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Show that the Hadamard gate is equivalent to a 180 degree rotation about the axis defined by $(\vec{e_x} - \vec{e_z}) / \sqrt{2}$ where $\vec{e_x}$ and $\vec{e_z}$ are unit vectors pointing along the x and z axes.

I can visualize that this is true based off the mapping of the computational basis vectors to the Hadamard basis vectors on the bloch sphere but I don't know how to show this mathematically.

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First confirm that $\vert \psi \rangle = (\vec{e_x} - \vec{e_z}) / \sqrt{2}$ is an eigenvector of $H$. Indeed, the eigenvalues of $H$ are $\pm 1$, and $H \vert \psi \rangle = (-1) \vert \psi \rangle$, confirming this axis is fixed by $H$. An equivalent statement is that $H \vert \psi \rangle = \vert \psi \rangle$, up to global phase.

Second, note that $H$ is an involution ($H^2=I$) with both eigenvectors fixed up to global phase, so it's action on the Bloch sphere must be equivalent to a rotation by $\pi$. It's worth pointing out that the Bloch sphere is a complex projective line, not a Euclidean 2-sphere where $\det(H)=-1$ would have signaled a reflection.

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  • $\begingroup$ How does showing that $| \psi \rangle$ is an eigenvector of $H$ confirm the axis is fixed by $H$. What does it mean for an axis to be "fixed by" $H$? $\endgroup$ – jmacuna Dec 30 '20 at 1:37
  • $\begingroup$ An axis fixed by $H$, means points on that axis don't move when operated on by $H$. In quantum computing, states that differ only by a global phase (i.e. a complex number with unit modulus) are equivalent (i.e. physically indistinguishable). So $\vert \psi \rangle = e^{i \theta} \vert \psi \rangle$ on the Bloch sphere. $\endgroup$ – Jonathan Trousdale Dec 30 '20 at 2:52
  • $\begingroup$ Eigenvalues of unitary matrices (such as $H$) are, by definition, complex numbers with unit modulus. So if $\vert \psi \rangle$ is an eigenvector of $H$, then $H \vert \psi \rangle = e^{i \theta} \vert \psi \rangle = \vert \psi \rangle$, up to global phase, i.e. $\vert \psi \rangle$ is either an axis of rotation or in a reflection plane. $\endgroup$ – Jonathan Trousdale Dec 30 '20 at 2:52
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A brute force way to do this is to use rotation matrices. You can rewrite the Hadamard matrix as \begin{align} H &= \frac{1}{\sqrt{2}}\begin{pmatrix} 1 & 1 \\ 1 & \text{-}1 \end{pmatrix} \\ &= \frac{1}{\sqrt{2}}\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} \begin{pmatrix} 1 & \text{-}1 \\ 1 & 1 \end{pmatrix} \\ &= R_x(\pi)R_y\left(\frac{\pi}{2}\right) \end{align}

So its just the composition of two rotation matrices. If you write out the $SO(3)$ matrix representations for these rotations you get

\begin{align} R_x(\pi)R_y\left(\frac{\pi}{2}\right) &\rightarrow \begin{pmatrix} 1 & 0 & 0 \\ 0 & \cos \pi & \text{-}\sin \pi \\ 0 & \sin \pi & \cos \pi \end{pmatrix} \begin{pmatrix} \cos \frac{\pi}{2} & 0 & \sin \frac{\pi}{2}\\ 0 & 1 & 0 \\ \text{-}\sin \frac{\pi}{2} & 0 & \cos \frac{\pi}{2}\end{pmatrix} \\ &= \begin{pmatrix} 0 & 0 & 1\\ 0 & \text{-}1 & 0 \\ 1 & 0 & 0\end{pmatrix} \end{align}

You can compare this to the rotation matrix for an arbitrary 3D rotation matrix $R_{\vec{n}}(\theta)$ around a unit vector $\vec{n}$ (see Wikipedia for example) to get an overconstrained system of equations. However in this case its easy to confirm that $\vec{n} = (1, 0, 1)/\sqrt{2}$ and $\theta=\pi$ is the correct choice.

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