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 transformation a 1 qubit operation, denoted by following transform, and performs the NOT a 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 classical channel Bob 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. goal: Alice has to determine, where and 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. If parties share entangled particles then can be computed by communicating bits. is 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 and sends the outcoming bit. in the basis 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 odd even otherwise. if, and 17
The RSA public key cryptosystem Eve public file Alice Bob private file plain message open channel plain message 18
private file prime numbers: and integres and such that (mod where least common multiple public file and mod mod It holds that mod 19
Example: then l.c.m. chose then mod public file: and If the plaintext is then the ciphertext is mod Bob uses mod plain text. to obtain the 20
Quantum Cryptography Eve Alice Bob plain message encryption decryption plain message open channel 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 0 0 0 1 1 0 1 1 1 1 Y N 1 N 1 N Y N Y= pass ; Bob observes N= fail. when he makes the projection. 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 $ $ where is orthogonal to. if if is constant, is balanced, 29
Step 4: apply the Hadamard transform qubits on the first 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 that for only one value. such 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 for, we define the unitary operator as $, if o 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