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.gitignore
View file @ 57d588e2
Makefile 0 → 100644
View file @ 57d588e2
# Directories for latex files and outputs
LATEX_DIR := content/latex
IMG_DIR := content/img
# Get all .tex files in the LaTeX source directory
TEX_FILES := $(wildcard $(LATEX_DIR)/*.tex)
# Define target PDF, DVI and SVG files in the image output directory
PDF_TARGETS := $(patsubst $(LATEX_DIR)/%.tex,$(IMG_DIR)/%.pdf,$(TEX_FILES))
DVI_TARGETS := $(patsubst $(LATEX_DIR)/%.tex,$(IMG_DIR)/%.dvi,$(TEX_FILES))
SVG_TARGETS := $(patsubst $(LATEX_DIR)/%.tex,$(IMG_DIR)/%.svg,$(TEX_FILES))
# Default removal command for Unix-like systems
RMDEL = rm -f
# Default command to create the output directory if it doesn't exist for Unix-like systems
MKDIR_IMG = mkdir -p $(IMG_DIR)
# Default command to delete all PDFs and SVGs to which a .tex file belongs
RM_CLEAN = \
@for texfile in $(TEX_FILES); do \
filename=$$(basename $$texfile .tex); \
pdf="$(IMG_DIR)/$$filename.pdf"; \
svg="$(IMG_DIR)/$$filename.svg"; \
[ -f "$$pdf" ] && $(RMDEL) "$$pdf"; \
[ -f "$$svg" ] && $(RMDEL) "$$svg"; \
done
# Generated files to clean up
AUX_FILES = $(IMG_DIR)/*.aux
LOG_FILES = $(IMG_DIR)/*.log
PDF_FILES = $(IMG_DIR)/*.pdf
DVI_FILES = $(IMG_DIR)/*.dvi
SVG_FILES = $(IMG_DIR)/*.svg
# Windows-specific settings: commands and backslash path separator
ifeq ($(OS),Windows_NT)
RMDEL = del /Q
MKDIR_IMG = if not exist $(subst /,\,$(IMG_DIR)) mkdir $(subst /,\,$(IMG_DIR))
RM_CLEAN = \
@for %%f in ($(TEX_FILES)) do ( \
if exist "$(IMG_DIR)\%%~nf.pdf" $(RMDEL) "$(IMG_DIR)\%%~nf.pdf" & \
if exist "$(IMG_DIR)\%%~nf.svg" $(RMDEL) "$(IMG_DIR)\%%~nf.svg" )
AUX_FILES = $(subst /,\\,$(IMG_DIR))\\*.aux
LOG_FILES = $(subst /,\\,$(IMG_DIR))\\*.log
DVI_FILES = $(subst /,\\,$(IMG_DIR))\\*.dvi
SVG_FILES = $(subst /,\\,$(IMG_DIR))\\*.svg
PDF_FILES = $(subst /,\\,$(IMG_DIR))\\*.pdf
endif
.PHONY: all preview clean clean-temp clean-dvi
# Default target: build all PDFs and SVGs, then clean .aux and .log files and run quarto render
all: $(PDF_TARGETS) $(SVG_TARGETS)
-@$(RMDEL) $(AUX_FILES)
-@$(RMDEL) $(LOG_FILES)
quarto render --to all
# Rule to live-preview the script in the browser
preview: $(PDF_TARGETS) $(SVG_TARGETS)
quarto preview index.qmd
# Rule to build PDFs: run pdflatex with output to IMG_DIR
$(IMG_DIR)/%.pdf: $(LATEX_DIR)/%.tex | $(IMG_DIR)
pdflatex -output-directory=$(IMG_DIR) $<
# Rule to build DVI files: run latex with output to IMG_DIR
$(IMG_DIR)/%.dvi: $(LATEX_DIR)/%.tex | $(IMG_DIR)
latex -output-directory=$(IMG_DIR) $<
# Rule to build SVG files from DVI files: run dvisvgm
$(IMG_DIR)/%.svg: $(IMG_DIR)/%.dvi
dvisvgm --no-fonts -o $@ $<
# Ensure output directory exists before building targets that depend on it
$(IMG_DIR):
$(MKDIR_IMG)
# Clean only final output files (SVGs and PDFs)
clean:
$(RM_CLEAN)
# Clean auxiliary files generated during LaTeX compilation
clean-temp:
-@$(RMDEL) $(AUX_FILES)
-@$(RMDEL) $(LOG_FILES)
# Clean intermediate DVI files
clean-dvi:
-@$(RMDEL) $(DVI_FILES)
README.md
View file @ 57d588e2
# Introduction to Quantum Computing - Lecture notes
Lecture note for the lecture "Introduction to Quantum Computing" by Dominique Unruh at RWTH Aachen.
## Installation guide
This guide is for VS Code. There may be other extensions for other development environments.
### 1. Install Quarto
Download and install Quarto from the official website:
<https://quarto.org/docs/get-started/>
### 2. Install the VS Code extension
Install the Quarto Extension for Visual Studio Code:
https://quarto.org/docs/get-started/
You can also find it in the Extensions tab by searching for `"Quarto"`.
### 3. Preview the script
To live-preview the project, use the following command in your terminal:
`"make preview"`
This will open your browser.
### 4. Render the script
To render the entire project to .pdf and .html use:
`"make"`
The output will be generated in the _book folder.
### 5. Add pictures to the lecture notes
Not all LaTeX packages are supported for output as HTML, so graphics and similar elements created with LaTeX must be included via external files. For HTML, the preferred format is **SVG** and for PDF documents, it is **PDF**.
The easiest way is to create LaTeX files in the `content/latex` folder. These are automatically converted to SVG and PDF by **make**.
The resulting files can then be integrated using:
`"![Figure description](img/filename){width=xxx%}"`
**Important:**
The SVGs and PDFs have the same name as the LaTeX files, and no file extension should be specified when including them.
## Project structure
The main components of the project are:
### `_quarto.yml`
This is the main configuration file for your project. It contains:
- **Title**, **authors**, **output formats** and other metadata
- The **structure** of the script is under the `chapters:` section
You can **toggle chapter visibility** by commenting/uncommenting lines using `#`. This is useful for revealing content gradually throughout the semester.
### `index.qmd`
This file contains the **introduction** of the script.
It must be located in the same directory as the `quarto.yml` file.
### `Makefile`
Generates all the required images and the lecture notes from them, regardless of the operating system.
The Makefile is explained with comments.
### `content/`
This folder contains all the **chapters** of the script, as individual `.qmd` files.
### `content/img/`
Subfolder that contains all **images** and **pdfs** used throughout the script.
### `content/latex/`
Subfolder that contains all **tex-files** for creating some PDFs and SVGs. Unfortunately, not all latex files are available anymore.
### `_extensions/`
This folder includes any **custom Quarto extensions** that add functionality or styling to your project.
### `_book/`
This is the **output folder** generated by Quarto after rendering.
It contains the final script in the formats **HTML** and **PDF**.
_book/Introduction-to-Quantum-Computing.epub deleted 100644 → 0
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File deleted
_book/Introduction-to-Quantum-Computing.pdf
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_book/bernstein_vazirani.svg deleted 100644 → 0
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<h1 class="title"><span class="chapter-number">9</span>&nbsp; <span class="chapter-title">Bernstein-Vazirani Algorithm</span></h1>
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......@@ -321,35 +277,40 @@ window.Quarto = {
<p>We will now look at a quantum algorithm which will find <span class="math inline">\(s\)</span> with only one evaluation of <span class="math inline">\(f\)</span>. This algorithm is sketched in the following circuit:</p>
<div class="quarto-figure quarto-figure-center">
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<p><img src="bernstein_vazirani.svg" class="img-fluid figure-img" style="width:90.0%"></p>
<p><img src="img/bernstein_vazirani.svg" class="img-fluid figure-img" style="width:90.0%"></p>
<figcaption>The quantum circuit for Bernstein-Vazirani</figcaption>
</figure>
</div>
<p>Note that <span class="math inline">\(U_f\)</span> is defined with the explanation below.</p>
<p>We start with <span class="math inline">\(n\)</span> qubits on the top wire. All of these qubits are in the state <span class="math inline">\(\ket{0}\)</span>, which we write <span class="math inline">\(\ket{0}^n = \ket{0} \otimes \dots \otimes \ket{0}\)</span>. The bottom wire is in the state <span class="math inline">\(\ket{1}\)</span>. Both wires composed together can be written as <span class="math inline">\(\psi_1 = \ket{0}^n \otimes \ket{1}\)</span>, which is the overall starting state of our algorithm (the state at slice 1). We now perform the following steps</p>
<p>We start with <span class="math inline">\(n\)</span> qubits on the top wire. All of these qubits are in the state <span class="math inline">\(\ket{0}\)</span>, which we write <span class="math inline">\(\ket{0}^n = \ket{0} \otimes \dots \otimes \ket{0}\)</span>. The bottom wire is in the state <span class="math inline">\(\ket{1}\)</span>. Both wires composed together can be written as <span class="math inline">\(\ket{\psi_0} = \ket{0^n1} = \ket{0}^n \otimes \ket{1}\)</span>, which is the overall starting state of our algorithm. We now perform the following steps</p>
<ol type="1">
<li><p>First we apply a Hadamard gate on all qubits. This is denoted for the first <span class="math inline">\(n\)</span> qubits by the <span class="math inline">\(H^{\otimes n}\)</span> gate and for the last qubit by the <span class="math inline">\(H\)</span> gate on the bottom wire. The resulting quantum state is calculated as follows: <span class="math display">\[
\begin{aligned}
\psi_2&amp;= \left(H^{\otimes n} \otimes H\right) \left(\ket{0}^n \otimes \ket{1}\right)\\
\ket{\psi_1}
&amp;= \left(H^{\otimes n} \otimes H\right) \left(\ket{\psi_0}\right)\\
&amp;= \left(H^{\otimes n} \otimes H\right) \left(\ket{0}^n \otimes \ket{1}\right)\\
&amp;=\left(H^{\otimes n}\ket{0}^n\right) \otimes H\ket{1}\\
&amp;=\left(\ket{+}\right)^{\otimes n} \otimes \ket{-}\\
&amp;=\ket{+}^{\otimes n} \otimes \ket{-}\\
&amp;=\left(\frac{1}{\sqrt{2}} \ket{0} + \frac{1}{\sqrt{2}} \ket{1} \right)^{\otimes n} \otimes \ket{-}\\
&amp;=\frac{1}{\sqrt{2^{n}}}\sum_{x\in \{0,1\}^n} \ket{x} \otimes \ket{-}
\end{aligned}
\]</span> Roughly speaking, we are now in the superposition over all classical possibilities on the top wire and in <span class="math inline">\(\ket{-}\)</span> on the bottom wire at slice 2.</p></li>
\]</span> Roughly speaking, we are now in the superposition over all classical possibilities on the top wire and in <span class="math inline">\(\ket{-}\)</span> on the bottom wire.</p></li>
<li><p>Next, we apply the unitary <span class="math inline">\(U_f\)</span> on both wires. This unitary is defined as <span class="math display">\[
U_f\,\ket{x,y} = \ket{x,y \oplus f(x)}
\]</span> This unitary represents the function <span class="math inline">\(f\)</span> and combines the output of <span class="math inline">\(f(x)\)</span> with the bottom wire <span class="math inline">\(y\)</span>. For our quantum states, this means that the state after <span class="math inline">\(U_f\)</span> at slice 3 can be calculated as follows: <span class="math display">\[
\]</span> This unitary represents the function <span class="math inline">\(f\)</span> and combines the output of <span class="math inline">\(f(x)\)</span> with the bottom wire <span class="math inline">\(y\)</span>. For our quantum states, this means that the state after <span class="math inline">\(U_f\)</span> can be calculated as follows: <span class="math display">\[
\begin{aligned}
\psi_3&amp;=U_f\frac{1}{\sqrt{2^{n}}}\sum_{x\in \{0,1\}^n} \ket{x} \otimes \ket{-}\\
\ket{\psi_2}
&amp;= U_f\ket{\psi_1}\\
&amp;= U_f\frac{1}{\sqrt{2^{n}}}\sum_{x\in \{0,1\}^n} \ket{x} \otimes \ket{-}\\
&amp;= U_f\frac{1}{\sqrt{2^{n}}}\sum_{x\in \{0,1\}^n} \ket{x} \otimes \ket{-}\\
&amp;=\frac{1}{\sqrt{2^{n}}}\sum_{x\in \{0,1\}^n} U_f\Bigl (\ket{x} \otimes \ket{-}\Bigr) \\
&amp;\stackrel{*}{=}\frac{1}{\sqrt{2^{n}}}\sum_{x\in \{0,1\}^n} (-1)^{f(x)}\ket{x} \otimes \ket{-} \\
&amp;=\left(\frac{1}{\sqrt{2^{n}}}\sum_{x\in \{0,1\}^n} (-1)^{f(x)}\ket{x} \right) \otimes \ket{-}
\end{aligned}
\]</span> Note that the <span class="math inline">\(*\)</span> holds since we can rewrite <span class="math inline">\(U_f(\ket{x}\otimes\ket{-})\)</span> as <span class="math display">\[
\]</span> Note that the <span class="math inline">\(*\)</span> holds since we can rewrite <span class="math inline">\(U_f\Bigl(\ket{x}\otimes\ket{-}\Bigr)\)</span> as <span class="math display">\[
\begin{aligned}
&amp;U_f\Bigl(\ket{x}\otimes\ket{-}\Bigr)\\
&amp;= \frac{1}{\sqrt{2}} U_f\ket{x,0} - \frac{1}{\sqrt{2}} \ket{x,1}\\
U_f\Bigl(\ket{x}\otimes\ket{-}\Bigr)
&amp;= \frac{1}{\sqrt{2}} U_f\ket{x,0} - \frac{1}{\sqrt{2}} U_f\ket{x,1}\\
&amp;= \frac{1}{\sqrt{2}} \ket{x,f(x)} - \frac{1}{\sqrt{2}} \ket{x,\overline{f(x)}}\\
&amp;= \begin{cases} \frac{1}{\sqrt{2}} \ket{x,0} - \frac{1}{\sqrt{2}} \ket{x,1} &amp; f(x)=0\\
\frac{1}{\sqrt{2}} \ket{x,1} - \frac{1}{\sqrt{2}} \ket{x,0} &amp; f(x)=1 \end{cases}\\
......@@ -358,9 +319,11 @@ U_f\,\ket{x,y} = \ket{x,y \oplus f(x)}
&amp;= (-1)^{f(x)} \ket{x} \otimes \ket{-}
\end{aligned}
\]</span> The bottom wire has not changed and is still <span class="math inline">\(\ket{-}\)</span>. But on the top wire, we now have <span class="math inline">\(f(x)\)</span> somehow encoded into our quantum state. The phenomenon that the output of <span class="math inline">\(f\)</span> is encoded as a <span class="math inline">\(-1\)</span> in the input register is called phase kickback. Measuring this quantum state would not give us any advantage, since we would just get one random <span class="math inline">\(x\)</span>. We therefore perform one final step before measuring.</p></li>
<li><p>As the final unitary, we perform another <span class="math inline">\(H^{\otimes n}\)</span> on the top wire. We hope that the result of this unitary transformation is the state <span class="math inline">\(\ket{s}\)</span>. To check, whether our hopes become reality, we can calculate <span class="math inline">\(\left(H^{\otimes n}\right)^\dagger \ket{s} \otimes \ket{-}\)</span> and check if it is equal to <span class="math inline">\(\psi_3\)</span>. We do it in this direction, since these calculations are a bit simpler: <span class="math display">\[
<li><p>As the final unitary, we perform another <span class="math inline">\(H^{\otimes n}\)</span> on the top wire. We hope that the result of this unitary transformation is the state <span class="math inline">\(\ket{\psi_3} = \ket{s} \otimes \ket{-}\)</span>. To check, whether our hopes become reality, we can calculate <span class="math inline">\(\left(H^{\otimes n}\right)^\dagger \ket{s} \otimes \ket{-}\)</span> and check if it is equal to <span class="math inline">\(\ket{\psi_2}\)</span>. We do it in this direction, since these calculations are a bit simpler: <span class="math display">\[
\begin{aligned}
\left( H^{\otimes n}\right )^\dagger \ket{s}\otimes \ket{-} &amp;= H^{\otimes n} \ket{s}\otimes \ket{-} \\
\left(\left( H^{\otimes n}\right )^\dagger \otimes I \right) \ket{\psi_3}
&amp;= \left( H^{\otimes n}\right )^\dagger \ket{s}\otimes \ket{-}\\
&amp;= H^{\otimes n} \ket{s}\otimes \ket{-} \\
&amp;= H^{\otimes n} (\ket{s_1} \otimes \dots \otimes \ket{s_n})\otimes \ket{-}\\
&amp;= H\ket{s_1} \otimes \dots \otimes H\ket{s_n}\otimes \ket{-}\\
&amp;= \bigotimes_{i=1}^n \left(\frac{1}{\sqrt{2}} \ket{0} + (-1)^{s_i}\frac{1}{\sqrt{2}} \ket{1}\right) \otimes \ket{-}\\
......@@ -369,9 +332,9 @@ U_f\,\ket{x,y} = \ket{x,y \oplus f(x)}
&amp;=\frac{1}{\sqrt{2^{n}}}\sum_{x\in\{0,1\}^n} \left( (-1)^{\sum_i x_is_i \bmod 2}\ket{x}\right) \otimes \ket{-}\\
&amp;=\frac{1}{\sqrt{2^{n}}}\sum_{x\in\{0,1\}^n} (-1)^{s\cdot x}\ket{x} \otimes \ket{-}\\
&amp;=\frac{1}{\sqrt{2^{n}}}\sum_{x\in \{0,1\}^n} (-1)^{f(x)}\ket{x} \otimes \ket{-}\\
&amp;= \psi_3
&amp;= \ket{\psi_2}
\end{aligned}
\]</span> This calculation shows that we have the quantum state <span class="math inline">\(\ket{s} \otimes \ket{-}\)</span> at slice 4.</p></li>
\]</span> This calculation shows that we have the quantum state <span class="math inline">\(\ket{s} \otimes \ket{-}\)</span> before the measurement.</p></li>
<li><p>We now perform a measurement on the top wire and measure <span class="math inline">\(s\)</span> as a result.</p></li>
</ol>
<p>This concludes the Bernstein-Vazirani algorithm.</p>
......@@ -381,24 +344,6 @@ U_f\,\ket{x,y} = \ket{x,y \oplus f(x)}
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......@@ -333,33 +289,40 @@ window.Quarto = {
</div>
<div class="callout-body-container">
<div id="def-composite-systems" class="definition theorem">
<p><span class="theorem-title"><strong>Definition 6.1 (Composite systems / Tensor product)</strong></span> Given two probabilistic or quantum systems <span class="math inline">\(A\)</span> and <span class="math inline">\(B\)</span> with the possibilities of <span class="math inline">\(A\)</span> given by <span class="math inline">\(1,\dots,N\)</span><a href="#fn1" class="footnote-ref" id="fnref1" role="doc-noteref"><sup>1</sup></a> and a distribution/state <span class="math inline">\(a\)</span> and with the possibilities of <span class="math inline">\(B\)</span> given by <span class="math inline">\(1,\dots,M\)</span> and a distribution/state <span class="math inline">\(b\)</span>, the <em>composite</em> system called <span class="math inline">\(AB\)</span> has the possibilities <span class="math display">\[
11,12,\dots,1M,21,22,\dots,2M,\dots,N1,N2,\dots,NM
\]</span></p>
<p>and the distribution/state <span class="math inline">\(ab\)</span> of <span class="math inline">\(AB\)</span> is given by the <em>tensor product</em> <span class="math display">\[
ab= a \otimes b = \begin{pmatrix} a_1 \cdot b\\ \vdots \\ a_N \cdot b \end{pmatrix}
\]</span> This vector has the size <span class="math inline">\(NM\)</span>.</p>
<p><span class="theorem-title"><strong>Definition 6.1 (Composite systems / Tensor product)</strong></span> Given two probabilistic or quantum systems <span class="math inline">\(A\)</span> and <span class="math inline">\(B\)</span> with the possibilities of <span class="math inline">\(A\)</span> given by <span class="math inline">\(x_1, \dots, x_N\)</span> and a distribution/state <span class="math inline">\(\mu_A\)</span> and with the possibilities of <span class="math inline">\(B\)</span> given by <span class="math inline">\(y_1, \dots, y_M\)</span> and a distribution/state <span class="math inline">\(\mu_B\)</span>, the <em>composite</em> system called <span class="math inline">\(AB\)</span> has the possibilities <span class="math display">\[
x_1y_1, x_1y_2, \dots, x_1y_M, x_2y_1, x_2y_2, \dots, x_2y_M,\dots, x_Ny_1, x_Ny_2, \dots, x_Ny_M
\]</span> and the distribution/state <span class="math inline">\(\mu_{AB}\)</span> of <span class="math inline">\(AB\)</span> is given by the <em>tensor product</em> <span class="math display">\[
\mu_{AB}
\coloneqq \mu_A \otimes \mu_B
= \begin{pmatrix} (\mu_A)_1 \cdot \mu_b \\ \vdots \\ (\mu_A)_N \cdot \mu_b \end{pmatrix}.
\]</span> This vector has the size <span class="math inline">\(NM\)</span>. Here <span class="math inline">\((\mu_A)_i\)</span> stands for the i-th entry of <span class="math inline">\(\mu_A\)</span>.</p>
</div>
</div>
</div>
</div>
<p>Note that the definition of combining a probabilistic and a quantum system are the same.</p>
<p>The definition of combining a probabilistic and a quantum system are the same.</p>
<p>Notice that the entry corresponding to the possibility <span class="math inline">\(x_i y_j\)</span> in the composite system is then <span class="math inline">\((\mu_{AB})_{x_i y_j} = (\mu_A)_{x_i} (\mu_B)_{y_j}\)</span>. Here we identify the classical possibility <span class="math inline">\(x_i\)</span> and <span class="math inline">\(y_j\)</span> with the indices <span class="math inline">\(1, \dots, N\)</span> and <span class="math inline">\(1, \dots, M\)</span>, respectively the classical possibility <span class="math inline">\(x_i y_j\)</span> with the indices <span class="math inline">\(1, \dots, NM\)</span>. (So <span class="math inline">\((\mu_{AB})_{x_i y_j}\)</span> has just one index, namely <span class="math inline">\(x_i y_j \in \{1, \dots NM \}\)</span>.)</p>
<div class="callout callout-style-simple callout-tip no-icon callout-titled">
<div class="callout-header d-flex align-content-center">
<div class="callout-icon-container">
<i class="callout-icon no-icon"></i>
</div>
<div class="callout-title-container flex-fill">
Example: Tensor product
Example: Composite system
</div>
</div>
<div class="callout-body-container callout-body">
<p>Given two vectors <span class="math inline">\(a\)</span> and <span class="math inline">\(b\)</span>, with <span class="math display">\[
<p>Let the distributions for the system <span class="math inline">\(A\)</span> with the possibilities <span class="math inline">\(1,2\)</span> and the system <span class="math inline">\(B\)</span> with the possibilities <span class="math inline">\(a,b,c\)</span> be given by <span class="math display">\[
\begin{aligned}
a &amp;= \begin{pmatrix} 1 \\ 2 \\ 4 \end{pmatrix} &amp; b &amp;= \begin{pmatrix} 10 \\ 100 \end{pmatrix}
\mu_A &amp;= \begin{pmatrix} \frac{1}{\sqrt{2}} \\ -\frac{1}{\sqrt{2}} \end{pmatrix},&amp;
\mu_B &amp;= \begin{pmatrix} \frac{1}{2} \\ 0 \\ \frac{\sqrt{3}}{2} \end{pmatrix}.
\end{aligned}
\]</span> the tensor product of those vectors is given by <span class="math display">\[
a\otimes b = \begin{pmatrix} 1 \\ 2 \\ 4 \end{pmatrix} \otimes \begin{pmatrix} 10 \\ 100 \end{pmatrix} = \begin{pmatrix} 1\cdot 10 \\ 1 \cdot 100 \\ 2\cdot 10 \\ 2 \cdot 100 \\ 4\cdot 10 \\ 4 \cdot 100 \end{pmatrix} = \begin{pmatrix} 10 \\ 100 \\ 20 \\ 200 \\ 40 \\ 400 \end{pmatrix}
\]</span> Then the composite system <span class="math inline">\(AB\)</span> has the possibilities <span class="math inline">\(1a, 1b, 1c, 2a, 2b, 2c\)</span> and the distribution <span class="math display">\[
\mu_{AB}
= \mu_A \otimes \mu_B
= \begin{pmatrix} \frac{1}{\sqrt{2}} \\ -\frac{1}{\sqrt{2}} \end{pmatrix} \otimes \begin{pmatrix} \frac{1}{2} \\ 0 \\ \frac{\sqrt{3}}{2} \end{pmatrix}
= \begin{pmatrix} \frac{1}{\sqrt{2}} \cdot \frac{1}{2} \\[3pt] \frac{1}{\sqrt{2}} \cdot 0 \\[3pt] \frac{1}{\sqrt{2}} \cdot \frac{\sqrt{3}}{2} \\[3pt] -\frac{1}{\sqrt{2}} \cdot \frac{1}{2} \\[3pt] -\frac{1}{\sqrt{2}} \cdot 0 \\[3pt] -\frac{1}{\sqrt{2}} \cdot \frac{\sqrt{3}}{2} \end{pmatrix}.
= \begin{pmatrix} \frac{1}{2\sqrt{2}} \\[3pt] 0 \\[3pt] \frac{\sqrt{3}}{2\sqrt{2}} \\[3pt] -\frac{1}{2\sqrt{2}} \\[3pt] 0 \\[3pt] -\frac{\sqrt{3}}{2\sqrt{2}} \end{pmatrix}.
\]</span></p>
</div>
</div>
......@@ -372,15 +335,45 @@ a\otimes b = \begin{pmatrix} 1 \\ 2 \\ 4 \end{pmatrix} \otimes \begin{pmatrix} 1
<div class="callout-body-container">
<div id="def-composite-systems-matrix" class="definition theorem">
<p><span class="theorem-title"><strong>Definition 6.2 (Composite matrices / Tensor product)</strong></span> Given two matrices <span class="math inline">\(S\)</span> and <span class="math inline">\(T\)</span> with <span class="math inline">\(S\)</span> of the size <span class="math inline">\(N \times N\)</span> and <span class="math inline">\(T\)</span> of the size <span class="math inline">\(M \times M\)</span>. The tensor product <span class="math inline">\(S \otimes T\)</span> of is given by <span class="math display">\[
S\otimes T = \begin{pmatrix} S_{11}T &amp; \dots &amp; S_{1N} T\\ \vdots &amp; \ddots &amp; \vdots\\ S_{N1} T &amp; \dots &amp; S_{NN}T \end{pmatrix}
S \otimes T
= \begin{pmatrix}
S_{11}T &amp; \dots &amp; S_{1N} T\\
\vdots &amp; \ddots &amp; \vdots\\
S_{N1} T &amp; \dots &amp; S_{NN}T
\end{pmatrix}.
\]</span></p>
</div>
</div>
</div>
</div>
<p>Overall we can say: If we apply <span class="math inline">\(S\)</span> to the system <span class="math inline">\(A\)</span> and <span class="math inline">\(T\)</span> to the system <span class="math inline">\(B\)</span>, we apply <span class="math inline">\(S\otimes T\)</span> to the composite system <span class="math inline">\(AB\)</span>.</p>
<p>If the distribution <span class="math inline">\(d_{AB}\)</span> of a given probabilistic system <span class="math inline">\(AB\)</span> can be written as a composite of two distributions <span class="math inline">\(d_A\)</span> and <span class="math inline">\(d_B\)</span>, we know that <span class="math inline">\(A\)</span> and <span class="math inline">\(B\)</span> are independent of each other. If we can not write <span class="math inline">\(d_{AB}\)</span> as two septate systems, the probabilities are depended on each other.</p>
<p>If the distribution <span class="math inline">\(d_{AB}\)</span> of a given probabilistic system <span class="math inline">\(AB\)</span> can be written as a composite of two distributions <span class="math inline">\(d_A\)</span> and <span class="math inline">\(d_B\)</span>, we know that <span class="math inline">\(A\)</span> and <span class="math inline">\(B\)</span> are independent of each other. If we cannot write <span class="math inline">\(d_{AB}\)</span> as two separate distributions, the probabilities are depended on each other.</p>
<p>If the quantum state <span class="math inline">\(\psi_{AB}\)</span> of a given quantum system <span class="math inline">\(AB\)</span> can be written as a composite of two different quantum states <span class="math inline">\(\psi_A\)</span> and <span class="math inline">\(\psi_B\)</span>, the quantum states of <span class="math inline">\(A\)</span> and <span class="math inline">\(B\)</span> are independent of each other. If we can not write <span class="math inline">\(\psi_{AB}\)</span> as a tensor product of two quantum systems, the quantum states depend on each other. We call this <em>entangled</em>.</p>
<div class="callout callout-style-simple callout-note no-icon">
<div class="callout-body d-flex">
<div class="callout-icon-container">
<i class="callout-icon no-icon"></i>
</div>
<div class="callout-body-container">
<div id="lem-math-composite" class="lemma theorem">
<p><span class="theorem-title"><strong>Lemma 6.1</strong></span> For the (unitary) matrices <span class="math inline">\(A, B, C\)</span> and <span class="math inline">\(D\)</span>, the vectors (quantum states) <span class="math inline">\(\psi, \phi\)</span> and <span class="math inline">\(\chi\)</span> and the constant <span class="math inline">\(c\)</span>, the following applies</p>
<ul>
<li><span class="math inline">\((A \otimes B)(\psi \otimes \phi) = A\psi \otimes B\phi\)</span>,</li>
<li><span class="math inline">\((A \otimes B)(C \otimes D) = AC \otimes BD\)</span>,</li>
<li><span class="math inline">\(\psi \otimes (\phi + \chi) = (\psi \otimes \phi) + (\phi \otimes \chi)\)</span>,</li>
<li><span class="math inline">\((\psi + \phi) \otimes \chi = (\psi \otimes \chi) + (\phi \otimes \chi)\)</span>,</li>
<li><span class="math inline">\(A \otimes (B + C) = (A \otimes B) + (A \otimes C)\)</span>,</li>
<li><span class="math inline">\((A + B) \otimes C = (A \otimes C) + (B \otimes C)\)</span>,</li>
<li><span class="math inline">\(c\phi \otimes \psi = c(\phi \otimes \psi)\)</span>,</li>
<li><span class="math inline">\(\phi \otimes c\psi = c(\phi \otimes \psi)\)</span>,</li>
<li><span class="math inline">\(A \otimes cB = c(A \otimes B)\)</span> and</li>
<li><span class="math inline">\(cA \otimes B = c(A \otimes B)\)</span>.</li>
</ul>
<p>These rules only apply if the dimensions of the matrices and vectors match.</p>
</div>
</div>
</div>
</div>
</section>
<section id="measuring-composite-systems" class="level2" data-number="6.2">
<h2 data-number="6.2" class="anchored" data-anchor-id="measuring-composite-systems"><span class="header-section-number">6.2</span> Measuring composite systems</h2>
......@@ -392,43 +385,56 @@ S\otimes T = \begin{pmatrix} S_{11}T &amp; \dots &amp; S_{1N} T\\ \vdots &amp; \
</div>
<div class="callout-body-container">
<div id="def-composite-systems-measure" class="definition theorem">
<p><span class="theorem-title"><strong>Definition 6.3 (Composite measurements)</strong></span> Given two systems <span class="math inline">\(A\)</span> and <span class="math inline">\(B\)</span> with possibilities <span class="math inline">\(1,\dots, N\)</span> and <span class="math inline">\(1,\dots, M\)</span> and two partial measurements <span class="math inline">\(M_A\)</span> and <span class="math inline">\(M_B\)</span> on systems <span class="math inline">\(A\)</span> and <span class="math inline">\(B\)</span> with alternatives <span class="math inline">\(A_1,\dots,A_N \subseteq \{1,\dots,N\}\)</span> and <span class="math inline">\(B_1,\dots,B_M \subseteq \{1,\dots,M\}\)</span>. The measurement <span class="math inline">\(M_A \otimes M_B\)</span> on <span class="math inline">\(AB\)</span> is a measurement with the alternatives <span class="math inline">\(C_{11},C_{12},\dots,C_{NM}\)</span> where <span class="math inline">\(C_{ij} = A_i \times B_j\)</span>.</p>
<p>If we only have a set of alternatives for system <span class="math inline">\(A\)</span>, we can do a measurement <span class="math inline">\(M_A \otimes I\)</span> with alternatives <span class="math inline">\(C_1, \dots ,C_N := A \otimes \{1,\dots, M \}\)</span>.</p>
<p><span class="theorem-title"><strong>Definition 6.3 (Composite measurements)</strong></span> Given two systems <span class="math inline">\(A\)</span> and <span class="math inline">\(B\)</span> with possibilities <span class="math inline">\(1,\dots, N\)</span> and <span class="math inline">\(1,\dots, M\)</span> and two partial measurements <span class="math inline">\(M_A\)</span> and <span class="math inline">\(M_B\)</span> on systems <span class="math inline">\(A\)</span> and <span class="math inline">\(B\)</span> with alternatives <span class="math inline">\(A_1, \dots, A_N \subseteq \{x_1, \dots, x_n\}\)</span> and <span class="math inline">\(B_1, \dots, B_M \subseteq \{y_1, \dots, y_M\}\)</span>. The measurement <span class="math inline">\(M_A \otimes M_B\)</span> on <span class="math inline">\(AB\)</span> is a measurement with the alternatives <span class="math inline">\(C_{11},C_{12},\dots,C_{NM}\)</span> where <span class="math inline">\(C_{ij} = A_i \times B_j\)</span>.</p>
<p>If we only have a set of alternatives for system <span class="math inline">\(A\)</span>, we can do a measurement <span class="math inline">\(M_A \otimes I\)</span> with alternatives <span class="math inline">\(C_1, \dots ,C_N \coloneqq A \otimes \{y_1, \dots, y_M\}\)</span>.</p>
</div>
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Example: Composite measurement (quantum)
</div>
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<div class="callout-body-container callout-body">
<p>Let <span class="math inline">\(A\)</span> be a system with the states <span class="math inline">\(1,2,3\)</span> and <span class="math inline">\(\mu_A\)</span> a measurement with the two alternatives <span class="math inline">\(A_{\text{low}} = \{1,2\}\)</span>, <span class="math inline">\(A_{\text{high}} = \{3\}\)</span>. The quantum state is <span class="math display">\[
\psi_A = \begin{pmatrix} \frac{2}{3} \\[2pt] \frac{1}{3} \\[2pt] -\frac{2}{3} \end{pmatrix}.
\]</span></p>
<p>Another system <span class="math inline">\(B\)</span> has the states <span class="math inline">\(a, b, c\)</span>. The measurement <span class="math inline">\(\mu_B\)</span> the two alternatives <span class="math inline">\(B_{\text{vocal}} = \{a\}\)</span>, <span class="math inline">\(B_{\text{consonant}} = \{b, c\}\)</span>. The quantum state is <span class="math display">\[
\psi_B = \begin{pmatrix} \frac{1}{2} \\[2pt] \frac{1}{2}i \\[2pt] \frac{1}{\sqrt{2}} \end{pmatrix}.
\]</span></p>
<p>So the composite system <span class="math inline">\(C = AB\)</span> has the classical possibilities <span class="math inline">\(1a, 1b, 1c, 2a, 2b, 2c, 3a, 3b\)</span> and <span class="math inline">\(3c\)</span>. The measurement <span class="math inline">\(\mu_C \coloneqq \mu_A \otimes \mu_B\)</span> has the alternatives <span class="math display">\[
\begin{aligned}
C_{\text{low, vocal}} &amp;= \{1a, 2a\}, &amp;&amp; C_{\text{low, consonant}} &amp;= \{1b, 1c, 2b, 2c\}, \\
C_{\text{high, vocal}} &amp;= \{3a\}, &amp;&amp; C_{\text{high, consonant}} &amp;= \{3b, 3c\}.
\end{aligned}
\]</span></p>
<p>The quantum state is <span class="math display">\[
\psi_C
= \psi_A \otimes \psi_B
= \begin{pmatrix} \frac{2}{3} \\[2pt] \frac{1}{3} \\[2pt] -\frac{2}{3} \end{pmatrix} \otimes \begin{pmatrix} \frac{1}{2} \\[2pt] \frac{1}{2}i \\[2pt] \frac{1}{\sqrt{2}} \end{pmatrix}
= \begin{pmatrix} \frac{2}{6} \\[4pt] \frac{2}{6}i \\[4pt] \frac{2}{3\sqrt{2}} \\[4pt] \frac{1}{6} \\[4pt] \frac{1}{6}i \\[4pt] \frac{1}{3\sqrt{2}} \\[4pt] -\frac{2}{6} \\[4pt] -\frac{2}{6}i \\[4pt] -\frac{2}{3\sqrt{2}} \end{pmatrix}.
\]</span></p>
<p>We get for the alternative <span class="math inline">\(C_{\text{low, vocal}}\)</span> the probability <span class="math display">\[
\left|\frac{2}{6}\right|^2 + \left|\frac{1}{6}\right|^2 = \frac{4}{36} + \frac{1}{36} = \frac{5}{36}
\]</span> and thus the post-measurement-state is <span class="math display">\[
\begin{pmatrix} \frac{2}{6} \\ 0 \\ 0 \\ \frac{1}{6} \\ 0 \\ 0 \\ 0 \\ 0 \\ 0 \end{pmatrix} / \sqrt{\frac{5}{36}}
= \begin{pmatrix} \frac{2}{6} \\ 0 \\ 0 \\ \frac{1}{6} \\ 0 \\ 0 \\ 0 \\ 0 \\ 0 \end{pmatrix} / \frac{\sqrt{5}}{6}
= \begin{pmatrix} \frac{2}{\sqrt{5}} \\ 0 \\ 0 \\ \frac{1}{\sqrt{5}} \\ 0 \\ 0 \\ 0 \\ 0 \\ 0 \end{pmatrix}.
\]</span></p>
</div>
</div>
</section>
<section id="footnotes" class="footnotes footnotes-end-of-document" role="doc-endnotes">
<hr>
<ol>
<li id="fn1"><p>Note that the possibilities are labels <span class="math inline">\(1,\dots,N\)</span> and do not have be the numbers <span class="math inline">\(1,\dots,N\)</span>. We could also label these e.g.&nbsp;<em>red</em>, <em>green</em> and <em>blue</em> (or any other label).<a href="#fnref1" class="footnote-back" role="doc-backlink">↩︎</a></p></li>
</ol>
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<li><a href="#definition-of-the-dft" id="toc-definition-of-the-dft" class="nav-link active" data-scroll-target="#definition-of-the-dft"><span class="header-section-number">10.1</span> Definition of the DFT</a></li>
<li><a href="#constructing-the-dft" id="toc-constructing-the-dft" class="nav-link" data-scroll-target="#constructing-the-dft"><span class="header-section-number">10.2</span> Constructing the DFT</a></li>
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<h1 class="title"><span class="chapter-number">10</span>&nbsp; <span class="chapter-title">Discrete Fourier Transformation</span></h1>
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<p>In <a href="shorsAlgorithm.html" class="quarto-xref"><span>Chapter 11</span></a>, we will discuss Shor’s algorithm, which requires the Discrete Fourier Transformation (DFT). Generally, a Fourier transformation is a mathematical technique that decomposes a function into its constituent frequencies. We use the DFT to find the period of a vector.</p>
<section id="definition-of-the-dft" class="level2" data-number="10.1">
<h2 data-number="10.1" class="anchored" data-anchor-id="definition-of-the-dft"><span class="header-section-number">10.1</span> Definition of the DFT</h2>
<p>The DFT is defined as follows:</p>
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<p><span class="theorem-title"><strong>Definition 10.1 (Discrete Fourier Transformation)</strong></span> The discrete Fourier transform (DFT) is a linear transformation on <span class="math inline">\(\mathbb{C}^N\)</span> represented by the matrix <span class="math display">\[
\operatorname{DFT}_N \coloneqq \frac{1}{\sqrt{N}} (\omega^{kl})_{kl=0,\dots,N-1} \in \mathbb{C}^{N\times N}
\]</span> with <span class="math inline">\(\omega = e^{2i\pi/N}\)</span>, which is the <span class="math inline">\(N\)</span>-th root of unity.</p>
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<p>It applies <span class="math inline">\(\omega^N=1\)</span> and <span class="math inline">\(\omega^i \neq 1\)</span> for all <span class="math inline">\(0 &lt; i &lt; N\)</span>.</p>
<p>This transformation is best imagined as a process that takes a periodic vector as an input and outputs the period of that vector. The DFT has some important properties which will help us later on.</p>
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<p><span class="theorem-title"><strong>Theorem 10.1 (Properties of the DFT)</strong></span> Here are some properties of the DFT which can be used without further proof.</p>
<ol type="1">
<li>The <span class="math inline">\(\operatorname{DFT}_N\)</span> is unitary.</li>
<li><span class="math inline">\(\omega^t = \omega^{t\mod N}\)</span> for all <span class="math inline">\(t \in \mathbb{Z}\)</span>.</li>
<li>Let <span class="math inline">\(t \mid N\)</span> and <span class="math inline">\(s\)</span> be integers and the quantum state <span class="math inline">\(\psi \in \mathbb{C}^N\)</span> be given by <span class="math display">\[
\psi_i \coloneqq
\begin{cases}
\sqrt{\frac{t}{N}} &amp; \text{if } i = at+s \text{ for some } a \\
0 &amp; \text{else.}
\end{cases}
\]</span> In other words <span class="math inline">\(\psi\)</span> is <span class="math inline">\(t\)</span>-periodic. Then for <span class="math inline">\(\phi \coloneqq \operatorname{DFT}_N\)</span> we have <span class="math display">\[
|\phi_i| =
\begin{cases}
\frac{1}{\sqrt{t}} &amp; \text{if } \frac{N}{t} \mid i \\
0 &amp; \text{else.}
\end{cases}
\]</span> The first peak of <span class="math inline">\(\phi\)</span> (besides <span class="math inline">\(\phi_0\)</span>) is at <span class="math inline">\(\frac{N}{t}\)</span>.</li>
</ol>
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<section id="constructing-the-dft" class="level2" data-number="10.2">
<h2 data-number="10.2" class="anchored" data-anchor-id="constructing-the-dft"><span class="header-section-number">10.2</span> Constructing the DFT</h2>
<p>So far we have described everything necessary for Shor’s algorithm, but only described the matrix representation of the <span class="math inline">\(\operatorname{DFT}_N\)</span>. We will now take a closer look into implementing the <span class="math inline">\(\operatorname{DFT}_M\)</span> as a quantum circuit. Since we only use the <span class="math inline">\(\operatorname{DFT}_N\)</span> for Shor’s algorithm so far, we will only look at <span class="math inline">\(N=2^n\)</span>, which is the <span class="math inline">\(\operatorname{DFT}\)</span> applied on <span class="math inline">\(n\)</span> qubits.</p>
<p>To start the circuit, we recall the definition of the <span class="math inline">\(\operatorname{DFT}_N\)</span> from <a href="#def-dft" class="quarto-xref">Definition&nbsp;<span>10.1</span></a>: <span class="math inline">\(\operatorname{DFT}_N \coloneqq \frac{1}{\sqrt{{2^n}}} (\omega^{kl})_{kl}\)</span> with <span class="math inline">\(\omega \coloneqq e^{2\pi i / 2^n}\)</span>. To apply the <span class="math inline">\(\operatorname{DFT}_N\)</span> to a quantum state <span class="math inline">\(\ket{j}\)</span> we calculate <span class="math display">\[
\begin{aligned}
\operatorname{DFT}_{N}\ket{j}
=&amp; \frac{1}{\sqrt{N}} \sum_{k=0}^{N-1} \omega^{jk} \ket{k}\\
=&amp; \frac{1}{\sqrt{N}} \sum_{k=0}^{N-1} e^{2\pi i j k / N} \ket{k}\\
=&amp; \frac{1}{\sqrt{N}} \sum_{k_1 \in \{0,1\}} \dots \sum_{k_n \in \{0,1\}} e^{2\pi i j (\sum_{l=1}^n k_l 2^{-l})} \ket{k_1 \dots k_n}\\
=&amp; \frac{1}{\sqrt{N}} \sum_{k_1 \in \{0,1\}} \dots \sum_{k_n \in \{0,1\}} \bigotimes_{l=1}^n e^{2\pi i j k_l 2^{-l}} \ket{k_l}\\
=&amp; \frac{1}{\sqrt{N}} \bigotimes_{l=1}^n \sum_{k_l \in \{0,1\}} e^{2\pi i j k_l 2^{-l}} \ket{k_l}\\
=&amp; \frac{1}{\sqrt{N}} \bigotimes_{l=1}^n (\ket{0} + e^{2\pi i j 2^{-l}} \ket{1})\\
=&amp; \frac{1}{\sqrt{N}} \bigotimes_{l=1}^n (\ket{0} + e^{2\pi i 0.j_{n-l+1} \dots j_{n}} \ket{1}).
\end{aligned}
\]</span> The expression <span class="math inline">\(0.j\)</span> expresses a binary fraction (e.g.&nbsp;<span class="math inline">\(0.101 = \frac{1}{2} + \frac{1}{8} = \frac{5}{8}\)</span>).</p>
<p>With this we have shown, that we can write <span class="math inline">\(\operatorname{DFT}_N \ket{j}\)</span> as the following tensor product of quantum states</p>
<p><span class="math display">\[
\frac{1}{\sqrt{2}}(\ket{0} + e^{2\pi i 0.j_n} \ket{1}) \otimes \frac{1}{\sqrt{2}}(\ket{0} + e^{2\pi i 0.j_{n-1}j_n} \ket{1}) \otimes \dots \otimes \frac{1}{\sqrt{2}}(\ket{0} + e^{2\pi i 0.j_1\dots j_n} \ket{1}).
\]</span></p>
<p>From this rewritten tensor product, we can get an idea on how to construct the quantum circuit for the <span class="math inline">\(\operatorname{DFT}_N\)</span>. Namely, we can construct a quantum circuit for each element of the tensor product and from this build the general circuit.</p>
<p>For this, we segment the tensor product into different elements <span class="math inline">\(\psi\)</span> as follows:</p>
<p><span class="math display">\[
\underbrace{\frac{1}{\sqrt{2}}(\ket{0} + e^{2\pi i 0.j_n} \ket{1})}_{\psi_1} \otimes \underbrace{\frac{1}{\sqrt{2}}(\ket{0} + e^{2\pi i 0.j_{n-1}j_n} \ket{1})}_{\psi_2} \otimes \dots \otimes \underbrace{\frac{1}{\sqrt{2}}(\ket{0} + e^{2\pi i 0.j_1\dots j_n} \ket{1})}_{\psi_n}.
\]</span></p>
<p>We also introduce a new gate <span class="math inline">\(R_k\)</span> which is defined by the following matrix:</p>
<p><span class="math display">\[
R_k:=\begin{pmatrix} 1 &amp; 0 \\ 0 &amp; e^{2 \pi i / 2^k} \end{pmatrix}.
\]</span></p>
<p>To understand the construction of the circuit from these elements, we will look at an example for <span class="math inline">\(n=3\)</span> first:</p>
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Example: Construction of the DFT circuit for <span class="math inline">\(n=3\)</span>
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<p>We start by building the tensor product for <span class="math inline">\(n=3\)</span>. The input for the DFT circuit is <span class="math inline">\(\ket{j} = \ket{j_1j_2j_3}\)</span>. Using the formula from above, we can write the result of <span class="math inline">\(\operatorname{DFT}_{2^3} \ket{j}\)</span> as the following tensor product:</p>
<p><span class="math display">\[
\underbrace{\frac{1}{\sqrt{2}}(\ket{0} + e^{2\pi i 0.j_3} \ket{1})}_{\psi_1} \otimes \underbrace{\frac{1}{\sqrt{2}}(\ket{0} + e^{2\pi i 0.j_2j_3} \ket{1})}_{\psi_2} \otimes \underbrace{\frac{1}{\sqrt{2}}(\ket{0} + e^{2\pi i 0.j_1j_2j_3} \ket{1})}_{\psi_3}.
\]</span></p>
<div class="quarto-figure quarto-figure-center">
<figure class="figure">
<p><img src="img/dft_3_qubits.svg" class="img-fluid figure-img" style="width:100.0%"></p>
<figcaption>The DTF for three qubits</figcaption>
</figure>
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<ol type="1">
<li>First we construct the <span class="math inline">\(\psi_3\)</span> element. Contrary to the intuition, we use the top wire containing <span class="math inline">\(\ket{j_1}\)</span> for this. We use a Hadamad-gate to bring <span class="math inline">\(\ket{j_1}\)</span> into the superposition <span class="math inline">\(\frac{1}{\sqrt{2}}(\ket{0} + (-1)^{j_1} \ket{1}) = \frac{1}{\sqrt{2}}(\ket{0} + e^{2 \pi i 0.j_1} \ket{1})\)</span>. This looks close to <span class="math inline">\(\psi_3\)</span> already, we now need to add the last two decimal places <span class="math inline">\(j_2j_3\)</span> to the state. For this we use <span class="math inline">\(R_2\)</span> and <span class="math inline">\(R_3\)</span>. We apply <span class="math inline">\(R_2\)</span> controlled by the wire <span class="math inline">\(j_2\)</span> and <span class="math inline">\(R_3\)</span> controlled by the wire <span class="math inline">\(j_3\)</span>. This means, that we only apply the <span class="math inline">\(R\)</span>-gate, if the corresponding wire contains a <span class="math inline">\(1\)</span>. You can see this written as a quantum circuit at the figure below. After applying <span class="math inline">\(R_2\)</span> we have the state <span class="math inline">\(\frac{1}{\sqrt{2}}(\ket{0} + e^{2 \pi i 0.j_1j_2} \ket{1})\)</span> and after applying <span class="math inline">\(R_3\)</span> we have the state <span class="math inline">\(\frac{1}{\sqrt{2}}(\ket{0} + e^{2 \pi i 0.j_1j_2j_3} \ket{1})\)</span> on the top wire. This is the same as <span class="math inline">\(\psi_3\)</span>, so we are done on the first wire (We are at the first slice in the figure).</li>
<li>The next step is to construct the <span class="math inline">\(\psi_2\)</span> state on the middle wire. We again use a Hadamad-gate to bring <span class="math inline">\(\ket{j_2}\)</span> into the superposition <span class="math inline">\(\frac{1}{\sqrt{2}}(\ket{0} + e^{2 \pi i 0.j_2} \ket{1})\)</span>. We now need to include the last decimal point <span class="math inline">\(j_3\)</span>, for which we use <span class="math inline">\(R_2\)</span> again, this time controlled by <span class="math inline">\(j_3\)</span>. The resulting superposition is now <span class="math inline">\(\frac{1}{\sqrt{2}}(\ket{0} + e^{2\pi i 0.j_2j_3} \ket{1})\)</span>, which is <span class="math inline">\(\psi_2\)</span>. (We are at the second slice in the figure).</li>
<li>On the bottom wire, we can just do a Hadamad-gate to bring <span class="math inline">\(\ket{j_3}\)</span> into the superposition <span class="math inline">\(\frac{1}{\sqrt{2}}(\ket{0} + e^{2 \pi i 0.j_3} \ket{1})\)</span>. We then have <span class="math inline">\(\psi_1\)</span> on the bottom wire. (We are at the third slice in the figure).</li>
<li>When applying this circuit, we get the state <span class="math inline">\(\psi_3 \otimes \psi_2 \otimes \psi_1\)</span> as a result. This very close to our desired state <span class="math inline">\(\psi_1 \otimes \psi_2 \otimes \psi_3\)</span>, just the order of the wires is flipped. To solve this, we apply a <span class="math inline">\(\operatorname{SWAP}\)</span> onto all wires, which flips the order of the wires an delivers the correct output for <span class="math inline">\(\operatorname{DFT}_{2^3}\)</span>.</li>
</ol>
</div>
</div>
<p>The more general approach to construct the <span class="math inline">\(\operatorname{DFT}_N\)</span> as a quantum circuit with <span class="math inline">\(n\)</span> qubits (<span class="math inline">\(N = 2^n\)</span>) works as follows:</p>
<ol type="1">
<li>Initialize wires with input <span class="math inline">\(\ket{j}\)</span>, so that <span class="math inline">\(\ket{j_1}\)</span> is on the top wire and <span class="math inline">\(\ket{j_n}\)</span> is on the bottom wire. Note, that this is not part of the circuit yet.</li>
<li>Start with the top wire. For each wire <span class="math inline">\(j_i\)</span> do the following:
<ol type="1">
<li>Apply a Hadamad-gate on the wire <span class="math inline">\(j_i\)</span>.</li>
<li>For each wire <span class="math inline">\(j_k\)</span> below the current wire <span class="math inline">\(j_i\)</span> (with <span class="math inline">\(i &lt; k \leq n\)</span>), add a <span class="math inline">\(R_{k-i+1}\)</span>-gate controlled by <span class="math inline">\(j_k\)</span>. Start with <span class="math inline">\(k = i+1\)</span> (if <span class="math inline">\(i&lt;n\)</span>, else stop).</li>
</ol></li>
<li>Perform a <span class="math inline">\(\operatorname{SWAP}\)</span> to flip all the wires. This means, that the first wire is swapped with the last wire, the second wire is swapped with the second to last wire and so on.</li>
</ol>
<p>Note: If the the output of the DFT circuit is measured right after applying it (as in Shor’s algorithm) or if the rest of the algorithm allows for it, it is more efficient to perform the <span class="math inline">\(\operatorname{SWAP}\)</span> classically, since this is considered to be the cheaper operation.</p>
<p>The more general layout of the quantum circuit for the <span class="math inline">\(\operatorname{DFT_N}\)</span> with the <span class="math inline">\(\operatorname{SWAP}\)</span> is shown in this figure.</p>
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<figure class="figure">
<p><img src="img/dft_n_qubits.svg" class="img-fluid figure-img" style="width:100.0%"></p>
<figcaption>The DTF for <span class="math inline">\(n\)</span> qubits</figcaption>
</figure>
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<li><a href="#constructing-operatornameflip_" id="toc-constructing-operatornameflip_" class="nav-link" data-scroll-target="#constructing-operatornameflip_"><span class="header-section-number">10.1.2</span> Constructing <span class="math inline">\(\operatorname{FLIP}_*\)</span></a></li>
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<li><a href="#constructing-operatornameflip_" id="toc-constructing-operatornameflip_" class="nav-link" data-scroll-target="#constructing-operatornameflip_"><span class="header-section-number">12.1.2</span> Constructing <span class="math inline">\(\operatorname{FLIP}_*\)</span></a></li>
</ul></li>
<li><a href="#the-algorithm-for-searching" id="toc-the-algorithm-for-searching" class="nav-link" data-scroll-target="#the-algorithm-for-searching"><span class="header-section-number">10.2</span> The algorithm for searching</a>
<li><a href="#the-algorithm-for-searching" id="toc-the-algorithm-for-searching" class="nav-link" data-scroll-target="#the-algorithm-for-searching"><span class="header-section-number">12.2</span> The algorithm for searching</a>
<ul class="collapse">
<li><a href="#understanding-the-algorithm-for-searching" id="toc-understanding-the-algorithm-for-searching" class="nav-link" data-scroll-target="#understanding-the-algorithm-for-searching"><span class="header-section-number">10.2.1</span> Understanding the algorithm for searching</a></li>
<li><a href="#understanding-the-algorithm-for-searching" id="toc-understanding-the-algorithm-for-searching" class="nav-link" data-scroll-target="#understanding-the-algorithm-for-searching"><span class="header-section-number">12.2.1</span> Understanding the algorithm for searching</a></li>
</ul></li>
</ul>
</nav>
......@@ -309,9 +264,9 @@ window.Quarto = {
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<div class="quarto-title">
<h1 class="title"><span class="chapter-number">10</span>&nbsp; <span class="chapter-title">Grover’s algorithm</span></h1>
<h1 class="title"><span class="chapter-number">12</span>&nbsp; <span class="chapter-title">Grover’s algorithm</span></h1>
</div>
......@@ -330,71 +285,98 @@ window.Quarto = {
<p>Another well known quantum algorithm is Grover’s algorithm for searching. It was developed by Lov Grover in 1996.</p>
<p>Grover’s algorithm takes a function <span class="math inline">\(f: \{0,1\}^n \rightarrow \{0,1\}\)</span>, where exactly one <span class="math inline">\(x_0\)</span> exists, such that <span class="math inline">\(f(x_0) = 1\)</span>. The goal is to find <span class="math inline">\(x_0\)</span>.</p>
<p>There are a number of interesting problems, which can be reduced to this general definition. One of these problems is the breaking of a (symmetric) encryption. The function <span class="math inline">\(f\)</span> would take a key as an input and output a <span class="math inline">\(1\)</span>, if the decryption is successful.</p>
<p>Classically, finding this <span class="math inline">\(x_0\)</span> takes approximately <span class="math inline">\(2^n\)</span> steps (when simply bruteforcing the function). Using Grover’s algorithm, we can reduce this runtime to approximately <span class="math inline">\(2^{\frac{n}{2}}\)</span> steps. As an example, a <span class="math inline">\(128\)</span>-bit encryption would only take about <span class="math inline">\(2^{64}\)</span> steps to break it, instead of about <span class="math inline">\(2^{128}\)</span> steps for the classical bruteforce.</p>
<section id="preparations" class="level2" data-number="10.1">
<h2 data-number="10.1" class="anchored" data-anchor-id="preparations"><span class="header-section-number">10.1</span> Preparations</h2>
<p>There are a number of interesting problems, which can be reduced to this general definition. One of these problems is the breaking of a (symmetric) encryption. The function <span class="math inline">\(f\)</span> would take a key as an input and output a <span class="math inline">\(1\)</span>, if the decryption is successful. Otherwise it will output a <span class="math inline">\(0\)</span>.</p>
<p>Classically, finding this <span class="math inline">\(x_0\)</span> takes approximately <span class="math inline">\(2^n\)</span> steps (when simply bruteforcing the function). Using Grover’s algorithm, we can reduce this runtime to approximately <span class="math inline">\(\sqrt{2^n}\)</span> steps. As an example, a <span class="math inline">\(128\)</span>-bit encryption would only take about <span class="math inline">\(2^{64}\)</span> steps to break it, instead of about <span class="math inline">\(2^{128}\)</span> steps for the classical bruteforce.</p>
<section id="preparations" class="level2" data-number="12.1">
<h2 data-number="12.1" class="anchored" data-anchor-id="preparations"><span class="header-section-number">12.1</span> Preparations</h2>
<p>To construct Grover’s algorithm, we first need to introduce two new gates <span class="math inline">\(V_f\)</span> and <span class="math inline">\(\operatorname{FLIP}_*\)</span>.</p>
<section id="constructing-the-oracle-v_f" class="level3" data-number="10.1.1">
<h3 data-number="10.1.1" class="anchored" data-anchor-id="constructing-the-oracle-v_f"><span class="header-section-number">10.1.1</span> Constructing the oracle <span class="math inline">\(V_f\)</span></h3>
<p>The uniform superposition <span class="math inline">\(\ket{*}\)</span> simply denotes the superposition over all classical possibilities <span class="math inline">\(\ket{*} = \frac{1}{\sqrt{2^n}} \sum_{x \in\{0,1\}^n} \ket{x}\)</span>.</p>
<section id="constructing-the-oracle-v_f" class="level3" data-number="12.1.1">
<h3 data-number="12.1.1" class="anchored" data-anchor-id="constructing-the-oracle-v_f"><span class="header-section-number">12.1.1</span> Constructing the oracle <span class="math inline">\(V_f\)</span></h3>
<p>In the previous algorithms, we have learned that we can implement a function <span class="math inline">\(f\)</span> as a unitary <span class="math inline">\(U_f\)</span> with <span class="math inline">\(U_f\ket{x,y} = \ket{x, y \oplus f(x)}\)</span>. We construct a different unitary called <span class="math inline">\(V_f\)</span> from this, which has the following behavior:</p>
<p><span class="math display">\[
V_f \ket{x} = \begin{cases} -\ket{x} &amp; \text{if } f(x) = 1\\ \ket{x} &amp; \text{else} \end{cases}
V_f \ket{x} \coloneqq
\begin{cases}
-\ket{x} &amp; \text{if } f(x) = 1 \\
\ket{x} &amp; \text{else}
\end{cases}
\]</span></p>
<p>We can construct <span class="math inline">\(V_f\)</span> from <span class="math inline">\(U_f\)</span> using the following circuit:</p>
<div class="quarto-figure quarto-figure-center">
<figure class="figure">
<p><img src="vf.svg" class="img-fluid figure-img" style="width:60.0%"></p>
<p><img src="img/vf.svg" class="img-fluid figure-img" style="width:60.0%"></p>
<figcaption>The circuit for <span class="math inline">\(V_f\)</span></figcaption>
</figure>
</div>
<p>The bottom wire can be discarded, since it always contains a <span class="math inline">\(\ket{-}\)</span> and thus is not entangled with the upper wire.</p>
</section>
<section id="constructing-operatornameflip_" class="level3" data-number="10.1.2">
<h3 data-number="10.1.2" class="anchored" data-anchor-id="constructing-operatornameflip_"><span class="header-section-number">10.1.2</span> Constructing <span class="math inline">\(\operatorname{FLIP}_*\)</span></h3>
<p>As a second ingredient for Grover’s algorithm, we define a unitary called <span class="math inline">\(\operatorname{FLIP}_*\)</span>. This unitary does nothing, if it is applied on the uniform superposition <span class="math inline">\(\ket{*}\)</span>. For any other quantum state <span class="math inline">\(\psi\)</span> with <span class="math inline">\(\psi\)</span> orthogonal to <span class="math inline">\(\ket{*}\)</span> it maps <span class="math inline">\(\psi\)</span> to <span class="math inline">\(-\psi\)</span>. The uniform superposition <span class="math inline">\(\ket{*}\)</span> simply denotes the superposition over all classical possibilities <span class="math inline">\(2^{-\frac{n}{2}}\sum_x \ket{x}\)</span>. We can construct this <span class="math inline">\(\operatorname{FLIP}_*\)</span> by the following quantum circuit:</p>
<section id="constructing-operatornameflip_" class="level3" data-number="12.1.2">
<h3 data-number="12.1.2" class="anchored" data-anchor-id="constructing-operatornameflip_"><span class="header-section-number">12.1.2</span> Constructing <span class="math inline">\(\operatorname{FLIP}_*\)</span></h3>
<p>As a second ingredient for Grover’s algorithm, we need a unitary <span class="math inline">\(\operatorname{FLIP}_*\)</span> (defined below). To realize this, we first need <span class="math inline">\(\operatorname{FLIP}_0\)</span>, which ist defined by the unitary <span class="math display">\[
\operatorname{FLIP}_0 \ket{x} \coloneqq
\begin{cases}
\ket{0} &amp; \text{if } x = 0 \\
-\ket{x} &amp; \text{else}.
\end{cases}
\]</span></p>
<p><span class="math inline">\(\operatorname{FLIP}_0\)</span> is implemented by the following quantum circuit:</p>
<div class="quarto-figure quarto-figure-center">
<figure class="figure">
<p><img src="flip.svg" class="img-fluid figure-img" style="width:60.0%"></p>
<figcaption>The circuit for <span class="math inline">\(\operatorname{FLIP}_*\)</span></figcaption>
<p><img src="img/flip_zero.svg" class="img-fluid figure-img" style="width:50.0%"></p>
<figcaption>The circuit for <span class="math inline">\(\operatorname{FLIP}_0\)</span></figcaption>
</figure>
</div>
<p><span class="math inline">\(\operatorname{FLIP}_0\)</span> is here definied by the unitary <span class="math display">\[
\operatorname{FLIP}_0 \ket{x} = \begin{cases} \ket{0} &amp; \text{if } x = 0\\ -\ket{x} &amp; \text{else} \end{cases}
<p><span class="math inline">\(Z\)</span> is the Pauli matrix from definition <a href="quantumCircuits.html#def-gates-pauli" class="quarto-xref">Definition&nbsp;<span>7.2</span></a>. The empty circles indicates a negative control wire. So <span class="math inline">\(Z\)</span> is only applied if the other wires are <span class="math inline">\(\ket{0}\)</span>.</p>
<p>Now we can define the unitary called <span class="math inline">\(\operatorname{FLIP}_*\)</span>. This unitary does nothing, if it is applied on the uniform superposition <span class="math inline">\(\ket{*}\)</span>. For any other quantum state <span class="math inline">\(\ket{\psi}\)</span> orthogonal to <span class="math inline">\(\ket{*}\)</span> it maps to <span class="math inline">\(-\ket{\psi}\)</span>. So <span class="math inline">\(\operatorname{FLIP}_*\)</span> is described by:</p>
<p><span class="math display">\[
\operatorname{FLIP}_* \ket{\psi} \coloneqq
\begin{cases}
\ket{*} &amp; \text{if } \ket{\psi} = \ket{*} \\
-\ket{x} &amp; \text{if } \braket{\psi | *} = 0 \text{ (orthogonal)}.
\end{cases}
\]</span></p>
<p>We can construct this <span class="math inline">\(\operatorname{FLIP}_*\)</span> by the following quantum circuit:</p>
<div class="quarto-figure quarto-figure-center">
<figure class="figure">
<p><img src="img/flip_star.svg" class="img-fluid figure-img" style="width:60.0%"></p>
<figcaption>The circuit for <span class="math inline">\(\operatorname{FLIP}_*\)</span></figcaption>
</figure>
</div>
</section>
</section>
<section id="the-algorithm-for-searching" class="level2" data-number="10.2">
<h2 data-number="10.2" class="anchored" data-anchor-id="the-algorithm-for-searching"><span class="header-section-number">10.2</span> The algorithm for searching</h2>
<section id="the-algorithm-for-searching" class="level2" data-number="12.2">
<h2 data-number="12.2" class="anchored" data-anchor-id="the-algorithm-for-searching"><span class="header-section-number">12.2</span> The algorithm for searching</h2>
<p>The actual algorithm takes a function <span class="math inline">\(f: \{0,1\}^n \rightarrow \{0,1\}\)</span> and outputs an <span class="math inline">\(x_0\)</span> with <span class="math inline">\(f(x_0)=1\)</span>. For simplicity, we assume that there is only one <span class="math inline">\(x_0\)</span> for which <span class="math inline">\(f(x_0) = 1\)</span> holds and for each other <span class="math inline">\(x \neq x_0\)</span> it holds that <span class="math inline">\(f(x)=0\)</span>.</p>
<p>With the two new unitaries <span class="math inline">\(V_f\)</span> and <span class="math inline">\(\operatorname{FLIP}_*\)</span> defined, we can construct the circuit for Grover’s algorithm, which is shown in the following figure:</p>
<div class="quarto-figure quarto-figure-center">
<figure class="figure">
<p><img src="grover.svg" class="img-fluid figure-img" style="width:70.0%"></p>
<p><img src="img/grover.svg" class="img-fluid figure-img" style="width:70.0%"></p>
<figcaption>The quantum circuit for Grover’s algorithm</figcaption>
</figure>
</div>
<p>The algorithm works as follows:</p>
<ol type="1">
<li>We start with a <span class="math inline">\(\ket{0}\)</span> entry on every qubit.</li>
<li>We bring the system into the superposition over all entries by applying <span class="math inline">\(H^{\otimes n}\)</span>. The quantum state is then <span class="math inline">\(2^\frac{-n}{2}\sum_x \ket{x}\)</span> which we also call <span class="math inline">\(\ket{*}\)</span>.</li>
<li>We bring the system into the superposition over all entries by applying <span class="math inline">\(H^{\otimes n}\)</span>. The quantum state is then <span class="math inline">\(\frac{1}{\sqrt{2^n}} \sum_{x \in \{0,1\}^n} \ket{x}\)</span> which we also call <span class="math inline">\(\ket{*}\)</span>.</li>
<li>We apply the unitary <span class="math inline">\(V_f\)</span>.</li>
<li>We apply the unitary <span class="math inline">\(\operatorname{FLIP}_*\)</span>.</li>
<li>We repeat steps 3 and 4 <span class="math inline">\(t\)</span> times, and then do a measurement.</li>
<li>We repeat steps 3 and 4 <span class="math inline">\(t\)</span> times.</li>
<li>We do a measurement.</li>
</ol>
<p>The measurement in step 5 will then give us <span class="math inline">\(x_0\)</span> with high probability.</p>
<section id="understanding-the-algorithm-for-searching" class="level3" data-number="10.2.1">
<h3 data-number="10.2.1" class="anchored" data-anchor-id="understanding-the-algorithm-for-searching"><span class="header-section-number">10.2.1</span> Understanding the algorithm for searching</h3>
<p>The measurement in step 6 will then give us <span class="math inline">\(x_0\)</span> with high probability.</p>
<section id="understanding-the-algorithm-for-searching" class="level3" data-number="12.2.1">
<h3 data-number="12.2.1" class="anchored" data-anchor-id="understanding-the-algorithm-for-searching"><span class="header-section-number">12.2.1</span> Understanding the algorithm for searching</h3>
<p>When looking at the quantum circuit, it is not completely intuitive why the algorithm gives the correct result. We therefore now look into what is happening in each step.</p>
<p>The desired quantum state after the algorithm finishes is <span class="math inline">\(\ket{x_0}\)</span>. At the beginning of the algorithm, we bring the system into the uniform superposition <span class="math inline">\(\ket{*} = 2^\frac{-n}{2}\sum_x \ket{x}\)</span>. We know that <span class="math inline">\(\ket{x_0}\)</span> is part of this superposition, therefore we can rewrite <span class="math inline">\(\ket{*}\)</span> as follows <span class="math display">\[
\ket{*} = 2^{-\frac{n}{2}}\sum_x \ket{x}
<p>The desired quantum state after the algorithm finishes is <span class="math inline">\(\ket{x_0}\)</span>. At the beginning of the algorithm, we bring the system into the uniform superposition <span class="math inline">\(\ket{*} = \frac{1}{\sqrt{2^n}} \sum_{x \in \{0,1\}^n} \ket{x}\)</span>. We know that <span class="math inline">\(\ket{x_0}\)</span> is part of this superposition, therefore we can rewrite <span class="math inline">\(\ket{*}\)</span> as follows <span class="math display">\[
\ket{*}
= \frac{1}{\sqrt{2^n}} \sum_{x \in \{0,1\}^n} \ket{x}
= \frac{1}{\sqrt{2^n}} \underbrace{\ket{x_0}}_{\textit{good}} + \sqrt{\frac{2^n-1}{2^n}} \underbrace{\sum_{x\neq x_0} \frac{1}{\sqrt{2^n-1}}\ket{x}}_{\textit{bad}}
\]</span></p>
<p>So the current state can be seen as a superposition of a “good” state <em>good</em> and a “bad” state <em>bad</em>.</p>
<p>The geometric interpretation of this superposition can be drawn as follows:</p>
<div class="quarto-figure quarto-figure-center">
<figure class="figure">
<p><img src="grover_plt1.svg" class="img-fluid figure-img" style="width:30.0%"></p>
<p><img src="img/grover_plot_1.svg" class="img-fluid figure-img" style="width:30.0%"></p>
<figcaption>Geometric interpretation of <span class="math inline">\(\ket{*}\)</span></figcaption>
</figure>
</div>
......@@ -407,7 +389,7 @@ V_f\ket{*}= -\frac{1}{\sqrt{2^n}} \underbrace{\ket{x_0}}_{\text{good}} + \sqrt{
<p>This looks like this in the geometric interpretation:</p>
<div class="quarto-figure quarto-figure-center">
<figure class="figure">
<p><img src="grover_plt2.svg" class="img-fluid figure-img" style="width:30.0%"></p>
<p><img src="img/grover_plot_2.svg" class="img-fluid figure-img" style="width:30.0%"></p>
<figcaption>Geometric interpretation after <span class="math inline">\(V_f\)</span></figcaption>
</figure>
</div>
......@@ -415,13 +397,13 @@ V_f\ket{*}= -\frac{1}{\sqrt{2^n}} \underbrace{\ket{x_0}}_{\text{good}} + \sqrt{
<p>After <span class="math inline">\(V_f\)</span>, we apply the <span class="math inline">\(\operatorname{FLIP}_*\)</span> operation on the quantum state. Since <span class="math inline">\(\operatorname{FLIP}_*\)</span> does nothing on the <span class="math inline">\(\ket{*}\)</span> entries and negates the amplitude of any vector orthogonal to it, <span class="math inline">\(\operatorname{FLIP}_*\)</span> mirrors the vector across <span class="math inline">\(\ket{*}\)</span>. This can be seen in the following figure:</p>
<div class="quarto-figure quarto-figure-center">
<figure class="figure">
<p><img src="grover_plt3.svg" class="img-fluid figure-img" style="width:30.0%"></p>
<p><img src="img/grover_plot_3.svg" class="img-fluid figure-img" style="width:30.0%"></p>
<figcaption>Geometric interpretation after <span class="math inline">\(\operatorname{FLIP}_*\)</span></figcaption>
</figure>
</div>
<p>All in all, we have seen that by applying <span class="math inline">\(V_f\)</span> and <span class="math inline">\(\operatorname{FLIP}_*\)</span>, we can increase the angle of the quantum state in relation to the “good” and “bad” states by <span class="math inline">\(2\theta\)</span>. Therefore two reflection give rotation. By repeating this step often enough, we can get the amplitude of <span class="math inline">\(\ket{x_0}\)</span> close to 1.</p>
<p>To be more precise: Since we know <span class="math inline">\(\theta\)</span> and we know that we will increase the <em>good</em>-ness of our quantum state by <span class="math inline">\(2\theta\)</span> each time, we can calculate that only <span class="math inline">\(t\)</span> iterations are necessary with <span class="math display">\[
t \approx \frac{\frac{\pi / 2}{\theta} - 1}{2} \approx \frac{\pi}{4 \theta} \approx \frac{\pi}{4} \cdot \sqrt{2^n}
t \approx \frac{\frac{\pi / 2}{\theta} - 1}{2} \approx \frac{\pi}{4 \theta} \approx \frac{\pi}{4} \cdot \sqrt{2^n}.
\]</span> Grover’s algorithm therefore takes <span class="math inline">\(O(\sqrt{2^n})\)</span> steps, where an evaluation of the circuit counts as one step.</p>
......@@ -431,24 +413,6 @@ t \approx \frac{\frac{\pi / 2}{\theta} - 1}{2} \approx \frac{\pi}{4 \theta} \app
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......@@ -840,14 +812,11 @@ window.document.addEventListener("DOMContentLoaded", function (event) {
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<i class="bi bi-arrow-left-short"></i> <span class="nav-page-text"><span class="chapter-number">9</span>&nbsp; <span class="chapter-title">Shor’s Algorithm</span></span>
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<i class="bi bi-arrow-left-short"></i> <span class="nav-page-text"><span class="chapter-number">11</span>&nbsp; <span class="chapter-title">Shor’s Algorithm</span></span>
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_book/content/img/double_slit_comic.png

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_book/content/img/double_slit_frames.png

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