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『Quantum Computing: A Gentle Introduction』の演習問題を解く 3.8

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Quantum Computing: A Gentle Introduction (Scientific and Engineering Computation) (English Edition)

Quantum Computing: A Gentle Introduction (Scientific and Engineering Computation) (English Edition)

目次はこちら

Exercise 3.8.

  • a. Show that  { \frac{1}{\sqrt{2}}\left(|0\rangle|0\rangle + |1\rangle|1\rangle\right) } and  { \frac{1}{\sqrt{2}}\left(|+\rangle|+\rangle + |-\rangle|-\rangle\right) } refer to the same quantum state.
  • b. Show that  { \frac{1}{\sqrt{2}}\left(|0\rangle|0\rangle - |1\rangle|1\rangle\right) } refers to the same state as  { \frac{1}{\sqrt{2}}\left(|\textbf{i}\rangle|\textbf{i}\rangle + |-\textbf{i}\rangle|-\textbf{i}\rangle\right) }.

補足
 { \textbf{i} }虚数単位です。 また  { |+\rangle,\,|-\rangle,\,|\textbf{i}\rangle,\,|-\textbf{i}\rangle } は以下で定義されます:

  { \displaystyle\begin{align*}
  |+\rangle &= \frac{1}{\sqrt{2}}\left(|0\rangle + |1\rangle\right) , &
  |-\rangle &= \frac{1}{\sqrt{2}}\left(|0\rangle - |1\rangle\right) \\
  |\textbf{i}\rangle &= \frac{1}{\sqrt{2}}\left(|0\rangle + \textbf{i}|1\rangle\right) , &
  |-\textbf{i}\rangle &= \frac{1}{\sqrt{2}}\left(|0\rangle - \textbf{i}|1\rangle\right)
\end{align*}}

a.
  { \displaystyle\begin{align*}
  \frac{1}{\sqrt{2}}\left(|+\rangle|+\rangle + |-\rangle|-\rangle\right)
    &= \frac{1}{2\sqrt{2}}\Big\{(|0\rangle + |1\rangle)(|0\rangle + |1\rangle)
      + (|0\rangle - |1\rangle)(|0\rangle - |1\rangle)\Big\} \\
    &= \frac{1}{\sqrt{2}}\left(|0\rangle|0\rangle + |1\rangle|1\rangle\right)
\end{align*}}

b.
  { \displaystyle\begin{align*}
  \frac{1}{\sqrt{2}}\left(|\textbf{i}\rangle|\textbf{i}\rangle + |-\textbf{i}\rangle|-\textbf{i}\rangle\right)
    &= \frac{1}{2\sqrt{2}}\Big\{(|0\rangle + \textbf{i}|1\rangle)(|0\rangle + \textbf{i}|1\rangle)
      + (|0\rangle - \textbf{i}|1\rangle)(|0\rangle - \textbf{i}|1\rangle)\Big\} \\
    &= \frac{1}{\sqrt{2}}\left(|0\rangle|0\rangle - |1\rangle|1\rangle\right)
\end{align*}}