Quantum Computing 2.3 -- Hadamard gate on the Bloch sphere

We first describe the geometric operations in three-dimensional space through matrices. The reflection at an origin plane with the normal vector $\hat n = (n_x, n_y, n_z)$ is generally described by the operation $\begin{aligned} \vec r & \mapsto \vec r - 2 (\vec r \cdot \hat n) \hat n = (I - 2 \hat n \otimes \hat n) \cdot \vec r =: \mathcal{R}_{\hat n} \cdot \vec r \\ \mathcal{R}_{\hat n} &= \left( \begin{array}{ccc} 1 - 2 n_x^2 & - 2 n_x n_y & - 2 n_x n_z \\ - 2 n_x n_y & 1 - 2 n_y^2 & - 2 n_y n_z \\ - 2 n_x n_z & - 2 n_y n_z & 1 - 2 n_z^2 \end{array} \right) \end{aligned}$ For the plane $x = z$ we have $\hat n = \frac{1}{\sqrt{2}} (-1,0,1)$ and therefore $\mathcal{R}_{x = z} = \left( \begin{array}{ccc} 0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0 \end{array} \right)$ A general rotation around the axis $\hat n$ with the angle $\alpha$ can be written in three-dimensional space as $\begin{aligned} \vec r & \mapsto (\hat n \cdot \vec r) \hat n + \cos \alpha (\hat n \times \vec r) \times \hat n + \sin\alpha (\hat n \times \vec r) \\ &= \left((1 - \cos \alpha) \hat n \otimes \hat n + I \cos (\alpha ) + \sin \alpha \sum_{i = x,y,z} (\hat n \times \hat i) \otimes \hat i \right) \cdot \vec r =: \mathcal{R}_{\hat n}(\alpha) \cdot \vec r \\ \mathcal{R}_{\hat n}(\alpha) &= \left( \begin{array}{ccc} n_x^2 (1- \cos \alpha) + \cos \alpha & n_x n_y ( 1 - \cos \alpha) - n_z \sin \alpha & n_x n_z ( 1 - \cos \alpha) + n_y \sin \alpha \\ n_x n_y ( 1 - \cos \alpha) - n_z \sin \alpha & n_y^2 (1 - \cos \alpha) + \cos \alpha & n_y n_z ( 1 - \cos \alpha) - n_x \sin \alpha \\ n_x n_z ( 1 - \cos \alpha) - n_y \sin \alpha & n_x n_y ( 1 - \cos \alpha) + n_z \sin \alpha & n_z^2 (1- \cos \alpha) + \cos \alpha \end{array} \right) \end{aligned}$ in our case, the rotation axes are $\hat n = (0,1,0)$ and $\hat n = \frac {1} {\sqrt {2}} (1,0,1)$ and the rotation angles are $\alpha = \pi / 2$ and $\alpha = \pi$ . Thus, the rotation matrices yield $\begin{aligned} \mathcal{R}_{\hat y} (\pi/2) &= \left( \begin{array}{ccc} 0 & 0 & 1 \\ 0 & 1 & 0 \\ -1 & 0 & 0 \end{array} \right) \\ \mathcal{R}_{\hat x + \hat z}(\pi) &= \left( \begin{array}{ccc} 0 & 0 & 1 \\ 0 & -1 & 0 \\ 1 & 0 & 0 \end{array} \right) \end{aligned}$ Obviously, these operations are different so you can try different example states to see which of these operations is the right one. You will find out that only the rotation around the $\hat x + \hat z$ -axis always gives the correct results. We now want to prove that this rotation is indeed equivalent to the Hadamard operation.

Application of the Hadamard gate yields $\begin{aligned} & & \left( \begin{array}{c} \alpha' \\ \beta' \end{array}\right) &= \frac{1}{\sqrt 2} \left(\begin{array}{cc} 1 & 1 \\ 1 & -1 \end{array} \right) \left( \begin{array}{c} \cos\frac{\theta}{2} \\ \sin\frac{\theta}{2} e^{i \phi} \end{array} \right) \\ & & &= \left( \begin{array}{c} \cos\frac{\theta}{2} + \sin\frac{\theta}{2} e^{i \phi} \\ \cos \frac{\theta}{2} - \sin\frac{\theta}{2} e^{i \phi} \end{array} \right) \\ & & &= e^{i \gamma'} \left( \begin{array}{c} \cos\frac{\theta'}{2} \\ \sin\frac{\theta'}{2} e^{i \phi'} \end{array} \right) \\ \Rightarrow & & \frac{\beta'}{\alpha'} = \tan \frac{\theta'}{2} e^{i \phi'} &= \frac{\cos \frac{\theta}{2} - \sin\frac{\theta}{2} e^{i \phi}}{\cos \frac{\theta}{2} + \sin\frac{\theta}{2} e^{i \phi}} \\ & & &= \frac{\cos \frac{\theta}{2} - \sin\frac{\theta}{2} e^{i \phi}}{\cos \frac{\theta}{2} + \sin\frac{\theta}{2} e^{i \phi}} \frac{\cos \frac{\theta}{2} + \sin\frac{\theta}{2} e^{-i \phi}}{\cos \frac{\theta}{2} + \sin\frac{\theta}{2} e^{-i \phi}} \\ & & &= \frac{\cos^2 \frac{\theta}{2} - \sin^2 \frac{\theta}{2} - \sin \frac{\theta}{2} \cos \frac{\theta}{2} (e^{i \phi} - e^{-i \phi})}{\cos^2 \frac{\theta}{2} + \sin^2 \frac{\theta}{2} + \sin \frac{\theta}{2} \cos \frac{\theta}{2} (e^{i \phi} + e^{-i \phi})} \\ & & &= \frac{\cos \theta - i \sin \theta \sin \phi}{1 + \sin \theta \cos \phi} \\ \Rightarrow & & \tan \phi' &= - \frac{\sin \theta \sin \phi}{\cos \theta} \\ \Rightarrow & & \tan \frac{\theta'}{2} &= \left|\frac{\cos \theta - i \sin \theta \sin \phi}{1 + \sin \theta \cos \phi}\right| \\ & & &= \frac{\sqrt{\cos^2 \theta + \sin^2 \theta \sin^2 \phi }}{1 + \sin \theta \cos \phi} \\ & & &= \frac{\sqrt{1 - \sin^2 \theta \cos^2 \phi }}{1 + \sin \theta \cos \phi} \\ & & &= \sqrt{\frac{1 - \sin \theta \cos \phi}{1 + \sin \theta \cos \phi}} \end{aligned}$ The rotation around the $\hat x + \hat z$ -axis yields $\begin{aligned} & & \left( \begin{array}{c} x' \\ y' \\ z' \end{array} \right) = \left( \begin{array}{c} \cos \phi' \sin \theta' \\ \sin \phi' \sin \theta' \\ \cos \theta' \end{array} \right) &= \left( \begin{array}{ccc} 0 & 0 & 1 \\ 0 & -1 & 0 \\ 1 & 0 & 0 \end{array} \right) \left( \begin{array}{c} \cos \phi \sin \theta \\ \sin \phi \sin \theta \\ \cos \theta \end{array} \right) \\ & & &= \left( \begin{array}{c} \cos \theta \\ -\sin \phi \sin \theta \\ \cos \phi \sin \theta \end{array} \right) \\ \Rightarrow & & \tan \phi' = \frac{y'}{x'} &= -\frac{\sin \phi \sin \theta}{\cos \theta} \\ \Rightarrow & & \tan \frac{\theta'}{2} &= \frac{ \sin \frac{\theta'}{2}}{\cos \frac{\theta'}{2} } \\ & & &= \frac{\sin \frac{\theta'}{2} \cos \frac{\theta'}{2}}{\cos^2 \frac{\theta'}{2} } \\ & & &= \frac{2 \sin \frac{\theta'}{2} \cos \frac{\theta'}{2}}{1 + \cos^2 \frac{\theta'}{2} - \sin^2 \frac{\theta'}{2}} \\ & & &= \frac{\sin \theta'}{1 + \cos \theta' } \\ & & &= \frac{\sqrt{x'^2 + y'^2 }}{1 + z'} \\ & & &= \frac{\sqrt{\cos^2 \theta + \sin^2 \phi \sin^2 \theta}}{1 + \cos \phi \sin \theta} \\ & & &= \frac{\sqrt{1 - \cos^2 \phi \sin^2 \theta}}{1 + \cos \phi \sin \theta} \\ & & &= \sqrt{\frac{1 - \cos \phi \sin \theta}{1 + \cos \phi \sin \theta} } \end{aligned}$ Thus, we get the same results for the angles $\theta '$ and $\phi'$ in both cases, so that $H \equiv \mathcal{R} _{\hat x + \hat z} (\pi)$ .