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トップ > 量子コンピューティング > 『Quantum Computing: A Gentle Introduction』の演習問題を解く 3.8

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

量子コンピューティング

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} } は虚数単位です。　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*}}