a success Similarly if Alice measures a 1 they are always successful In the second case XA 0 XB 1 XA XB 0 If Alice measures a 0 then the new state of the system is 01 Bob s qubit is in the state 1 In the rotated basis Bob measures a 1 and outputs its complement 0 with probability cos2 8 The probability of failure is therefore 1 cos2 8 sin2 8 Similarly if Alice measures a …