Richard Cleve DC 3524 cleve@cs.uwaterloo

1. Introduction to Quantum Information ProcessingCS 467 / CS 667Phys 667 / Phys 767C&O 481 / C&O 681 Richard Cleve DC 3524 cleve@cs.uwaterloo.ca Lecture 2 (2005)

2. Superdense coding

3. How much classical information in n qubits? • 2n1 complex numbers apparently needed to describe an arbitrary n-qubit pure quantum state: • 000000 + 001001 + 010010 +  + 111111 • Does this mean that an exponential amount of classical information is somehow stored in n qubits? • Not in an operational sense ... • For example, Holevo’s Theorem (from 1973) implies: one cannot convey more than n classical bits of information in n qubits

4. b1 b1 U U ψ b2 ψ b2 b3 b3 n qubits n qubits bn bn 0 bn+1 0 bn+2 m qubits 0 bn+3 0 bn+4 0 bn+m Holevo’s Theorem Easy case: Hard case (the general case): b1b2 ... bncertainly cannot convey more than n bits! The difficult proof is beyond the scope of this course

5. Superdense coding (prelude) Suppose that Alice wants to convey two classical bits to Bob sending just one qubit ab Alice Bob ab By Holevo’s Theorem, this is impossible

6. Superdense coding In superdense coding, Bob is allowed to send a qubit to Alice first ab Alice Bob ab How can this help?

7. if a=1then apply X to qubit if b=1then apply Z to qubit send the qubit back to Bob • Alice: How superdense coding works • Bob creates the state 00+11 and sends the first qubit to Alice Bell basis • Bob measures the two qubits in the Bell basis

8. H Measurement in the Bell basis Specifically, Bob applies to his two qubits ... and then measures them, yielding ab This concludes superdense coding

9. Teleportation

10. 2 2 span of 0 and 1 0 1 Incomplete measurements (I) Measurements up until now are with respect to orthogonal one-dimensional subspaces: The orthogonal subspaces can have other dimensions: (qutrit)

11. Incomplete measurements (II) Such a measurement on 0 0 +1 1+2 2 (renormalized) results in 00+11 with prob 02+12 2with prob 22

12. Result is the mixture 0000+0101 with prob 002+012 1010+1111 with prob 102+112 Measuring the first qubit of a two-qubit system 0000+0101 +1010 +1111 • Defined as the incomplete measurement with respect to the two dimensional subspaces: • span of 00 & 01 (all states with first qubit 0), and • span of 10 & 11 (all states with first qubit 1)

13. Easy exercise: show that measuring the first qubit and then measuring the second qubit gives the same result as measuring both qubits at once

14. 0+1 0+1 Teleportation (prelude) Suppose Alice wishes to convey a qubit to Bob by sending just classical bits If Alice knows and , she can send approximations of them ―but this still requires infinitely many bits for perfect precision Moreover, if Alice does not know  or , she can at best acquire one bit about them by a measurement

15. Teleportation scenario In teleportation, Alice and Bob also start with a Bell state (1/2)(00+11) 0+1 and Alice can send two classical bits to Bob Note that the initial state of the three qubit system is: (1/2)(0+1)(00+11) = (1/2)(000+ 011+ 100+ 111)

16. Initial state: (0+1)(00+11)(omitting the 1/2 factor) = 000+ 011+ 100+ 111 = ½(00+ 11)(0 +1) + ½(01+ 10)(1 +0) + ½(00−11)(0 −1) + ½(01−10)(1 −0) How teleportation works Protocol: Alice measures her two qubits in the Bell basis and sends the result to Bob (who then “corrects” his state)

17. ½00(0 +1) + ½01(1 +0) +½10(0 − 1) + ½11(1 −0) (00, 0 +1) with prob. ¼ (01, 1 +0) with prob. ¼ (10, 0 −1) with prob. ¼ (11, 1 −0) with prob. ¼ H What Alice does specifically Alice applies to her two qubits, yielding: Then Alice sends her two classical bits to Bob, who then adjusts his qubit to be 0 +1 whatever case occurs

18. 00, 0 +1 01, X(1 +0)=0 +1 10, Z(0 −1) =0 +1 11, ZX(1 −0)= 0 +1 Bob’s adjustment procedure Bob receives two classical bits a, b from Alice, and: if b=1he applies X to qubit if a=1he applies Z to qubit yielding: Note that Bob acquires the correct state in each case

19. Summary of teleportation 0+1 H Alice a 00+11 b Bob 0+1 X Z Quantum circuit exercise: try to work through the details of the analysis of this teleportation protocol

20. No-cloning theorem

21. a a a 0 ψ ψ ψ 0 Classical information can be copied a a a 0 What about quantum information? ?

22. Candidate: works fine for ψ =0and ψ =1 ... but it fails for ψ =(1/2)(0+1) ... ... where it yields output (1/2)(00+11) instead of ψψ =(1/4)(00+01+10+11)

23. ψ ψ U 0 ψ 0 g No-cloning theorem Theorem: there is no valid quantum operation that maps an arbitrary state ψ to ψψ Proof: Let ψ and ψ′ be two input states, yielding outputs ψψg and ψ′ψ′g′ respectively Since U preserves inner products: ψψ′ = ψψ′ψψ′gg′ so ψψ′(1− ψψ′gg′) = 0 so ψψ′ =0or1

24. THE END