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| 1 | +\documentclass{beamer} |
| 2 | + |
| 3 | +\usepackage{amsmath, amssymb, braket, tikz} |
| 4 | +\usetikzlibrary{quantikz} |
| 5 | + |
| 6 | +\title{Quantum Fourier Transform (QFT) on Two Qubits} |
| 7 | +\author{MHJ} |
| 8 | +\date{\today} |
| 9 | + |
| 10 | +\begin{document} |
| 11 | + |
| 12 | +\begin{frame} |
| 13 | + \titlepage |
| 14 | +\end{frame} |
| 15 | + |
| 16 | +% Slide 1: Introduction |
| 17 | +\begin{frame}{Introduction} |
| 18 | + \textbf{Quantum Fourier Transform (QFT)} |
| 19 | + \begin{itemize} |
| 20 | + \item Quantum analogue of the Discrete Fourier Transform (DFT). |
| 21 | + \item Used in quantum algorithms like Shor's algorithm and phase estimation. |
| 22 | + \item Converts computational basis states into a superposition of phases. |
| 23 | + \end{itemize} |
| 24 | +\end{frame} |
| 25 | + |
| 26 | +% Slide 2: Definition |
| 27 | +\begin{frame}{Definition of QFT} |
| 28 | + The QFT on an \( n \)-qubit system is defined as: |
| 29 | + \[ |
| 30 | + \text{QFT} \ket{x} = \frac{1}{\sqrt{2^n}} \sum_{y=0}^{2^n-1} e^{2\pi i xy / 2^n} \ket{y} |
| 31 | + \] |
| 32 | + where \( x \) is the input integer in binary representation. |
| 33 | +\end{frame} |
| 34 | + |
| 35 | +% Slide 3: QFT on Two Qubits |
| 36 | +\begin{frame}{QFT on Two Qubits} |
| 37 | + For a two-qubit system (\( n=2 \)), the QFT transformation is: |
| 38 | + \[ |
| 39 | + \text{QFT} \ket{x_1 x_0} = \frac{1}{2} \sum_{y=0}^{3} e^{2\pi i x y / 4} \ket{y} |
| 40 | + \] |
| 41 | + where \( x = x_1 x_0 \) (binary representation). |
| 42 | +\end{frame} |
| 43 | + |
| 44 | + |
| 45 | + |
| 46 | +% Slide 5: Matrix Representation |
| 47 | +\begin{frame}{Matrix Representation} |
| 48 | + The QFT on two qubits has the matrix: |
| 49 | + \[ |
| 50 | + U_{\text{QFT}} = \frac{1}{2} |
| 51 | + \begin{bmatrix} |
| 52 | + 1 & 1 & 1 & 1 \\ |
| 53 | + 1 & i & -1 & -i \\ |
| 54 | + 1 & -1 & 1 & -1 \\ |
| 55 | + 1 & -i & -1 & i |
| 56 | + \end{bmatrix} |
| 57 | + \] |
| 58 | +\end{frame} |
| 59 | + |
| 60 | +% Slide 6: Example Calculation |
| 61 | +\begin{frame}{Example Calculation} |
| 62 | + Applying QFT to \( \ket{10} \) (binary 2): |
| 63 | + \[ |
| 64 | + \text{QFT} \ket{10} = \frac{1}{2} (\ket{0} + i\ket{1} - \ket{2} - i\ket{3}) |
| 65 | + \] |
| 66 | +\end{frame} |
| 67 | + |
| 68 | +% Slide 7: Conclusion |
| 69 | +\begin{frame}{Conclusion} |
| 70 | + \begin{itemize} |
| 71 | + \item The QFT transforms basis states into Fourier superpositions. |
| 72 | + \item Two-qubit QFT involves Hadamards, phase gates, and a SWAP. |
| 73 | + \item It is essential for quantum algorithms like phase estimation. |
| 74 | + \end{itemize} |
| 75 | +\end{frame} |
| 76 | + |
| 77 | +\end{document} |
| 78 | +% Slide 4: Quantum Circuit |
| 79 | +\begin{frame}{Quantum Circuit for QFT (2 Qubits)} |
| 80 | + \begin{center} |
| 81 | + \begin{quantikz} |
| 82 | + \lstick{\ket{x_1}} & \gate{H} & \ctrl{1} & \qw & \swap{1} & \qw \\ |
| 83 | + \lstick{\ket{x_0}} & \gate{H} & \gate{R_2} & \qw & \targX{} & \qw |
| 84 | + \end{quantikz} |
| 85 | + \end{center} |
| 86 | + \begin{itemize} |
| 87 | + \item Hadamard $H$ applies equal superposition. |
| 88 | +% \item $R_2$ is the controlled phase gate: $R_2 = \begin{bmatrix} 1 & 0 \\ 0 & e^{i\pi/2} \end{bmatrix}$. |
| 89 | +% \item The final SWAP reorders the qubits. |
| 90 | + \end{itemize} |
| 91 | +\end{frame} |
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