quantum computing talk

8/14/2019 Quantum Computing Talk

1/16

A gentle introduction to quantum computing

Mark Przepiora

University of Calgary

July 9, 2009

Mark Przepiora CUMC 2009
http://find/http://goback/


2/16

What is a computation?

Intuitively, you start with n bits, say 011011101.

Gears turn, electrons flow, and you modify the bits in someway, until you end up with an m-bit string, say 101.

So a computation implements some functionf : {0, 1}n {0, 1}mThis doesnt explain how the computation is performed,though.

Strategy: interpret a simple computation algebraically, and

see how this interpretation generalizes to the quantum world.



3/16

A simple, classical computation

Say our computer has exactly one bit of storage, so the stateof our machine is always simply either 0 or 1.

We want to compute the NOT function, i.e. NOT(0) = 1and NOT(1) = 0.

Lets identify states in our machine with basis vectors:

0 10 1 01

Then it makes sense to identify

NOT 0 11 0 (1)The above gives us an algebraic way of looking atcomputation without needing to worry about the physicalprocess performing it.



4/16

And now for something completely different

Lets interpret the above in the context of quantum mechanics.

According to quantum mechanics, every physical system canbe represented by a Hilbert space (i.e. Cn in thefinite-dimensional case)

Any nonzero vector | Cn is a possible state of the systemAnd a mechanism can (in principle) be built to implement anyunitary transformation on Cn

Only unitary transformations can be realized. In particular, alloperations are invertible. (Information never destroyed.)

Given a state

|

and any orthonormal basis

{|x

}, we can

build a device to measure | = x ax |x in this basis,which will result in |x being observed with (relative)probability |ax|2 and the system will collapse into this state.So if | is already a basis vector, it can be measured exactly.



5/16

Reinterpreting the first slide

Consider a system with n = 2 (e.g., the spin of an electron.)

A system whose state space is C2 is called a qubit.

By convention we denote |0 =

10

and |1 =

01

Note that the NOT matrix X = 0 1

1 0

is unitary

If we could prepare the system in one of the states |0 or |1and construct a device to implement the above matrix, wecould apply it to compute NOT of a single bit.

This method is actually much more general, though,because we could prepare the system in any superpositiona |0 + b|1 and compute its quantum NOT



6/16

A more interesting one-qubit operator

Consider the unitary transformation H = 12

1 11 1

(the

Hadamard matrix.)

Easy to see it is unitary.

H |0 = |0

+

|1

2 andH |1 = |

0

|1

2Generates a qubit thats half 0 and half 1.

Measuring such a qubit would result in either a 0 or 1 beingobserved, each with 50% probability.

This transformation is a fundamental part of many quantumalgorithms.



7/16

Quantum circuits

A quantum computation is the path a qubit takes (timepassing from left to right) along a series of quantum gates(each of which implements a single unitary transformation.)

For example, the circuit X H implements

the transformation HX.

A circuit may also involve a measurement denoted by

NM

, which measures that qubit in the computational

basis (according to the laws given earlier.)

For example, the circuit|0 H NM

b

implements a random number generator.

We will use this model later to discuss a quantum algorithm.



8/16

Tensor products

More generally, two qubits (which are physically separate)may be considered to be a single physical system via tensor

products.

Their combined state space is C4, and a basis for it is givenby {|00 , |01 , |10 , |11}For example, a two-qubit system may be in the state

|01

, i.e.

the first qubit is a |0 and the second qubit is a |1. Here itmakes sense to talk about the state of each qubit individually.

Or it could be in the state |00 + |11 which is an example ofan entangled state.

The state |01 may also be written |0 |1 or |0 |1 or |0, 1to emphasize that it comes from two separate physicalsystems.

This product behaves exactly like youd expect.(Distributivity, etc.)



9/16

Entanglement

What about this state |00 + |11?

It no longer makes sense to talk about each qubit being insome state on its own.

If we measure the two-qubit system all at once, by the laws ofmeasurement we would expect to observe the results 00 and11 with equal probability.

But what if we measure a qubit individually? Remember theyare separate physical systems, so we should be able to do this!

We havent discussed the laws governing with this situation,but intuitively we can see that if we measure the first qubit to

be a 0, then any subsequent measurement of the second qubitmust also show that it is a 0.

The qubits may be miles apart and this will still hold. (Whydoes this not allow for FTL information transfer?)


C ll d NOT


10/16

Controlled NOT

So we can see that the two-qubit case is genuinely interesting!

What kind of interesting two-qubit operators can weimplement?

Controlled NOT is a quantum analog of the XOR function,mapping

|00 |00 , |01 |01|10 |11 , |11 |10

The above mapping can be restated as

|a, b |a, a bor leave the first bit alone, and flip the second bit if the firstbit is a 1.


F i l


11/16

Function oracles

Quantum computation should be a generalization of classicalcomputation, so it is reasonable to ask how we can compute a

classical function f : {0, 1} {0, 1} in the quantum world.Ideally, wed like an operator that performs the following map:

|a, |a, f(a)

(In which the first qubit is the input, and we place the outputin the second qubit.) However, it is clear that this mapping isnot invertible.

We can use a generalization of the CNOT operator instead.

LetUf denote the two-qubit operator that performs thefollowing map:

|a, b |a, f(a) bAn operator of this form is called the quantum oracle of f,and we can think of it as a black box.


L i b l f ti


12/16

Learning a boolean function

Forget about the quantum world, and say we have a black boxthat computes some unknown f : {0, 1} {0, 1} classically.f could be one of four functions. Say wed like to learn whichit is, without being allowed to look inside the black box. How

many queries are required?Its easy to prove that we need two queries classically.

In the quantum case, what if were given f as a bit oracle Uf?

Here, the answer is still two queries! So we see that quantum

computation doesnt allow us to speed up all computationaltasks, only some.


L i t


13/16

Learning a property

So we cant speed up the process of learning f exactly, but what if

we simply want to determine the valuef

(0) f(1), i.e. whether ornot f is a constant function?Classically, it is again easy to convince yourself that twoqueries are required.

However, only a single query is required to the quantum bitoracle!

This is the Deutsch Algorithm, which is displayed below.

|0 H

Uf

HNM

b

|1 HThe result is b = f(0) f(1).


Q t thb sti


14/16

Quantum mythbusting

I often read things like this:

A quantum computer could split into millions of identicalcopies, unleashing immense parallel computing power...New Scientist

Quantum machines may one day be capable of massivelyparallel computing, in which billions of calculations happen at

oncea feat that will never be possible with silicon chips.Popular Science

Why is it deceptive?

The billions of calculations interfere with each other.Finding a quantum algorithm to determine even a simpleproperty is hard!

Can we really use a quantum computer to find a needle in ahaystack?


Quantum search


15/16

Quantum search

Say we have n items, exactly one of which has an interestingproperty. How many items do we need to check before we findthe needle?

Classically, n

1 in the worst case.

With a quantum computer, amazingly only n queries need tobe made in the black box model. This is Grovers algorithm.

Just as importantly,

n is the best we can do.

This is the difference between 1,000,000 queries and 1,000

queries, but not the speedup promised by magazines.


Addendum 1 destroying information


16/16

Addendum 1 - destroying information

Question: Doesnt measurement destroy information?Answer: Not really, because of deferred measurement.Intermediate measurements can be moved to the very end of the

computation, and the intermediate steps can be replaced byquantum operations.In other words, we can worry about measurements when theuniverse ends.


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