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=0N1ajj=(a0aN1) |\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~=UQFTψ \begin{aligned} |\tilde{\psi}\rangle=U_{Q F T}|\psi\rangle \end{aligned} the QFT map can be represnt by a matrix UNQFT=1N(111111ωω2ω3ωN11ω2ω4ω6ω2(N1)1ω3ω6ω9ω3(N1)1ωN1ω2(N1)ω3(N1)ω(N1)(N1)) 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 012k=01ei2π0k/2k=12(0+1)112k=01ei2π1k/2k=12(01) \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=α0+β1|\psi\rangle=\alpha|0\rangle+\beta|1\rangle UQFTψ=UQFT(α0+β1)=12(α+β)0+12(αβ)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}

UQFTψ=UQFT(α0+β1)=α12[1111](10)+β12[1111](01)=12(α+β)(10)+12(αβ)(01)=12(α+β)0+12(αβ)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=0100\left.|0\rangle=\left|0_1 \rangle \right| 0_0\right\rangle
  • 1=0110|1\rangle =| 0_1 \rangle | 1_0\rangle
  • 2=1100|2\rangle =| 1_1 \rangle | 0_0 \rangle
  • 3=1110|3\rangle=\left|1_1\rangle \right| 1_0 \rangle

the orthonormal basis transformed into 014k=03ei2π0k/4k=12(0+1+2+3)114k=03ei2π1k/4k=12(0+i12i3)214k=03ei2π2k/4k=12(01+23)314k=03ei2π3k/4k=12(0i12+i3) \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(11111i1i11111i1i) \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

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