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<meta name="generator" content="DocOnce: https://github.com/doconce/doconce/" />
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<meta name="description" content="April 1-5, 2024: Quantum Computing, Quantum Machine Learning and Quantum Information Theories">
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<title>April 1-5, 2024: Quantum Computing, Quantum Machine Learning and Quantum Information Theories</title>
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<meta name="description" content="Quantum Computing, Quantum Machine Learning and Quantum Information Theories">
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<title>Quantum Computing, Quantum Machine Learning and Quantum Information Theories</title>
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<!-- reveal.js: https://lab.hakim.se/reveal-js/ -->
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<section>
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<!-- ------------------- main content ---------------------- -->
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<center>
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<h1 style="text-align: center;">April 1-5, 2024: Quantum Computing, Quantum Machine Learning and Quantum Information Theories</h1>
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<h1 style="text-align: center;">Quantum Computing, Quantum Machine Learning and Quantum Information Theories</h1>
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</center> <!-- document title -->
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<!-- author(s): Morten Hjorth-Jensen -->
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</center>
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<br>
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<center>
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<h4>April 3, 2024</h4>
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<h4>April 2, 2025</h4>
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</center> <!-- date -->
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<br>
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</section>
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<section>
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<h2 id="plans-for-the-week-of-april-1-5-2024">Plans for the week of April 1-5, 2024 </h2>
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<h2 id="plans-for-the-week-of-march-31-april-4-2025">Plans for the week of March 31-April 4, 2025 </h2>
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<div class="alert alert-block alert-block alert-text-normal">
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<b></b>
@@ -205,8 +205,8 @@ <h2 id="plans-for-the-week-of-april-1-5-2024">Plans for the week of April 1-5, 2
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<p><li> Quantum Fourier transforms (QFTs), basic mathematical expressions</li>
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<p><li> Setting up circuits for QFT</li>
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<p><li> Reading recommendation Hundt, Quantum Computing for Programmers, sections 6.1-6.4 on QFT.</li>
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<p><li> <a href="https://youtu.be/XxifXp4M2Fk" target="_blank">Video of lecture</a></li>
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<p><li> <a href="https://github.com/CompPhysics/QuantumComputingMachineLearning/blob/gh-pages/doc/HandWrittenNotes/2024/NotesApril3.pdf" target="_blank">Whiteboard notes</a></li>
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<p><li> <a href="https://youtu.be/" target="_blank">Video of lecture TBA</a>
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<!-- o <a href="https://github.com/CompPhysics/QuantumComputingMachineLearning/blob/gh-pages/doc/HandWrittenNotes/2024/NotesApril3.pdf" target="_blank">Whiteboard notes</a> --></li>
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</ol>
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</div>
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</section>
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</ol>
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<p>
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<p><li> Study the solution of quantum mechanical eigenvalue problems with systems from atomic/molecular physics and quantum chemistry</li>
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<p><li> Quantum machine learning projects, quantum boltzmann machines or related topics</li>
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<p><li> Quantum machine learning projects, quantum boltzmann machines or related topics
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<!-- For project 2, in order to be time efficient, you can use software like Qiskit, Pennylane, qBraid and/or other --></li>
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</ul>
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<p>
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<p>For project 2, in order to be time efficient, you can use software like Qiskit, Pennylane, qBraid and/or other</p>
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</section>
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<section>
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</section>
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<section>
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<h2 id="plans-for-the-week-of-april-8-12">Plans for the week of April 8-12 </h2>
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<h2 id="qft">QFT </h2>
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<ol>
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<p><li> Finalize our discussion of QFTs</li>
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<p><li> Implementing the QFT algorithm</li>
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<p><li> Implementing the phase estimation algorithm for finding eigenvalues</li>
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<p><li> Discussions of Simon's algorithm</li>
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</ol>
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<p>The Quantum Fourier Transform (QFT) has mathematically the same equation as starting point
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but the notation is generally different. We generally compute the
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quantum Fourier transform on a set of orthonormal basis state vectors
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\(|0\rangle, |1\rangle, ..., |N-1\rangle\). The linear operator defining
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the transform is given by the action on basis states
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</p>
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<p>&nbsp;<br>
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$$
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|j\rangle \mapsto \frac{1}{\sqrt{N}} \sum_{k=0}^{N-1} e^{2\pi ijk/N}|k\rangle.
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$$
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<p>&nbsp;<br>
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</section>
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<section>
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<h2 id="in-terms-of-arbitrary-states">In terms of arbitrary states </h2>
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<p>This can be written on arbitrary states,</p>
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<p>&nbsp;<br>
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$$
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\sum_{j=0}^{N-1}x_j|j\rangle \mapsto \sum_{k=0}^{N-1} y_k|k\rangle
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$$
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<p>&nbsp;<br>
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<p>where each amplitude \(y_k\) is the discrete Fourier transform of
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\(x_j\).
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</p>
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</section>
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<section>
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<h2 id="unitarity">Unitarity </h2>
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<p>The quantum Fourier transform is unitary. Taking \(N = 2^n\),
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for \(n\) qubits gives us the orthonormal (computational) basis
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</p>
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<p>&nbsp;<br>
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$$
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|0\rangle, |1\rangle, ..., |2^{n}-1\rangle.
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$$
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<p>&nbsp;<br>
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</section>
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<section>
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<h2 id="binary-representation">Binary representation </h2>
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<p>Each of the computational basis states can be represented in binary</p>
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<p>&nbsp;<br>
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$$
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j = j_1j_2 \cdots j_n
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$$
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<p>&nbsp;<br>
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<p>where each \(j_k\) is either \(0\) or \(1\), and the corresponding
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binary vector is \(|j_1j_2 \cdots j_n\rangle\).
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</p>
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</section>
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<section>
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<h2 id="rewriting-the-qft">Rewriting the QFT </h2>
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<p>The quantum Fourier
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transform on one of these \(n\)-qubit vectors can be written as,
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</p>
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<p>&nbsp;<br>
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$$
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\begin{align*}
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|j_1j_2 \cdots j_n \rangle = \frac{\left(|0\rangle +e^{2\pi i0.j_n}|1\rangle\right) \otimes \left(|0\rangle +e^{2\pi i0.j_{n-1}j_n}|1\rangle\right) \otimes \cdots \otimes \left(|0\rangle +e^{2\pi i0.j_1j_2 \cdots j_n}|1\rangle\right)}{2^{n/2}}
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\end{align*}
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$$
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<p>&nbsp;<br>
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</section>
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<section>
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<h2 id="short-hand-notation">Short hand notation </h2>
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<p>In the above, we use the notation</p>
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<p>&nbsp;<br>
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$$
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0.j_lj_{l+1} \cdots j_n = \frac{j_l}{2} + \frac{j_{l+1}}{2^2} + \cdots + \frac{j_n}{2^{n-l+1}}
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$$
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<p>&nbsp;<br>
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<p>If we start counting from zero, that is we have instead</p>
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<p>&nbsp;<br>
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$$
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\begin{align*}
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|j_0j_1 \cdots j_{n-1} \rangle = \frac{\left(|0\rangle +e^{2\pi i0.j_{n-1}}|1\rangle\right) \otimes \left(|0\rangle +e^{2\pi i0.j_{n-2}j_n}|1\rangle\right) \otimes \cdots \otimes \left(|0\rangle +e^{2\pi i0.j_0j_1 \cdots j_{n-1}}|1\rangle\right)}{2^{n/2}},
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\end{align*}
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$$
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<p>&nbsp;<br>
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<p>we get</p>
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<p>&nbsp;<br>
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$$
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0.j_lj_{l+1} \cdots j_{n-1} = \frac{j_l}{2} + \frac{j_{l+1}}{2^2} + \cdots + \frac{j_{n-1}}{2^{n-l}}
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$$
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<p>&nbsp;<br>
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</section>
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<section>
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<h2 id="four-qubit-system">Four qubit system </h2>
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<p>The basis states are</p>
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<p>&nbsp;<br>
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$$
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|j_1j_2j_3j_4 \rangle
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$$
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<p>&nbsp;<br>
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<p>where \(j_k\) is either \(0\) or \(1\). We have</p>
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<p>&nbsp;<br>
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$$
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\begin{align*}
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0.j_3 &= \frac{j_3}{2} \\
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0.j_2j_3 &= \frac{j_2}{2} + \frac{j_3}{4} \\
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0.j_1j_2j_3 &= \frac{j_1}{2} + \frac{j_2}{4} + \frac{j_3}{8} \\
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0.j_0j_1j_2j_3 &= \frac{j_0}{2} + \frac{j_1}{4} + \frac{j_2}{8} + \frac{j_3}{16} \\
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\end{align*}
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$$
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<p>&nbsp;<br>
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</section>
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<section>
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<h2 id="qft-acts-as-follows">QFT acts as follows </h2>
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<p>The quantum Fourier transform acts as follows:</p>
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<p>&nbsp;<br>
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$$
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\begin{align*}
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|j_1j_2j_3j_4 \rangle \mapsto \frac{1}{\sqrt{2^{4/2}}}
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\left(|0\rangle + e^{2 \pi i 0.j4}|1\rangle \right) \otimes
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\left(|0\rangle + e^{2 \pi i 0.j_3j4}|1\rangle \right) \otimes
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\left(|0\rangle + e^{2 \pi i 0.j_2j_3j4}|1\rangle \right) \otimes
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\left(|0\rangle + e^{2 \pi i 0.j_1j_2j_3j4}|1\rangle \right)
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\end{align*}
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$$
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<p>&nbsp;<br>
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</section>
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<section>
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<h2 id="quantum-circuits">Quantum circuits </h2>
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<p>To compose a quantum circuit that calculates the quantum Fourier
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transform we use the operators
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</p>
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<p>&nbsp;<br>
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$$
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\begin{align*}
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R_k =
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\begin{bmatrix}
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1 & 0 \\
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0 & e^{2\pi i/2^k}
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\end{bmatrix}.
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\end{align*}
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$$
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<p>&nbsp;<br>
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</section>
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<section>
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<h2 id="rotation-gates">Rotation gates </h2>
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<p>In this example, the \(R_k\) gates are:</p>
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<p>&nbsp;<br>
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$$
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\begin{align*}
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R_1 = \begin{bmatrix}
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1 & 0 \\
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0 & e^{2 \pi i /2^0}
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\end{bmatrix} =
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\begin{bmatrix}
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1 & 0 \\
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0 & 1
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\end{bmatrix}, \quad
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R_2 = \begin{bmatrix}
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1 & 0 \\
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0 & e^{2 \pi i /2^2}
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\end{bmatrix}, \quad
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R_3 = \begin{bmatrix}
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1 & 0 \\
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0 & e^{2 \pi i /2^3}
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\end{bmatrix}, \quad
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R_4 = \begin{bmatrix}
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1 & 0 \\
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0 & e^{2 \pi i /2^4}
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\end{bmatrix}.
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\end{align*}
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$$
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<p>&nbsp;<br>
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</section>
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<section>
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<h2 id="using-the-hadamard-gate">Using the Hadamard gate </h2>
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<p>The Hadamard
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gate on a single qubit creates an equal superposition of its basis
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states, assuming it is not already in a superposition, such that
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</p>
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<p>&nbsp;<br>
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$$
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H\vert 0 \rangle = \frac{1}{\sqrt{2}} \left(\vert 0 \rangle + \vert 1\rangle\right), \quad H\vert 1\rangle = \frac{1}{\sqrt{2}} \left(\vert 0 \rangle - \vert 1\rangle\right)
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$$
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<p>&nbsp;<br>
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<p>The \( R_k \)&#160;gate simply adds a phase if the qubit it acts on is in the state \( \vert 1\rangle \)</p>
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<p>&nbsp;<br>
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$$
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R_k\vert 0 \rangle = \vert 0 \rangle, \quad R_k\vert 1\rangle = e^{2\pi i/2^{k}}\vert 1\rangle
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$$
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<p>&nbsp;<br>
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<p>Since all this gates are unitary, the quantum Fourier transfrom is also unitary.</p>
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</section>
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<section>
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<h2 id="algorithm">Algorithm </h2>
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<p>Assume we have a quantum register of \( n \) qubits in the state \( \vert j_1 j_2 \dots j_n\rangle \).
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Applying the Hadamard gate to the first qubit
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produces the state
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</p>
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<p>&nbsp;<br>
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$$
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H\vert j_1 j_2 \dots j_n\rangle = \frac{\left(\vert 0 \rangle + e^{2\pi i 0.j_1}\vert 1\rangle\right)}{2^{1/2}} \vert j_2 \dots j_n\rangle.
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$$
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<p>&nbsp;<br>
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</section>
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<section>
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<h2 id="binary-fraction">Binary fraction </h2>
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<p>Here we have made use of the binary fraction to represent the action of the Hadamard gate </p>
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<p>&nbsp;<br>
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$$
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\exp{2\pi i 0.j_1} = -1,
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$$
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<p>&nbsp;<br>
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<p>if \( j_1 = 1 \) and \( +1 \) if \( j_1 = 0 \).</p>
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</section>
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<section>
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<h2 id="controlled-rotation-gate">Controlled rotation gate </h2>
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<p>Furthermore we can apply the controlled-\( R_k \)&#160;gate, with all the other qubits \( j_k \) for \( k>1 \) as control qubits to produce the state</p>
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<p>&nbsp;<br>
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$$
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\frac{\left(\vert 0 \rangle + e^{2\pi i 0.j_1j_2\dots j_n}\vert 1\rangle\right)}{2^{1/2}} \vert j_2 \dots j_n\rangle
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$$
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<p>&nbsp;<br>
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<p>Next we do the same procedure on qubit \( 2 \)&#160;producing the state</p>
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<p>&nbsp;<br>
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$$
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\frac{\left(\vert 0 \rangle + e^{2\pi i 0.j_1j_2\dots j_n}\vert 1\rangle\right)\left(\vert 0 \rangle + e^{2\pi i 0.j_2\dots j_n}\vert 1\rangle\right)}{2^{2/2}} \vert j_2 \dots j_n\rangle
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$$
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<p>&nbsp;<br>
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</section>
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<section>
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<h2 id="applying-to-all-qubits">Applying to all qubits </h2>
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<p>Doing this for all \( n \) qubits yields state</p>
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<p>&nbsp;<br>
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$$
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\frac{\left(\vert 0 \rangle + e^{2\pi i 0.j_1j_2\dots j_n}\vert 1\rangle\right)\left(\vert 0 \rangle + e^{2\pi i 0.j_2\dots j_n}\vert 1\rangle\right)\dots \left(\vert 0 \rangle + e^{2\pi i 0.j_n}\vert 1\rangle\right)}{2^{n/2}} \vert j_2 \dots j_n\rangle
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$$
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<p>&nbsp;<br>
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<p>At the end we use swap gates to reverse the order of the qubits</p>
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<p>&nbsp;<br>
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$$
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\frac{\left(\vert 0 \rangle + e^{2\pi i 0.j_n}\vert 1\rangle\right)\left(\vert 0 \rangle + e^{2\pi i 0.j_{n-1}j_n}\vert 1\rangle\right)\dots\left(\vert 0 \rangle + e^{2\pi i 0.j_1j_2\dots j_n}\vert 1\rangle\right) }{2^{n/2}} \vert j_2 \dots j_n\rangle
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$$
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<p>&nbsp;<br>
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<p>This is just the product representation from earlier, obviously our desired output.</p>
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</section>
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