Tutorial on Quantum Computing Vwani P. …helper.ipam.ucla.edu/publications/nanotut/nanotut_1865.pdfQuantum Teleportation There are two players: Alice and Bob. Alice is given a quantum

Tutorial on Quantum Computing

Vwani P. Roychowdhury

Lecture 1: Introduction

1

Fundamentals

Qubits

A single qubit is a two–state system, such as a two–level atom

we denote two orthogonal states of a single qubit as�� Any state of this system can be in arbitrary superposition:

� ��

� qubits

more generally, a system of � qubits represents a Hilbert space of

dimension "! , and the state of the “system” is a superposition of the

basic states # �%$&('*) � & �,+ & �-�# �%$&('*) � � & � � �

�,+ & �*� �/. ) . �*01010 . !�� . &32 4 �5�687

2

Entangeled States

Example 1: product state��

��

��

��

Example 2: EPR (Einstein–Podolsky–Rosen) pair��

��

3

Unitary Operations

��

Hadamard transformationa 1–qubit operation, denoted by � , and performs the

following transform

��

� � � � � � ��

NOTa 1–qubit operation

��

4

Measurement

Example 1:��

� ��

measurement ��

with probability)��

with probability��

Example 2: partial measurement�

��5� � �� first system measured ��

with probability)� and the system collapses to

��

with probability)� and the system collapses to

��

Example 3: �� )

�� )

�� first system measured ��

the system collapses to) �

�� the system collapses to

) � ��

5

Quantum Teleportation

There are two players: Alice and Bob.

Alice is given a quantum system�� unknown to her.

Alice Bob�

��

classical channel

Alice wants to send sufficient information about��

so that Bob is able to make an accurate copy of it.

Alice is not allowed to transform the original particle.

Alice Bob

��

6

Suppose Alice and Bob each has a particle of an EPR

pair:

Alice Bob�

��

��

��

EPR pair ��

��

EPR pair:��

� ��

7

the state of the three particles:��

� � ��

��

��

It can be written as

��

��

8

where

��

��

��

��

The four states

��

are a complete orthonormal basis for particles 1 and 2.

9

Alice measures in Bell operator basis of� �� the joint system

consists of the particle 1 (i.e.,�� ) and the particle 2

(her EPR particle).

Regardless of the unknown state�� , the four

measurment outcomes are equally likely.

After Alices’s measurment, Bob’s particle 3 will be

projected into one of the following four pure states:

result of Alice’s measurment Bob’s particle 3��

� � ��

Alice, via the classical channel, sends the result of the

measurement. Then Bob performs the corresponding

unitary opertaor on his particle.

10

Communication Complexity

Alice Bob�

��

�

Alice and Bob are provided with binary strings � and� .

goal: Alice has to determine � � � � � � , where

� � � � � ��

with as little communication between Alice and Bob as

possible.

11

trivial solution: Bob sends all his � bits to Alice.

nontrivial question: Can this be done with

communicating less than � bits?

Two–party communication complexity of the

function � is the minimum number of bits that must be

communicated between Alice and Bob in order for

Alice to compute � � � � � � .

Generalization to more than two parties is

straightforward.

12

Example 1: The parity function

� � � � � � � � ��

: addition modulo two.

It suffices for Bob to send a single bit (i.e.,� �� ) to Alice.

Example 2: The inner product modulo two

��

It is proved that � bits of communication is necessary.

That is, there is no communication protocol that

allows Alice to compute�� if the total number

of bits communicated is less than � .

Question: If Alice and Bob share quantum entangled

particles, is it possible to reduce the communication

complexity?

13

There are � parties. The party��

receives� ��

. The inputs satisfy�� mod� � � � �

The goal is to compute

� � � � � � � � � � ��

�� mod� �

which is always is 0 or 1.

The classical communication complexity of�

is� �� .

If parties share entangled particles then�

can be

computed by communicating � bits.

14

The parties share the cat state

��

where the party� �

holds the qubit� �

.

15

the protocol:

(1) Each party� �

applies the phase–change operator

��

� � � � � �� $ � � �

on his qubit� �

.

(2) Each� �

applies the Hadamard transform

��

� � � � ��

on his qubit� �

.

(3)� �

measures his qubit� �

in the basis� ��

,

and sends the outcoming bit� �

to the others.

Then� � � � � � � � � � � �� .

16

Why is this protocol correct?

After applying the phase–change operator, the

entangled state becomes�� $��

The Hadamard transform results in (with�

� $�� factor)

�� $�� $��

�� $� � � � � � � ��

or

�� $��

�� $� � � � � � � ��

which is � even

�� if� � � � � � � � � � �

, and

� odd

�� otherwise.

17

The RSA public key cryptosystem

plain message � ��

Alice

� � � plain message

Bob

� �private file

�

� �public file

�open channel

�Eve

18

private file

prime numbers: � and ��

integres � and � such that � � �� (mod � � � � �where � � � � � least common multiple � � � � � � � � �

public file�

and �

� � � � � � � � mod � �

� � � � � � � � mod � �

It holds that

� � � � � � � � � � � � � mod � � � �

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Example:

� � � � � � � � then � � ��

� � � � � l.c.m.��

chose� � � � � then

� �� mod

� � � ��

public file: � � �� and� � � �

If the plaintext is� � �� then the ciphertext is

� � ��

mod ��

Bob uses� � � � � ��

mod �� to obtain the

plain text�

.

20

Quantum Cryptography

plain message � encryption

Alice

decryption � plain message

Bob

�open channel

�Eve

� �

key

Quantum channel

21

Generating quantum key distribution:

Alice and Bob generate their own independent sets of

random numbers:

Alice 0 0 1 0 1 1 0 1 0 1��

Bob 1 0 0 0 1 0 1 1 0 0��

They try to find a suset of identical bits in their sets of

random bits.

Alice * 0 * * 1 * * 1 * *��

Bob * 0 * * 1 * * 1 * *��

The result is their common key:

� � � ��

22

How to find the common bits securely?

Alice sends, through a quantum channel, the following

quantum states for each bit in her set

bit state� ��

� ��

Bob makes a measurement on each state he receives.

The measurement depends on his corresponding bit.

bit measurement� ��

where��

� ��

� �� where��

23

Measurement� � � � is a projection opertor

��

The result of this measurement on state��

is

� � � �� , with probability� � � � � � � ��

,

� � � � � � � �� , with probability� � � � � � � � � ��

,

24

The probability of Bob’s measurement results:

Alice’s bit Bob’s bit measurement result probability decision

0 0�� Y

0 0�� N

0 1��

1 N

1 0��

1 N

1 1�� Y

1 1�� N

Y=“pass”; Bob observes��

when he makes the projection��

.

N=“fail”.

If the bits are different, the decision is always “N”. If Bob and Alice

have the same bit then with probability)� the decision is “Y”.

25

Bob sends the sequence of Y–N decisions to Alice over

a public channel.

They keep only the bits for which the decision is “Y”.

The result becomes the shared key.

26

Deutsch’s Algorithm: Fault Detection (?)

A Boolean function � � � � � �� is given.

It is known that either � is constant or balanced (i.e.,

takes values 0 and 1 an equal number of times). The

problem is to decide whether � is constant or

balanced.

� � �

��

��

��

��

��

��

��

measurement

� � � � � ��

� ��

� � ��

27

Step 1: prepare the state

��

where the first register consists of � qubits.

Step 2: perform the Hadamard transform on the first �

qubits

� �� $� � � ��

�� $

� � �� $

� � � � � ��

28

Step 3: apply� � � � � ��

� �� $� � ��

� �� $� � ��

note that for� � � � � ��

��

so the result of this step is�� $

� � � � � ��

note that

� �� $� � � � � ��

�� #

� �� $ � � � if � is constant,�� if � is balanced,

where��

is orthogonal to��

.

29

Step 4: apply the Hadamard transform � on the first

� qubits

note that

� ��

because � � � � ��

is orthogonal to � �� ,

because the unitary operation � maps orthogonal

states to orthogonal ones.

the final result

if � is constant � ��

if � is balanced��

where��

is orthogonal to��

measure the first � qubits

� is constant if and only if we observe � zeros

30

Grover’s algorithm: Quantum search

A Boolean function ��

such

that ��

for only one value � � � � .Find: � � .We assume that � can be evaluated in unit time.

Any classical algorithm (deterministic or probabilistic)

for this problem needs to check at least�� vectors of� � � ��

.

Grover’s algorithm shows the quantum computer can

solves this problem in time �� .

31

Fix � � � � � �� , we define the unitary operator

��as

for� � � � � ��

� $,�� if � � � � � � � otherwise.

We will apply� �

, for � � �, and

� � � . The operator� �can be computed efficiently (in time polynomial in

� );

In the beginning we do not know � � , but still it is

possible to compute� � � efficiently, if we assume that

�� can be computed efficiently on each given input

� � � � � �� .

The algorithm is based on iteration of the following

transform:

� � � � � � � � � �

32

First we prepare the uniform superposition

�� $

� � �

Then we apply the transform

on��

repeatedly

The operator

maps each superposition of the form

� � � � ��

��

to itself.

33

Initially � � � � � ��

� � and after � iteration of

, for ��

� � $ ,

��

��

After�

iterations the probability of measuring � � ,i.e.�� , is very close to 1 if

� � ��

34

��

��

� � �

��

� � �

��

� � � � � � ��

�

� �

Computing the unitary operator� � �

35

� � �

�

�

�

�

�

��

Computing the unitary operator

36

�

�

�

��

��

��

Search algorithm

37

Documents

Tutorial on Quantum Computing Vwani P. …helper.ipam.ucla.edu/publications/nanotut/nanotut_1865.pdfQuantum Teleportation There are two players: Alice and Bob. Alice is given a quantum