Quantum Country exercise 4

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 4. Consider a quantum circuit in which the Hadamard gate H is applied to a quantum state \vert \psi \rangle and then the outout is measured in the computational basis.  Show that when the state

Answer: First consider the case when the state

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

So the result of applying H to the state is the state \vert 0 \rangle.

In the general case the state \alpha \vert 0 \rangle + \beta \vert 1 \rangle will produce the result 0 with probability |\alpha|^2 and the value 1 with probability |\alpha|^2 when measured in the computational basis. Since the result in this case is \vert 0 \rangle we have \alpha = 1 and \beta = 0 so the result will be m = 0 with probability |1|^2 = 1, with the result m = 1 having probability |0|^2 = 0. So the output as measured will always be m = 0.

When the state

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

So the result of applying H to the state is the state \vert 1 \rangle, which when measured in the computational basis will always produce the value m = 1.

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