Quantum Country exercise 2

This is one in a series of posts working through the exercises in the Quantum Country online introduction to quantum computing and related topics. The exercises in the original document are not numbered; I have added my own numbers for convenience in referring to them.

Exercise 2. Suppose that instead of the Hadamard matrix H we’d defined a matrix J = \frac{1}{\sqrt{2}} \begin{bmatrix} 1&1 \\ 1&1 \end{bmatrix}. Explain why J would not make a suitable quantum gate by applying it to the quantum state \frac{| 0 \rangle - | 1 \rangle}{\sqrt{2}}.

Answer: We have

\frac{\vert 0 \rangle - \vert 1 \rangle}{\sqrt{2}} = \frac{1}{\sqrt{2}} \left( \begin{bmatrix} 1 \\ 0 \end{bmatrix} - \begin{bmatrix} 0 \\ 1 \end{bmatrix} \right) = \frac{1}{\sqrt{2}} \begin{bmatrix} 1 \\ -1 \end{bmatrix}

so that applying J to \frac{\vert 0 \rangle - \vert 1 \rangle}{\sqrt{2}} gives us

\frac{1}{\sqrt{2}} \begin{bmatrix} 1 & 1 \\ 1& 1 \end{bmatrix} \frac{1}{\sqrt{2}} \begin{bmatrix} 1 \\ -1 \end{bmatrix} = \frac{1}{2} \begin{bmatrix} 1 - 1 \\ 1 - 1 \end{bmatrix}

= \frac{1}{2} \begin{bmatrix} 0 \\ 0 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix} = 0

So J applied to \frac{\vert 0 \rangle - \vert 1 \rangle}{\sqrt{2}} gives us the zero state. That is not a result we’d want from applying a quantum gate.

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