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

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. Holevo’s Theorem U U ψ ψ
Easy case:
Hard case (the general case): U
ψ
n qubits b1 b2 b3 bn
ψ
n qubits b1 b2 b3 bn U
0
m qubits
bn+1
bn+2
bn+3
bn+4
bn+m
b1b bn certainly 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 ab ab Alice Bob
In superdense coding, Bob is allowed to send a qubit to Alice first ab
Alice
Bob ab
How can this help?

7. How superdense coding works
Bob creates the state 00 + 11 and sends the first qubit to Alice
Alice:
if a = 1 then apply X to qubit if b = 1 then apply Z to qubit send the qubit back to Bob ab
state 00
00 + 11 01
00 − 11 10
01 + 10 11
01 − 10
Bell basis
Bob measures the two qubits in the Bell basis

8. Measurement in the Bell basis
Specifically, Bob applies
input
output
00 + 11
00
01 + 10
01
00 − 11
10
01 − 10
11 H
to his two qubits ...
and then measures them, yielding ab
This concludes superdense coding

9. Teleportation

10. Incomplete measurements (I)
Measurements up until now are with respect to orthogonal one-dimensional subspaces:
The orthogonal subspaces can have other dimensions:
0
1
2
span of 0 and 1
2
(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. 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)
Result is the mixture 0000 + 0101 with prob 002 + 012
1010 + 1111 with prob 102 + 112

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. Teleportation (prelude)
Suppose Alice wishes to convey a qubit to Bob by sending just classical bits
0 + 1
0 + 1
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. How teleportation works
(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)
Initial state:
Protocol: Alice measures her two qubits in the Bell basis and sends the result to Bob (who then “corrects” his state)

17. What Alice does specifically
Alice applies
to her two qubits, yielding:
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
Then Alice sends her two classical bits to Bob, who then adjusts his qubit to be 0 + 1 whatever case occurs

18. Bob’s adjustment procedure
Bob receives two classical bits a, b from Alice, and:
if b = 1 he applies X to qubit if a = 1 he applies Z to qubit
00, 0 + 1
01, X(1 + 0) = 0 + 1
10, Z(0 − 1) = 0 + 1
11, ZX(1 − 0) = 0 + 1
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. Classical information can be copieda a
What about quantum information? ?
ψ
0

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. No-cloning theorem U ψ 0 g ψψ′ = ψψ′ψψ′gg′ so
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
ψ
0
g U
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′ = 0 or 1

24. THE END