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<!DOCTYPE HTML>
<html>
<head>
<meta charset="utf-8">
<meta name="viewport" content="initial-scale=1.0, width=device-width" />
<title>An Interactive Introduction To Quantum Computing Part 2</title>
<link rel="stylesheet" href="main.css" />
</head>
<body>
<article>
<h1>An Interactive Introduction To Quantum Computing Part 2</h1>
<div class="subtitle">
<b>Quantum Search</b>
</div>
<p class="aside">
This article was originally written in 2014,
but has had some minor improvements in December 2017 and January 2018
when I received some useful feedback after the article
had unexpectedly been posted on
<a href="https://news.ycombinator.com/">Hacker News</a>
in December 2017.
</p>
<p>
This is Part 2 of a two part series.
In <a href="index.html">Part 1</a>
I explained some of the basics of qubits,
including the mysterious notions of phase and quantum interference.
Here in Part 2, you will learn
how quantum computers may one day be able to solve search problems
faster than conventional computers.
</p>
<h2>Multiple qubits</h2>
<p>
<b>Important</b>: There is something really important I feel the need to emphasize
as we start dealing with multiple qubits.
</p>
<p>The blue disks introduced in Part 1 do <em>not</em> correspond to single qubits.</p>
<p>
<b>Each blue disk corresponds to a distinct state of the entire quantum system.</b>
If you are not sure what I mean by that,
then please hang in there,
it will hopefully become aparent very soon.
</p>
<h2>Controlled NOT</h2>
<p>
The <em>Controlled NOT</em> operator is a very simple but powerful
two-qubit operator.
</p>
<p>
The Controlled NOT operator modifies the value of one qubit (the target qubit)
based on the value of another qubit (the control qubit).
</p>
<p>
Specifically, <b>the Controlled NOT operator flips the value of the target qubit
if the control qubit has a value of 1.</b>
</p>
<p>
For example:
Suppose we have two qubits with
an initial value of “11”.
Let's use the left qubit as the control qubit
and the right qubit as the target.
Then the result of applying a Controlled NOT
will be “10”.
As usual, we represent this as
“11 → 10”
</p>
<p>
Continuing with our choice of left qubit as control qubit
and right qubit as target,
there are four scenarios to consider for
the Controlled NOT operator.
</p>
<ul>
<li>00 → 00 (i.e. no effect)</li>
<li>01 → 01 (i.e. no effect)</li>
<li>10 → 11 (i.e. target bit flips)</li>
<li>11 → 10 (i.e. target bit flips)</li>
</ul>
<h2>Quantum Entanglement</h2>
<p>
A combination of the Hadamard followed by a Controlled NOT
can put two qubits into a state of “quantum entanglement”.
</p>
<p>
In what follows, we take a two qubit system in the state “00”,
apply the Hadamard operation to the left qubit,
and then use that left qubit as the control in a Controlled NOT
of the right qubit:
</p>
<h3>Initial State</h3>
<div class="animationBlock" id="cnot-0">
<div class="qstate">State: <span class="qstateValue"></span></div>
<svg class="qstateSvg qstateSvg__hideBitLabels"></svg>
</div>
<h3>After applying a Hadamard to the left qubit</h3>
<p>
When you apply a Hadamard to the left qubit,
that qubit goes into
a superposition of 0 and 1.
The right qubit is not touched and continues to have the value 0.
So the pair of qubits goes into
a superposition of 00 and 10.
</p>
<div class="animationBlock" id="cnot-1">
<div class="qstate">State: <span class="qstateValue"></span></div>
<svg class="qstateSvg qstateSvg__hideBitLabels"></svg>
</div>
<h3>After a Controlled NOT of the right qubit controlled by the left qubit</h3>
<p>
The Controlled NOT will leave the 00 state unchanged as the control qubit (left qubit) is 0.
</p>
<p>
When acting on the 10 state, the control qubit is 1,
and so the target bit is flipped resulting in 11.
So the pair of qubits goes into
a superposition of 00 and 11.
</p>
<div class="animationBlock" id="cnot-2">
<div class="qstate">State: <span class="qstateValue"></span></div>
<svg class="qstateSvg qstateSvg__hideBitLabels"></svg>
</div>
<p>
Note that, if you were to measure the qubits while they are in this state,
you would find that they are either both 0 or both 1.
This sort of correlation between qubit values
within a superposition is called
“quantum entanglement”.
</p>
<p>
Why quantum entanglement is so special deserves an entire article of its own.
For now, I am simply using entanglement to
emphasise how quantum superposition can apply to
the combined state of a number of qubits.
</p>
<h2>Arbitrary Logic Functions</h2>
<p>
In addition to the NOT operator that we have already seen,
a quantum computer can implement the standard
logic functions of AND and OR.
From these, one can compute any function over a fixed number of bits.
</p>
<p>
The details are surprisingly fascinating
and not as straight forward as you might expect.
However,
for the purposes of this article,
I want you to trust me when I say it can be done.
This will allow us to move more quickly to
the even more fascinating topic of quantum search.
</p>
<p>
Before we can cover quantum search,
there is one more quantum operator I need to introduce.
</p>
<h2>Controlled Phase Flip</h2>
<p>
Recall that the term <em>phase</em>,
introduced in <a href="index.html">Part 1</a>,
refers to the arrow direction of a state.
</p>
<p>
The controlled phase flip is a multi qubit operator.
It flips the phase of the states where
all of the specified qubits have the value of 1.
</p>
<p>
More useful to us, however,
is a generalised form of the controlled phase flip.
It will flip the phase of the states where the target qubits
have any specified combination of 1's and 0's.
</p>
<p>Try it yourself:</p>
<div class="animationBlock" id="generalisedControlledPhaseFlipExample">
<div class="animationBlock_buttons">
<div>
<button class="operator" data-operator="f00"
data-skipInterferenceSteps
data-qubits="0">Phase Flip 00</button>
<button class="operator" data-operator="f01"
data-skipInterferenceSteps
data-qubits="0">Phase Flip 01</button>
<button class="operator" data-operator="f10"
data-skipInterferenceSteps
data-qubits="0">Phase Flip 10</button>
<button class="operator" data-operator="f11"
data-skipInterferenceSteps
data-qubits="0">Phase Flip 11</button>
</div>
</div>
<svg class="qstateSvg qstateSvg__hideBitLabels"></svg>
</div>
<h2>Quantum Search</h2>
<p>
Imagine you are trying to crack an n-bit encryption key by brute force:
i.e. by trying every one of the 2<sup>n</sup> possible values.
</p>
<p>
This is a classic needle in a haystack search problem.
</p>
<p>
In other words,
it belongs to a class of problems for which
there exists an efficient function,
called an <em>oracle</em> (or <em>verification</em>) function,
that takes a candidate value as input
and returns true if the candidate is the correct value.
However,
the only way to find the correct value
is to test the oracle function on practically every possible value.
</p>
<p>
A conventional computer will need to evaluate the oracle function,
on average,
<span class="math">2<sup>n-1</sup></span> times.
</p>
<p>
By using
<a href="http://wikipedia.org/wiki/Grover's_algorithm">Grover's algorithm</a>,
a quantum computer only needs to evaluate the oracle function less than
<span class="math">√(2<sup>n</sup>)</span> times.
</p>
<p>
Even if a conventional computer can compute the oracle function in a nanosecond,
if the input was 64 bits long,
then it would probably take hundreds of years
for that conventional computer
to find the answer.
A quantum computer might possibly take just a few seconds.
<span class="help">
A conventional computer would have to evaluate the oracle function an average of
<span class="math">2<sup>63</sup></span> times,
compared to just over four billion times for a quantum computer.
It is likely however,
at least in the early days of quantum computing,
that individual operations on a quantum computer may be vastly slower
than on a conventional computer.
It may take a while before quantum computers ever get
to solve a problem any faster than conventional computers.
</span>
</p>
<p>
For the extremely simple case where the input is only 2 bits,
there are four possible answers.
A conventional computer may need to test up to three
of those possible values.
(If it isn't the first three,
then you can infer it must be fourth one without testing it.)
As you will see below,
a quantum computer can find the answer
using a single invocation of the oracle function.
</p>
<h2>Quatum Search over 2 Qubits</h2>
<p>
First recall that
the objective is to find the value of <span class="math">x</span>
for which the oracle function is true.
For Grover's Search Algorithm to work,
it is essential that the oracle function is true
for exactly one value of <span class="math">x</span>.
</p>
<p>
In the conventional descriptions of Grover's Search Algorithm,
it is assumed that the the oracle function
will flip the value of an “output” qubit
when the input matches the desired result.
It is a lot more convenient if we do away with the output bit,
and instead flip the <em>phase</em>
when the input matches the desired result.
</p>
<p>
Here are the steps to Grover's Search Algorithm
for the very simple case of two bits.
</p>
<p>
For example, the problem may be to
<em>find factors of 15 that are less than 4</em>.
This is obviously trivial for such small values,
but the problem of finding factors for very large numbers
(e.g. numbers with over 100 digits)
is currently a very computationally expensive problem.
</p>
<ol>
<li>
Start with the qubits in a superposition of all possible states:
00, 01, 10, and 11 (all with the same phase).
</li>
<li>
Apply the oracle function once only.
Note that,
because the system starts in a superposition of all possible states,
the oracle function is effectively applied to all possible states,
but it will only flip the phase of the state for which we are searching.
If you were to perform a measurement now,
all the possible outcomes are still equally likely
and so we don't seem to have made any progress.
However, the next three steps use quantum interference to find the correct answer.
</li>
<li>Apply a Hadamard operation to both qubits.</li>
<li>Flip the phase of the 00 state.</li>
<li>Apply a Hadamard operation to both qubits again.</li>
<li>
Constructive and destructive quantum interference will ensure that,
of the four possible states,
a measurement is guaranteed to result in only the desired state.
</li>
</ol>
<p>
You can try it yourself below.
Your task is to find the value for which the oracle function
<span class="math">f</span>
is "true".
i.e. the state for which
<span class="math">f</span>
will flip the phase.
On a classical computer you would typically end up testing
at least half the states before finding the answer.
On a quantum computer you can test all the states simultaneously.
</p>
<div class="animationBlock" id="SimpleSearchExample"
data-bit-labels="bit-a,bit-b">
<div class="animationBlock_buttons">
<div>
<div data-step="1">
Apply the oracle function (to all states):
<button class="operator" data-operator="f10"
data-skipInterferenceSteps
data-qubits="0"><span class="math">f<sub>oracle</sub></span></button>
</div>
<div>
<button class="operator" data-operator="hadamard"
data-qubits="1" data-step="2">Hadamard qubit-a</button>
<button class="operator" data-operator="hadamard"
data-qubits="0" data-step="3">Hadamard qubit-b</button>
</div>
<div>
<button class="operator" data-operator="f00" data-skipInterferenceSteps
data-qubits="0" data-step="4">Phase flip 00</button>
</div>
<div>
<button class="operator" data-operator="hadamard"
data-qubits="1" data-step="5">Hadamard qubit-a</button>
<button class="operator" data-operator="hadamard"
data-qubits="0" data-step="6">Hadamard qubit-b</button>
</div>
<div class="aside">
Only the searched-for state now has a non-zero probability
and a measurement can be safely applied.
</div>
</div>
</div>
<svg class="qstateSvg"></svg>
<div class="animationBlock_buttons">
<button class="reset">Reset</button>
</div>
</div>
<p>
Don't forget,
unlike what happens in the animation,
each operation is effectively applied to
all the states simultaneously.
</p>
<h2>Beyond Quantum Search</h2>
<p>
The two bit search is a bit of a special case
where the oracle function only needs to be invoked once.
As the number of bits increases,
it becomes necessary to invoke the oracle function multiple times.
In fact it needs to be invoked
<span class="math">¼π√(2<sup>n</sup>)</span>
times.
</p>
<p>
While general quantum search techniques may seem quite impressive,
even <span class="math">¼π√(2<sup>n</sup>)</span> gets very large very quickly.
</p>
<p>
All you have to do is double the number of bits to wipe out any advantages.
For example,
over a 128 bit search space,
a quantum computer would need to evaluate the oracle function
almost as many times as a conventional computer does
over a 64 bit search space.
Even at one evaluation per nanosecond,
it would take hundreds of years to crack
a 128 bit encryption key.
</p>
<p>
However,
encryption keys
commonly used for securing internet traffic
(<a href="http://wikipedia.org/wiki/RSA_(algorithm)">RSA</a> keys)
can be broken using a special purpose quantum algorithm.
These encryption schemes rely on the difficulty of
factoring very large numbers,
something that can be easily done on a quantum computer using
<a href="http://wikipedia.org/wiki/Shor's_algorithm">Shor's Algorithm</a>.
</p>
<p>
Shor's Algorithm has a “asymptotic” behaviour of
<span class="math">O((log N)<sup>3</sup>)</span>.
I won't try to explain
the exact meaning of this statement,
but it is quite possible that
current encryption keys could be broken in sub-second time frames.
Also, simply doubling the length of the key would have a barely noticeable
effect on the time taken.
The limiting factor is likely to be the number of qubits required
since you need at least as many qubits as the key length.
</p>
<h2>D-Wave</h2>
<p>
You may have heard of
<a href="http://www.dwavesys.com/">D-Wave</a>
quantum computers.
</p>
<p>
These are currently the only commercially available quantum computers.
D-Wave are building quantum computers containing 512 qubits,
which in theory could be in a super position of
<span class="math">2<sup>512</sup></span> states.
<span class="aside">
Update Jan 2017:
<a href="https://www.nature.com/news/d-wave-upgrade-how-scientists-are-using-the-world-s-most-controversial-quantum-computer-1.21353">
D-Wave have now built a 2000 qubit system
and have plans for a 4000 qubit system.
</a>
</span>
</p>
<p>
Despite having such a large number of qubits,
there is little evidence that they have been solving
problems any faster than conventional computers.
</p>
<p>
D-Wave computers are special purpose
“quantum annealing” computers.
The quantum computers that I have been describing so far are
known as “quantum gate” computers.
Describing the differences are beyond the scope of this
(already long) article.
</p>
<p>
Before they can start out-performing conventional computers,
it is possible that D-Wave simply needs to use more qubits,
or perhaps they may need to figure out how to make
more effective use of their existing qubits.
However,
it quite possibly will not be long before
we start seeing impressive results.
</p>
<h2>Some philosophy</h2>
<p>
Describing quantum computing
in terms of superpositions of states
that “collapse”
to the observed state
upon “measurement”
is a very conventional approach
to explaining the physics
behind quantum computing.
However, there are some serious
deficiencies with this approach.
</p>
<p>
Probably the most fundamental problem is:
<b>
What is so special about measurement that it causes
quantum superpositions to collapse?
</b>
</p>
<p>
The measuring equipment itself is made of the same stuff
as the qubits being measured
(protons, neutrons, electrons, and photons).
</p>
<p>
When measuring a qubit that is in a superposition of
two different states,
there is nothing in quantum physics that says why
the measuring equipment
might cause the superposition to collapse to a single state
instead of the measuring equipment
becoming entangled with the qubit.
Instead of the quantum state collapsing,
the result could be a superposition of the following two states:
</p>
<ul>
<li>the qubit being 0 and the measuring equipment detecting 0</li>
<li>the qubit being 1 and the measuring equipment detecting 1</li>
</ul>
<p>
If you consider your conscious awareness to be
nothing more than patterns of activity in your brain,
and that your brain is governed by the laws of physics,
then there is nothing in quantum physics that says why,
when you look at the measurement outcome,
you yourself do not become entangled with
the quantum qubit and
the measuring equipment.
The result would be a superposition of the following two states:
</p>
<ul>
<li>the qubit being 0, the measuring equipment detecting 0,
and you being conscious of the measurement being 0</li>
<li>the qubit being 1, the measuring equipment detecting 1,
and you being conscious of the measurement being 1</li>
</ul>
<p>
<b>Think of it as two alternative realities existing.</b>
In one reality you see a 0,
and in the other you see a 1.
Quantum interference is the only ways that these realities interact.
</p>
<p>
Variations on this interpretation of quantum physics
are taken quite seriously
by many physicists and physics philosophers, and is
commonly referred to as the “Many Worlds”
interpretation of quantum physics.
Much has been written on the topic,
and an internet search will point you to
many good (and not so good) articles.
</p>
<p>
Quantum computing
provides a nice way of looking at some of these issues.
It may even provide a way of simulating ideas by,
for example,
representing different states of an observer using qubits.
</p>
<p>
This article has already covered more topics than I originally planned.
However, I feel the need to include a little bit of
a mathematical explanation to some of the material.
For those that are not interested in the mathematics,
feel free to skip to the
<a href="#usefulLinks">Useful Links Section</a>
at the end.
</p>
<h2 id="maths">Some mathematics</h2>
<p>
In this article
I used little blue disks with red arrows
to represent the probabilities
of a set of bits or qubits
being in different states.
It is likely that I caused some confusion
by quietly using the disks in subtly different ways
when describing
two quite different types of uncertainty.
</p>
<p>
When describing ordinary non-quantum bits,
the system is in a definite state,
but which state that is may be unknown.
In this situation,
it is most convenient if
the arrow lengths are proportional to the probabilities
of the respective states
so that they can be added head to tail.
If there were two scenarios, S1 and S2, both ending in the same state,
where S1 has a probability of p1
and S2 has a probability of p2,
then the probability of that final state is
<span class="math">p1 + p2</span>.
</p>
<p>
When describing qubits
in a superposition of different states,
it is no longer simply a case of not knowing which state they are in,
the system actually seems to be in multiple states at once.
In this situation,
it is most convenient if
the arrow lengths are proportional to the <em>square roots</em> of the probabilities
of the respective states
(the disk <em>areas</em> now effectively represent probabilities
except for a factor of π).
The arrow directions represent their phases.
When two scenarios end in the same state,
you need to use vector addition to add the arrows.
</p>
<p>
<b>The arrows represent what physicists call “amplitudes”</b>.
</p>
<p>
Conventionally, an amplitude is represented using a complex number
whose value on the complex number plane coincides with the end of our arrow.
If our arrow has a length r and a phase θ,
then it is represented as the complex number:
</p>
<p>
<span class="math">r (cos(θ) + i sin(θ))</span>
</p>
<p>
If a state had this amplitude,
then the probability of that state being the outcome
of a measurement is <span class="math">r<sup>2</sup></span>.
</p>
<p>
When we talk of “adding arrows head to tail”,
we can simply add the corresponding complex numbers using
ordinary complex number arithmetic.
</p>
<p>
Given that the probabilities of all possible outcomes have to add up to 1,
it follows that the sum of the squares of the amplitude magnitudes
also have to add up to 1.
All quantum operators preserve this property.
</p>
<p>
If the amplitudes for the 0 and 1 states of a qubit are
the (possibly complex) numbers
<span class="math">a<sub>0</sub></span>
and
<span class="math">a<sub>1</sub></span>
respectively,
then applying a Hadamard results in
the amplitudes of 0 and 1 becoming
<span class="math">(1/√2)(a<sub>0</sub> + a<sub>1</sub>)</span>
and
<span class="math">(1/√2)(a<sub>0</sub> - a<sub>1</sub>)</span>
respectively.
</p>
<p>
For the simple case of a qubit being 100% in
the 1 state with a phase of 0,
applying a Hadamard results in
the amplitudes of 0 and 1 being
1/√2 and -1/√2 respectively.
Notice that if you square and add these together that you get 1.
</p>
<p>
That is already more mathematics as I wanted to cover in this article.
For more, see some of the useful links listed in the next section.
</p>
<h2 id="usefulLinks">Useful links</h2>
<ul>
<li>
An excellent learning resource is
<a href="https://cs.uwaterloo.ca/~watrous/CPSC519/LectureNotes/all.pdf">
John Watrous' Introduction to Quantum Computing Lecture Notes
</a>.
</li>
<li>
If you want to experiment with the various qubit operators,
then have a play with the
<a href="http://davidbkemp.github.io/quantum-gate-playground/">
Quantum Computer Gate Playground
</a>.
</li>
<li>
Try doing some quantum programming using
the <a href="http://davidbkemp.github.io/jsqubits/jsqubitsRunner.html">
jsqubits runner
</a>,
or the
<a href="http://qcplayground.withgoogle.com/#/home">
Quantum Computing Playground
</a>
</li>
<li>
The
<a href="http://machinelevel.com/qc/">QCEngine</a>
is definitely worth exploring.
</li>
</ul>
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