To be continued

Baisc Algorithms

building block

[TOC]

Reference

  1. Mikio Nakahara, Tetsuo Ohmi.《Quantum Computing. From Linear Algebra to Physical Realizations 》
  2. David McMahon. 《Quantum Computing Explained 》
  3. Michael A Nielsen and Isaac Chuang.《Quantum computation and quantum information》, 2002
  4. Ivan B. Djordjevic《Quantum Information Processing, Quantum Computing, and Quantum Error Correction: An Engineering Approach》

1. Quantum Fourier transform

Given a state ∣ψ⟩=∑j=0N−1aj∣j⟩=(a0⋮aN−1) |\psi\rangle=\sum_{j=0}^{N-1} a_j|j\rangle=\left(\begin{array}{c} a_0 \\ \vdots \\ a_{N-1} \end{array}\right) the quantum Fourier transform UQFTU_{Q F T} acts on a quantum state ∣ψ⟩|\psi\rangle and maps it to the quantum state ∣ψ~⟩|\tilde{\psi}\rangle ∣ψ~⟩=UQFT∣ψ⟩ \begin{aligned} |\tilde{\psi}\rangle=U_{Q F T}|\psi\rangle \end{aligned} the QFT map can be represnt by a matrix UNQFT=1N(1111⋯11ωω2ω3⋯ωN−11ω2ω4ω6⋯ω2(N−1)1ω3ω6ω9⋯ω3(N−1)⋮⋮⋮⋮⋮1ωN−1ω2(N−1)ω3(N−1)⋯ω(N−1)(N−1)) U_{N}^{Q F T}=\frac{1}{\sqrt{N}}\left(\begin{array}{cccccc} 1 & 1 & 1 & 1 & \cdots & 1 \\ 1 & \omega & \omega^2 & \omega^3 & \cdots & \omega^{N-1} \\ 1 & \omega^2 & \omega^4 & \omega^6 & \cdots & \omega^{2(N-1)} \\ 1 & \omega^3 & \omega^6 & \omega^9 & \cdots & \omega^{3(N-1)} \\ \vdots & \vdots & \vdots & \vdots & & \vdots \\ 1 & \omega^{N-1} & \omega^{2(N-1)} & \omega^{3(N-1)} & \cdots & \omega^{(N-1)(N-1)} \end{array}\right)

1.1.1. Example: 1-qubit QFT

the orthonormal basis {∣j⟩}={∣0⟩,∣1⟩}\{|j\rangle\}=\{|0\rangle ,| 1\rangle \} transformed into ∣0⟩→12∑k=01ei2π⋅0⋅k/2∣k⟩=12(∣0⟩+∣1⟩)∣1⟩→12∑k=01ei2π⋅1⋅k/2∣k⟩=12(∣0⟩−∣1⟩) \begin{aligned} |0\rangle & \rightarrow \frac{1}{\sqrt{2}} \sum_{k=0}^1 e^{i 2 \pi \cdot 0 \cdot k / 2}|k\rangle=\frac{1}{\sqrt{2}}(|0\rangle+|1\rangle)\\ |1\rangle & \rightarrow \frac{1}{\sqrt{2}} \sum_{k=0}^1 e^{i 2 \pi \cdot 1 \cdot k / 2}|k\rangle=\frac{1}{\sqrt{2}}(|0\rangle-|1\rangle) \end{aligned} acts on a single qubit state ∣ψ⟩=α∣0⟩+β∣1⟩|\psi\rangle=\alpha|0\rangle+\beta|1\rangle UQFT∣ψ⟩=UQFT(α∣0⟩+β∣1⟩)=12(α+β)∣0⟩+12(α−β)∣1⟩ \begin{aligned} U_{Q F T}|\psi\rangle=& U_{Q F T} (\alpha|0\rangle+\beta|1\rangle) \\=&\frac{1}{\sqrt{2}}(\alpha+\beta)|0\rangle+\frac{1}{\sqrt{2}}(\alpha-\beta)|1\rangle \end{aligned}

UQFT∣ψ⟩=UQFT(α∣0⟩+β∣1⟩)=α12[111−1](10)+β12[111−1](01)=12(α+β)(10)+12(α−β)(01)=12(α+β)∣0⟩+12(α−β)∣1⟩ \begin{aligned} U_{Q F T}|\psi\rangle=& U_{Q F T} (\alpha|0\rangle+\beta|1\rangle) \\=&\alpha\frac{1}{\sqrt{2}}\left[\begin{array}{cc} 1 & 1 \\ 1 & -1 \end{array}\right] \left(\begin{array}{c} 1 \\ 0 \end{array}\right)+\beta\frac{1}{\sqrt{2}}\left[\begin{array}{cc} 1 & 1 \\ 1 & -1 \end{array}\right] \left(\begin{array}{c} 0 \\ 1 \end{array}\right) \\=&\frac{1}{\sqrt{2}}(\alpha+\beta)\left(\begin{array}{l} 1 \\ 0 \end{array}\right)+\frac{1}{\sqrt{2}}(\alpha-\beta)\left(\begin{array}{l} 0 \\ 1 \end{array}\right) \\=&\frac{1}{\sqrt{2}}(\alpha+\beta)|0\rangle+\frac{1}{\sqrt{2}}(\alpha-\beta)|1\rangle \end{aligned}

1.1.2. Examples 2-qubit QFT

the orthonormal basis of 2-qbit system

  • ∣0⟩=∣01⟩∣00⟩\left.|0\rangle=\left|0_1 \rangle \right| 0_0\right\rangle
  • ∣1⟩=∣01⟩∣10⟩|1\rangle =| 0_1 \rangle | 1_0\rangle
  • ∣2⟩=∣11⟩∣00⟩|2\rangle =| 1_1 \rangle | 0_0 \rangle
  • ∣3⟩=∣11⟩∣10⟩|3\rangle=\left|1_1\rangle \right| 1_0 \rangle

the orthonormal basis transformed into ∣0⟩→14∑k=03ei2π⋅0⋅k/4∣k⟩=12(∣0⟩+∣1⟩+∣2⟩+∣3⟩)∣1⟩→14∑k=03ei2π⋅1⋅k/4∣k⟩=12(∣0⟩+i∣1⟩−∣2⟩−i∣3⟩)∣2⟩→14∑k=03ei2π⋅2k⋅/4∣k⟩=12(∣0⟩−∣1⟩+∣2⟩−∣3⟩)∣3⟩→14∑k=03ei2π⋅3⋅k/4∣k⟩=12(∣0⟩−i∣1⟩−∣2⟩+i∣3⟩) \begin{aligned} |0\rangle & \rightarrow \frac{1}{\sqrt{4}} \sum_{k=0}^3 e^{i 2 \pi \cdot 0 \cdot k / 4}|k\rangle=\frac{1}{2}(|0\rangle+|1\rangle+|2\rangle+|3\rangle) \\ |1\rangle & \rightarrow \frac{1}{\sqrt{4}} \sum_{k=0}^3 e^{i 2 \pi \cdot 1 \cdot k / 4}|k\rangle=\frac{1}{2}(|0\rangle+i|1\rangle-|2\rangle-i|3\rangle) \\ |2\rangle & \rightarrow \frac{1}{\sqrt{4}} \sum_{k=0}^3 e^{i 2 \pi \cdot 2 k \cdot / 4}|k\rangle=\frac{1}{2}(|0\rangle-|1\rangle+|2\rangle-|3\rangle) \\ |3\rangle & \rightarrow \frac{1}{\sqrt{4}} \sum_{k=0}^3 e^{i 2 \pi \cdot 3 \cdot k / 4}|k\rangle=\frac{1}{2}(|0\rangle-i|1\rangle-|2\rangle+i|3\rangle) \end{aligned} 2-qubit QFT in matrix form: F=12(11111ωω2ω31ω21ω21ω3ω2ω)=12(11111i−1−i1−11−11−i−1i) \begin{aligned} F=&\frac{1}{2}\left(\begin{array}{cccc} 1 & 1 & 1 & 1 \\ 1 & \omega & \omega^2 & \omega^3 \\ 1 & \omega^2 & 1 & \omega^2 \\ 1 & \omega^3 & \omega^2 & \omega \end{array}\right) \\=& \frac{1}{2}\left(\begin{array}{cccc} 1 & 1 & 1 & 1 \\ 1 & \mathrm{i} & -1 & -\mathrm{i} \\ 1 & -1 & 1 & -1 \\ 1 & -\mathrm{i} & -1 & \mathrm{i} \end{array}\right) \end{aligned} where

  • ω=eπi/2\omega=\mathrm{e}^{\pi \mathrm{i} / 2}

1.1.3. binary number expression

To be continued

results matching ""

    No results matching ""