Superdense coding. How much classical information in n qubits? Observe that 2 n 1 complex numbers apparently needed to describe an arbitrary n -qubit.

Superdense Superdense codingcoding

How much classical information in How much classical information in nn qubits?qubits?

• Observe that 2n1 complex numbers apparently needed to describe an arbitrary n-qubit pure quantum state:

because 000000 + 001001 + 010010 + + 111111

• 2n is exponential so 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 one cannot convey more than nn classical bits of information in classical bits of information in nn qubits qubits

Holevo’s TheoremHolevo’s Theorem

Uψn qubits

b1

b2

b3

bn

Easy case:

b1b2 ... bn certainly

cannot convey more than n bits!

Hard case (the general case):

ψn qubits

b1

b2

b3

bn

U00

000

m qubits

bn+1

bn+2

bn+3

bn+4

bn+m

The difficult proof is beyond the scope of this course

measurementmeasurement

We can use only n classical bits

Info only here

Superdense Superdense codingcoding

Superdense coding (prelude)Superdense coding (prelude)

By Holevo’s Theorem, this is impossible

Alice Bob

ab

Suppose that Alice wants to convey two classical bits to Bob sending just one qubit

ab

Can we convey two classical bits by sending just one qubit?

Superdense codingSuperdense coding

How can this help?

Alice Bob

ab

In superdense coding, Bob is allowed to send a qubit to Alice first

ab

The idea is to use entanglement!

H

HX Z

AliceAlice

Bob measures in Bell Base

Bob Creates EPR entanglement

0 0

In this scheme the measurement is only here

General idea of General idea of Superdense Coding Superdense Coding can be explained by this distributed can be explained by this distributed

quantum circuitquantum circuit

How superdense coding works?How superdense coding works?1. Bob creates the state the state 0000 ++ 1111 and sends the firstfirst qubit qubit

to Alice

Alice:

01

10X

10

01Zif a = 1 then apply X to qubit

if b = 1 then apply Z to qubit send the qubit back to Bob

ab state00 00 +

11

01 00 − 11

10 01 + 10

11 01 − 10

3. Bob measures the two qubits in the Bell basis

Bell basis

2. Alice wants to send two bits a and b

So let us analyze what Alice sends Alice sends back to Bobback to Bob?

No change

Alice applies X to first qubit

To analyze this communication scheme we need to use our known methods for To analyze this communication scheme we need to use our known methods for quantum circuit analysisquantum circuit analysis

See last slide

Bob creates the state the state 0000 ++ 1111 and sends the firstfirst qubitqubit to Alice

ab state00 00 +

11

01 00 − 11

10 01 + 10

11 01 − 10

Bob

0000 + + 1111X

Alice changes

0 1 1 0

1 0 0 1

x=

0 0 1 0 0 0 0 1 1 0 0 0 0 1 0 0

0 0 1 0 0 0 0 1 1 0 0 0 0 1 0 0

1 0 0 1

=

01 10

Similarly other cases can be calculated

Bob does Bob does measurementmeasurementIn Bell basisIn Bell basis

( ) ( )

Three stages of communication

Example of calculation the operator executed by Alice

We can say that Alice changes one entanglement to another entanglement

Measurement in the Bell basisMeasurement in the Bell basis

H

Specifically, Bob applies

to his two qubits ...

input output00 + 11

00

01 + 10

01

00 − 11

10

01 − 10

11 and then measures them, yielding ab

Homework or exam

H

to his two qubits ...

input output00 + 11

00

01 + 10

01

00 − 11

10

01 − 10

11

and then measures them, yielding ab

This concludes superdense coding

H

X Z

AliceAlice

Bob measures in Bell Base

Bob Creates EPR entanglement

Observe that Bob knows what Alice has done.This is used in Quantum Games. Bob knows more than Alice if he sends her quantum entangled info.

0 0

In this scheme the measurement is only here

Results of Bell Base measurement

Incomplete measurementsIncomplete measurements• von Neumann Measurement: associated with a partition of

the space into mutually orthogonal subspaces

When the measurement is performed, the state collapses to each subspace with probability the square of the length of its projection on that subspace

00 01

10

span of 00 and 01

10

Incomplete measurementsIncomplete measurementsMeasurements up until now are with respect to orthogonal one-dimensional subspaces:

0 1

2

The orthogonal subspaces can have other can have other dimensions:dimensions:

span of 0 and 1

2

(qutrit)

Such a measurement on 0 0 + 1 1 + 2 2

results in 00 + 11 with prob 02 + 12

2 with prob 22

(renormalized)

TeleportationTeleportation

Measuring the Measuring the first qubitfirst qubit of a of a two-qubit systemtwo-qubit system

Result is the mixture 0000 + 0101 with prob 002 + 012

1010 + 1111 with prob 102 + 112

Measuring this first qubit is 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)

0000 + 0101 + 1010 + 1111

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

Homework or exam

Hint: continue calculations from last slide

Teleportation (prelude)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

Teleportation scenarioTeleportation scenario

0 + 1(1/2)(00 + 11)

In teleportation, Alice and Bob also start with a Bell state

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)

teleportation circuitteleportation circuit

H

X Z

0 + 1

00 + 11

0 + 1ba

Alice

Bob

Measurement by Alice in Bell Basis

These two qubits are entangled so Alice changing her private qubit changes the qubit in Bob’s possession

Bob

input output

00 + 11 00

01 + 10 01

00 − 11 10

01 − 10 11

Results of Bell Base measurement

(0 + 1)(00 + 11) (omitting the 1/2

factor)

= 000 + 011 + 100 + 111 == ½ ½ 000000 + ½ + ½ 000000 + ½ ½ 011011 + + ½ ½ 011011 ++½ ½

100100 + + ½ ½ 100100 + + ½ ½ 111111 + + ½ ½ 111111 + ½ + ½ 110110 + ½ + ½

001001 + ½ + ½ 011011 + ½ + ½ 100100 …..….. etc factors = after

factorization =

= 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)

This is state of the qubit in Bob possession which is changed by entanglement with changes in Alice circuit

These qubits are measured by Alice

16 terms after multiplication. Some cancel

You can multiply the lower formula to get the upper formula

How teleportation worksHow 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)

This is state of the qubit in Bob possession which is changed by entanglement with changes in Alice circuit

These are classical bits that Alice sends to Bob

What Alice does specificallyWhat Alice does specificallyAlice 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

½00(0 + 1)+ ½01(1 + 0)+ ½10(0 − 1)+ ½11(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

HBefore measurement After

measurement

This is what Alice sends to Bob

Bob’s adjustment procedureBob’s adjustment procedure

01

10X

10

01Z

if b = 1 he applies X to qubit

if a = 1 he applies Z to qubit

Bob receives two classical bits a, b from Alice, and:

00, 0 + 101, X(1 + 0) = 0 + 110, Z(0 − 1) = 0 + 111, ZX(1 − 0) = 0 + 1

yielding:

Note that Bob acquires the correct state in each case

Summary of teleportation: circuitSummary of teleportation: circuit

H

X Z

0 + 1

00 + 11

0 + 1ba

Alice

Bob

Quantum circuit exercise: try to work through the details of the analysis of this teleportation protocol

Homework or exam

measurement

This circuit works correctly regardless the randomness of measurements

No-cloning No-cloning theoremtheorem

ClassicalClassical information can be copied information can be copied

0

a a

a 0

a a

a

What about quantum information?

ψ

0

ψ

ψ ?

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 ψψ = (1/4)(00 + 01 + 10 + 11)

Candidate:

No-cloning theoremNo-cloning theoremTheorem: there is no valid quantum operation that maps an arbitrary state ψ to ψψ

Proof:

Let ψ and ψ′ be two input

states, yielding outputs ψψg and ψ′ψ′g′ respectively

Since U preserves inner products:

ψψ′ = ψψ′ψψ′gg′ so

ψψ′(1− ψψ′gg′) = 0 so

ψψ′ = 0 or 1

ψ

0

0

ψ

ψ

g

U

Homework or exam

Introduction to Introduction to Quantum Information ProcessingQuantum Information Processing

CS 467 / CS 667CS 467 / CS 667Phys 667 / Phys 767Phys 667 / Phys 767C&O 481 / C&O 681C&O 481 / C&O 681

Richard Cleve DC [email protected]

Lecture 2 (2005)

Source of slidesSource of slides

• Used in 2007, easy to explain, students can do detailed calculations for circuits to analyze them. Good for similar problems.