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As we know Hadamard gates are used to bring quantum bits into superposition states.

I’m trying to understand how identities $HXH = Z$ & $HZH = X$ w.r.t rotation.

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2 Answers 2

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Consider the matrix representations of these operations and with little algebra, things will become obvious.

$$H = \frac{1}{\sqrt{2}}\begin{bmatrix} 1 & 1\\ 1 & -1 \end{bmatrix}$$ $$X = \begin{bmatrix} 0 & 1\\ 1 & 0 \end{bmatrix}$$ $$Z = \begin{bmatrix} 1 & 0\\ 0 & -1 \end{bmatrix}$$

Then it is clear that $$ HXH = \frac{1}{\sqrt{2}}\begin{bmatrix} 1 & 1\\ 1 & -1 \end{bmatrix} \begin{bmatrix} 0 & 1\\ 1 & 0 \end{bmatrix} \frac{1}{\sqrt{2}}\begin{bmatrix} 1 & 1\\ 1 & -1 \end{bmatrix} $$ $$= \begin{bmatrix} 1 & 0\\ 0 & -1 \end{bmatrix} = Z$$

Things will follow similarly for $HZH$.

You can think of this operation as change of basis in linear algebra. (Keep in mind that $H=H^{-1}=H^{†}$)

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$X$ is a 180$^{\circ}$ rotation about the x-axis,

$Z$ is a 180$^{\circ}$ rotation about the z-axis,

Since $H$ swaps the x-axis with the z-axis, (and a subsequent $H$ swaps them back to their original orientation)

then: $HXH = Z$

Also because $H=H^{-1}$ and so:

$X = H^{-1}ZH^{-1} = HZH$

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