summaryrefslogtreecommitdiff
path: root/source/know/concept/deutsch-jozsa-algorithm/index.md
blob: 223877a3704f08dd0232e2a6de63a213feb6957c (plain)
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
100
101
102
103
104
105
106
107
108
109
110
111
112
113
114
115
116
117
118
119
120
121
122
123
124
125
126
127
128
129
130
131
132
133
134
135
136
137
138
139
140
141
142
143
144
145
146
147
148
149
150
151
152
153
154
155
156
157
158
159
160
161
162
163
164
165
166
167
168
169
170
171
172
173
174
175
176
177
178
179
180
181
182
183
184
185
186
187
188
189
190
191
192
193
194
195
196
197
198
199
200
201
202
203
204
205
206
207
208
209
210
211
212
213
214
215
216
217
218
219
220
221
222
223
224
225
226
227
228
229
230
---
title: "Deutsch-Jozsa algorithm"
sort_title: "Deutsch-Jozsa algorithm"
date: 2021-04-08
categories:
- Quantum information
- Algorithms
layout: "concept"
---

The **Deutsch algorithm** and its extension, the **Deutsch-Jozsa algorithm**,
were first to prove that quantum computers can
solve certain problems more efficiently
than any classical system.

Given an unknown "black box" binary function $$f(x)$$ of one or more bits $$x$$,
the goal is determine whether $$f$$ is
**constant** (i.e. $$f(x)$$ is the same for all $$x$$)
or **balanced** (i.e. exactly 50% of all $$x$$-values yield $$f(x) = 0$$,
and the other 50% yield $$f(x) = 1$$).
We can query $$f$$ as many times as we want with inputs of our choice,
but we want to solve the problem using as few queries as possible.

The problem is extremely artificial and of no practical use,
but quantum computers can solve it with a single query,
while classical computers need up to $$2^{N - 1} + 1$$ queries
for an $$N$$-bit $$x$$.



## Deutsch algorithm

The Deutsch algorithm handles the simplest case,
where $$x$$ is only a single bit.
Only four $$f$$ exist:

+ **Constant**: $$(f(0) = f(1) = 0)$$ or $$(f(0) = f(1) = 1)$$.
+ **Balanced**: $$(f(0) = 0, f(1) = 1)$$, or $$(f(0) = 1, f(1) = 0)$$.

In other words, we only need to determine if $$f(0) = f(1)$$ or $$f(0) \neq f(1)$$.
To do this, we use the following quantum circuit,
where $$U_f$$ is the oracle we query:

{% include image.html file="deutsch-circuit.png" width="48%"
    alt="Deutsch circuit" %}

Due to unitarity constraints,
the action of $$U_f$$ is defined to be as follows,
with $$\oplus$$ meaning XOR:

$$\begin{aligned}
    \Ket{x} \Ket{y}
    \quad \to \boxed{U_f} \to \quad
    \Ket{x} \Ket{y \oplus f(x)}
\end{aligned}$$

Starting on the left from two qubits $$\Ket{0}$$ and $$\Ket{1}$$,
we apply the [Hadamard gate](/know/concept/quantum-gate/) $$H$$ to both:

$$\begin{aligned}
    \Ket{0} \Ket{1}
    \quad \to \boxed{H^{\otimes 2}} \to \quad
    \Ket{+} \Ket{-}
    = \frac{1}{2} \Big( \Ket{0} + \Ket{1} \Big) \Big( \Ket{0} - \Ket{1} \Big)
\end{aligned}$$

Feeding this result into the oracle $$U_f$$ then leads us to:

$$\begin{aligned}
    \to \boxed{U_f} \to \quad
    \frac{1}{2} \Ket{0} \Big( \Ket{0 \oplus f(0)} - \Ket{1 \oplus f(0)} \Big)
    + \frac{1}{2} \Ket{1} \Big( \Ket{0 \oplus f(1)} - \Ket{1 \oplus f(1)} \Big)
\end{aligned}$$

The parenthesized superpositions can be reduced:
let us suppose that $$f(b) = 0$$, then:

$$\begin{aligned}
    \Ket{0 \oplus f(b)} - \Ket{1 \oplus f(b)}
    = \Ket{0 \oplus 0} - \Ket{1 \oplus 0}
    = \Ket{0} - \Ket{1}
\end{aligned}$$

On the other hand, if we assume that $$f(b) = 1$$,
we get the opposite result:

$$\begin{aligned}
    \Ket{0 \oplus f(b)} - \Ket{1 \oplus f(b)}
    = \Ket{0 \oplus 1} - \Ket{1 \oplus 1}
    = - \big(\Ket{0} - \Ket{1}\big)
\end{aligned}$$

We can thus combine both cases, $$f(b) = 0$$ or $$f(b) = 1$$,
into the following expression:

$$\begin{aligned}
    \Ket{0 \oplus f(b)} - \Ket{1 \oplus f(b)}
    = (-1)^{f(b)} \big(\Ket{0} - \Ket{1}\big)
\end{aligned}$$

Using this, we rewrite the intermediate state of the quantum circuit like so:

$$\begin{aligned}
    \Ket{0} \Ket{1}
    \quad \to \boxed{H^{\otimes 2}} \to \boxed{U_f} \to \quad
    \frac{1}{2} \Big( (-1)^{f(0)} \Ket{0} + (-1)^{f(1)} \Ket{1} \Big) \Big( \Ket{0} - \Ket{1} \Big)
\end{aligned}$$

The second qubit in state $$\Ket{-}$$ is garbage (i.e. no longer of interest).
The first qubit is:

$$\begin{aligned}
    \frac{1}{\sqrt{2}} \Big( (-1)^{f(0)} \Ket{0} + (-1)^{f(1)} \Ket{1} \Big)
    = \frac{(-1)^{f(0)}}{\sqrt{2}} \Big( \Ket{0} + (-1)^{f(0) \oplus f(1)} \Ket{1} \Big)
\end{aligned}$$

If $$f$$ is constant, then $$f(0) \oplus f(1) = 0$$,
meaning this state is $$(-1)^{f(0)} \Ket{+}$$.
On the other hand, if $$f$$ is balanced, then $$f(0) \oplus f(1) = 1$$,
meaning this state is $$(-1)^{f(0)} \Ket{-}$$.
Taking the Hadamard transform of this qubit therefore yields:

$$\begin{aligned}
    \to \boxed{H} \to \quad
    (-1)^{f(0)} \Ket{f(0) \oplus f(1)}
\end{aligned}$$

Depending on whether $$f$$ is constant or balanced,
the measurement outcome of this state will be $$\Ket{0}$$ or $$\Ket{1}$$
with 100% probability. We have solved the problem!

Note that we only consulted the oracle (i.e. applied $$U_f$$) once.
A classical computer would need to query it twice,
once with input $$x = 0$$, and again with $$x = 1$$.



## Deutsch-Jozsa algorithm

The Deutsch-Jozsa algorithm generalizes the above to $$N$$-bit inputs $$x$$.
We are promised that $$f(x)$$ is either constant or balanced;
other possibilities are assumed to be impossible.
This algorithm is then implemented by the following quantum circuit:

{% include image.html file="deutsch-jozsa-circuit.png" width="52%"
    alt="Deutsch-Jozsa circuit" %}

There are $$N$$ qubits in initial state $$\Ket{0}$$, and one in $$\Ket{1}$$.
The oracle $$U_f$$ performs this action:

$$\begin{aligned}
    \Ket{x_1} \Ket{x_2} \cdots \Ket{x_N} \Ket{y}
    \quad \to \boxed{U_f} \to \quad
    \Ket{x_1} \cdots \Ket{x_N} \Ket{y \oplus f(x_1, ..., x_N)}
\end{aligned}$$

Applying the $$N + 1$$ Hadamard gates to the initial state
yields the following superposition:

$$\begin{aligned}
    \Ket{0}^{\otimes N} \Ket{1}
    \quad \to \boxed{H^{\otimes N + 1}} \to \quad
    \Ket{+}^{\otimes N} \Ket{-}
    = \frac{1}{\sqrt{2^N}} \sum_{x = 0}^{2^N - 1} \Ket{x} \Ket{-}
\end{aligned}$$

Where $$\Ket{x} = \Ket{x_1} \cdots \Ket{x_N}$$ denotes a classical binary state.
For example, if $$x = 5 = 2^0 + 2^2$$ in the summation,
then $$\Ket{x} = \Ket{1} \Ket{0} \Ket{1} \Ket{0}^{\otimes N-3}$$
(from least to most significant digit).

We give this state to the oracle,
and, by the same logic as for the Deutsch algorithm,
get back:

$$\begin{aligned}
    \to \boxed{U_f} \to \quad
    \frac{1}{\sqrt{2^N}} \sum_{x = 0}^{2^N - 1} (-1)^{f(x)} \Ket{x} \Ket{-}
\end{aligned}$$

The last qubit $$\Ket{-}$$ is garbage.
Next, applying the Hadamard transform to the other $$N$$ gives:

$$\begin{aligned}
    \to \boxed{H^{\otimes N}} \to \quad
    \frac{1}{\sqrt{2^N}} \sum_{x = 0}^{2^N - 1} (-1)^{f(x)}
    \bigg( \frac{1}{\sqrt{2^N}} \sum_{y = 0}^{2^N - 1} (-1)^{x \cdot y} \Ket{y} \bigg)
\end{aligned}$$

Where $$x \cdot y$$ is the bitwise dot product of the binary representations of $$x$$ and $$y$$,
so, for example, if $$N = 2$$, then $$x \cdot y = x_1 y_1 + x_2 y_2$$.
Note that the above expression has not been reduced at all;
it follows from the definition of the Hadamard transform.
We can rewrite it like so:

$$\begin{aligned}
    \frac{1}{2^N} \sum_{x = 0}^{2^N - 1} \sum_{y = 0}^{2^N - 1} (-1)^{f(x) + x \cdot y} \Ket{y}
    = \sum_{y = 0}^{2^N - 1} \bigg( \frac{1}{2^N} \sum_{x = 0}^{2^N - 1} (-1)^{f(x) + x \cdot y} \bigg) \Ket{y}
    = \sum_{y = 0}^{2^N - 1} c_y \Ket{y}
\end{aligned}$$

The parenthesized expression can be interpreted as the coefficients
of a superposition of several $$y$$-values.
Therefore, the probability that a measurement yields $$y = 0$$,
i.e. $$\Ket{y} = \Ket{0}^{\otimes N}$$, is:

$$\begin{aligned}
    |c_0|^2
    = \bigg| \frac{1}{2^N} \sum_{x = 0}^{2^N - 1} (-1)^{f(x)} \bigg|^2
\end{aligned}$$

The summation always contains an even number of terms, for all values of $$N$$.
Consequently, if $$f$$ is constant, then $$|c_0|^2 = |\!\pm\! 2^N / 2^N|^2 = 1$$.
Otherwise, if $$f$$ is balanced, all the terms cancel out, so we are left with $$|c_0|^2 = 0$$.
In other words, we reach the same result as the Deutsch algorithm:
we only need to measure the $$N$$ qubits once;
$$f$$ is constant if and only if all are zero.

The Deutsch-Jozsa algorithm needs only one oracle query to give an error-free result,
whereas a classical computer needs $$2^{N-1} + 1$$ queries in the worst case.
A revolutionary discovery!


## References
1.  J.S. Neergaard-Nielsen,
    *Quantum information: lectures notes*,
    2021, unpublished.
2.  S. Aaronson,
    *Introduction to quantum information science: lecture notes*,
    2018, unpublished.