Quantum Teleportation Quantum Superdense Coding Quantum Parallelism and the Deutsch Algorithm

Lecture 23

Quantum Information Processing and Computation - II Quantum No-Cloning Theorem Quantum Teleportation Quantum Superdense Coding Quantum Parallelism and the Deutsch Algorithm

In this lecture you will learn:

• Can one copy qubits? • Quantum teleportation • Quantum information, coding, and superdense coding • Quantum parallelism, quantum computing and the Deutsch algorithm • Quantum parallelism, quantum computing and the Bernstein-Vazirani algorithm

ECE 3030 – Summer 2009 – Cornell University Classical Digital Circuits: Bit Copying A B Copying of bits is very common in digital circuits

We make copies of digital data and store it all the time every day:

Magnetic hard drive Electronic flash drive Cloud data storage

Can quantum bits or qubits be copied?

ECE 3030 – Summer 2009 – Cornell University Copying (or Cloning) Quantum Bits (or Qubits)

Suppose one has a qubit  A of system A and one needs to make a copy of this qubit

Qubit to Qubit to  A  be copied A be copied Input ˆ Output Initialized U 0  Copy (or clone) system B B B of the qubit input Qubit copying gate Suppose one has realized this quantum copying or cloning device. We write its operation as: ˆ AB  U AB0

Suppose e now use this machine to clone another qubit  A of system A:  A  A Input Uˆ Output 0 B  B Qubit copying gate ˆ AB  U AB0

ECE 3030 – Summer 2009 – Cornell University Copying (or Cloning) Quantum Bits (or Qubits) Satisfied with your success, you decide to make a copy of the superposition state: 1  2 AA Input Uˆ Output 0 B Qubit copying gate We can work out the output state as follows: 11 UUˆˆ  000    22AA B  A BA B  11 UUUˆˆˆ00 0 0 22AB AB   AB AB 1 Entangled state! 2 AB  AB

But we wanted this output: 11    22AA BB No linear quantum operation can perfectly clone any arbitrary qubit

ECE 3030 – Summer 2009 – Cornell University Quantum No-Cloning Theorem: Formal Proof by Contradiction Suppose we have a cloning machine that can clone two arbitrary states of system A:  A and  A : ˆ AB  U AB0

ˆ AB  U AB0

Take the inner product of the states appearing on the left and right sides of the above two equations: ˆˆ† AB ABAB  00UU AB 00UUˆˆ† AABBAB    AB 00 AABBAABB   2   This means that either   0 or   1 . Which means that the two states we considered  and cannot be arbitrary – they are either the same or A  A orthogonal and hence our initial assumption that any arbitrary state of system A can be cloned was wrong

ECE 3030 – Summer 2009 – Cornell University Quantum Teleportation: The Setting Consider the following scenario:

Alice has a qubit in some unknown state  A and she wants to send it to Bob, who is sitting on the planet Zorg with his own qubit initialized in state . Alice 0 B and Bob want 0 to become B  B

B 0 B  A A

 B

In the most general case:  A 01

Note: Alice cannot “look” at her qubit and then convey the information to Bob over the telephone so that Bob can convert his own qubit to match that of Alice. “Looking” will collapse the qubit.

The only way, it seems, this could be possible is if Alice were to send her own qubit physically on a starship all the way to Bob ………..

ECE 3030 – Summer 2009 – Cornell University Quantum Teleportation: Entanglement as a Resource

A long long time ago, when Bob was still on planet Earth, they generated an entangled qubit pair ……….

1 0 01 A A 2 AA A H 1 00 11 A 2 AB AB B B 0 B 0 B B And then Bob went away to planet Zorg and took his entangled qubit with him …….

ECE 3030 – Summer 2009 – Cornell University Quantum Teleportation: Entanglement as a Resource

A  A Teleportation

B A

Alice and Bob, now on distant planets, share an entangled qubit pair: 1 S 00 11 2 AB AB

This existing shared entangled qubit pair can be used to “teleport” Alice’s new qubit  A onto Bob’s qubit B

ECE 3030 – Summer 2009 – Cornell University Quantum Teleportation: Local Operations of Alice

 A 01 A B Shared entangled qubit pair 1 S 00 11 A 2 AB AB

A  A H A S B

  S 1 in A  00011 AABAB  1 2 00  11 AABAB 1 2  1 1 0  0 1 A AB AB 1  2 00011   AABAB  1 2 01 0011   1 22AA ABAB   1 0 0  1 1 A  2 AB AB  1  0 1 1 0 0 1  22A AABAB 

ECE 3030 – Summer 2009 – Cornell University Quantum Teleportation: Local Operations of Alice A Measurement  A H A Measurement S B

 1 01 0011   22AA ABAB 0 or 1  1  0 1 1 0 0 1  0 or 1 22A AABAB     11 01 0011   01  1001  22A AABABAAABAB 22 11  00 0 1 01  1 0 22AA B B AA  B B 11  1 0 0 1 1 1  1 0 22AA B B AA  B B Alice’s Results Probability Bob’s Collapsed Qubit

0 and 0 0.25  01BB  0 and 1 0.25  10BB 

1 and 0 0.25  01BB  1 and 1 0.25  10BB 

ECE 3030 – Summer 2009 – Cornell University Quantum Teleportation: Local Operations of Bob

Telephone call A Alice provides just two bits of classical B information to Bob A based on her measurement results Alice’s Results Bob’s Qubit after Local Operations

01 0 and 0 BB

B B  10  01 0 and 1 BB X BB

1 and 0 B B  Z 01BB  01BB  

B B Y i 01 1 and 1  10BB  BB

ECE 3030 – Summer 2009 – Cornell University Quantum Teleportation: End Result

B A

When the dust has settled the quantum state of the three qubits is any one of the following – each with a-priori probability 1/4

00 0 1 AA B B

01 0 1 AA B B

10 0 1 A ABB 

i 11 0 1 AA B B  Entanglement has vanished – but the initial qubit A of Alice has been “teleported” to qubit B of Bob

ECE 3030 – Summer 2009 – Cornell University Classical Information in Quantum States Consider a quantum system whose states belong to a N-dimensional Hilbert space with the following basis states: N ˆ  eejj1 ee jk  jk j 1 Any quantum state of the system can be written as: N    aejj j 1

If Alice wants to encode classical information in bits in the quantum state  , and then send the quantum state  to Bob, how much information in bits can Alice send to Bob which Bob can gain by making the best possible measurements on  ?

 01001100 …..  

ECE 3030 – Summer 2009 – Cornell University Classical Information in Quantum States: Alice’s Coding  010011001 …..  

Suppose Alice chooses the following coding scheme: Classical  Info

000 e1 Hilbert space dimension: N=8

001 e2 8 ˆ e  eejj1 ee jk  jk 010 3 j 1 011 e4 Three classical bits are mapped to one of 100 e5 eight different states in the Hilbert space of 101 e6 the quantum system

110 e7

111 e8

ECE 3030 – Summer 2009 – Cornell University Classical Information in Quantum States: Bob’s Measurements  010011001 …..   Measurement

We assume that the quantum system has a CSCO and that each basis state e j can be associated with a unique set of eigenvalues of the operators in the CSCO

So if each observable in the CSCO is measured for the quantum state  by Bob, then these measurements will let Bob figure out which one of the states e j was sent by Alice

Result:

For each state  sent by Alice, Bob obtains 3 bits of classical information after making his measurements

ECE 3030 – Summer 2009 – Cornell University Classical Information in Quantum States: Generalization Consider a quantum system whose states belong to a N-dimensional Hilbert space with the following basis states: N ˆ  eejj1 ee jk  jk j 1 N Any quantum state of the system can be written as:    aejj j 1

 01001100 …..  

Answer:

Measurements on the state  will give Bob log2(N) bits of classical information

Therefore, a state in a 2-dimensional Hilbert space (i.e. a qubit) carries just one bit of classical information

ECE 3030 – Summer 2009 – Cornell University Quantum Superdense Coding

Question: Can Alice send Bob more than one bit of classical information by sending just one qubit?

ECE 3030 – Summer 2009 – Cornell University Quantum Superdense Coding: Alice’s Local Operations Suppose Alice and Bob, on distant planets, share an entangled qubit pair: 1 S 00 11 2 AB AB A B

Alice’s Classical Alice’s Local Operations Information

A A 1 1 00 11 00 11 1ˆ AB AB 00 2 AB AB 2

1 A A 1 01 00 11 X 10 01 2 AB AB 2 AB AB

1 A A 1 10 00 11 Z 00 11 2 AB AB 2 AB AB

1 A A i 00 11 Y 10 01 11 2 AB AB 2 AB AB

ECE 3030 – Summer 2009 – Cornell University Quantum Superdense Coding: Bob’ Local Operations

Alice then sends qubit to Bob

Bob uses a reverse Bell circuit on both the qubits B

Alice’s Bob’s Local Operations A A Classical H Information B B 1 A A 00 11 00 00 AB AB A B Reverse Bell Circuit 2 B B

A A Alice is able to 1 01 10 01 01 send two classical AB AB A B 2 B B bits of information A A by sending just 10 1 00 11  10A B AB AB one qubit by using 2 B B the entangled i A A resource !!! 10 01 i 11 11 AB AB A B 2 B B

ECE 3030 – Summer 2009 – Cornell University Quantum Parallelism and the Deutsch Algorithm Suppose we have a classical function of one input classical bit:

x F . Fx

There are four possibilities for the function F:

FF00 10  Constant function FF00 11  One-two-one function   FF01 10 One-two-one function   FF01 11 Constant function

Question: Given a function black box, how many times do we need to evaluate the function F(.) to figure out if the function F(.) is a constant or a one-to-one function??

Answer: At least two times! Once with a “0” input and once with a “1” input.

ECE 3030 – Summer 2009 – Cornell University Quantum Parallelism and the Deutsch Algorithm x A A F x A xy,0,1  B y yFx  B B

Suppose we have a two-qubit unitary quantum gate that does not do anything to the A input qubit, but it changes the input B qubit such that it contains the XOR of B’s original value “y” and the function F(x), where “x” is the value of A

ECE 3030 – Summer 2009 – Cornell University Quantum Parallelism and the Deutsch Algorithm A 0 A H F H Measurement 1 B B H

01AB 1 01 01 2 AA BB 1 00FFFF 0 01 0 10 1 11 1 2 AAAABBBB

Suppose F is a constant then the above state is one of the following two: 1 00 01 10 11 2 AB AB AB AB

Suppose F is a one-to-one then the above state is one of the following two: 1 00 01 11 10 2 A B AB AB A B

ECE 3030 – Summer 2009 – Cornell University Quantum Parallelism and the Deutsch Algorithm A 0 A H F H Measurement 1 B B H Output Suppose F is a constant then the output is one of the following two: 1 001  ABB2  Suppose F is a one-to-one then the output is one of the following two: 1 101  ABB2  Therefore, if a measurement of qubit A at the end yields “1” then F is one-to-one, otherwise F is a constant

It follows that quantum mechanics allows one to determine if a function is a constant or one-to-one using only a single evaluation of the function

ECE 3030 – Summer 2009 – Cornell University Quantum Parallelism and the Bernstein-Vazirani Algorithm Suppose we have a classical function of N classical input bits and one output bit:

x1 x2 x F . 3 N Fx123, x , x .... xNjj  sx mod 2 j 1 xN 

Question: Given a function black box, how many times do we need to evaluate the function F(.) to figure out the function (i.e. figure out the string sss 123 , , .... s N )

Answer: At least N times! Each time with one of the following inputs:

x1 1 x1 0 x1 0 x1 0  x 0 x 1 x 0 x 0 2  2  2  2  x3  0 x3  0 x3  1 x3  0               xN 0 xN 0 xN 0 xN 1

ECE 3030 – Summer 2009 – Cornell University Quantum Parallelism and the Bernstein-Vazirani Algorithm A x1 1 x1 A1 A1 x A2 2 A x2 2 A2 xyj ,0,1 F

A xN N xN AN AN

B y yFxx , ,... x B 12 N B

Suppose we have a (N+1)-qubit unitary quantum gate that does not do anything to the input A qubits, but it changes the input B qubit such that it contains the XOR of B’s original value “y” and the function Fx12,,... x xN

ECE 3030 – Summer 2009 – Cornell University Quantum Parallelism and the Bernstein-Vazirani Algorithm A 0 H 1 H A1 s1 0 A2 A2 H H s2 F

0 AN AN H H sN

B 1 H yFxx , ,... x B 12 N B

A measurement of all qubits A at the end yields the string sss123, , .... sN

It follows that quantum mechanics allows one to determine the string sss 123 , , .... sN using only a single evaluation of the function !! This is N times faster than the best classical algorithm!

ECE 3030 – Summer 2009 – Cornell University