# How to represent the Hadamard gate as a rotations on the Bloch sphere?

I am new to Quantum Computing, and I have decided to try and learn the quantum gates. I am trying to understand how to represent some basic gates as rotations on the Bloch Sphere. I was able to represent the Pauli gates in terms of rotations. For example: \begin{align} \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} \begin{pmatrix} \cos(\frac{\theta}{2}) \\ \sin(\frac{\theta}{2}) e^{i\phi} \end{pmatrix} & = \begin{pmatrix} \sin(\frac{\theta}{2})e^{i\phi} \\ \cos(\frac{\theta}{2}) \end{pmatrix} \\ & = e^{i\phi} \begin{pmatrix} \cos(\frac{\pi - \theta}{2}) \\ \sin(\frac{\pi - \theta}{2})e^{-i\phi} \end{pmatrix} \end{align} \\

Therefore $$X|\psi⟩$$ is the same as taking a point $$(\theta, \phi)$$ on the Bloch Sphere and rotating it it, expressed by $$(\pi - \theta, -\phi)$$. I tried to do the same process with the Hadamard Gate, but I was unable to resolve it in the same way. Is there a way to find the rotation of the Hadamard gate on the Bloch sphere? What are these rotations?

• – glS
Mar 16 at 11:36

I would suggest that the easiest way to do this in general is to take the unitary $$U$$ that you want to understand, and find its eigenvectors. These, by definition, are the vectors that are not changed by the rotation. In other words, they define the axis about which the rotation occurs.
For example, if I take the Hadamard matrix, it has eigenvectors $$\frac{1}{\sqrt{2(2\pm\sqrt{2})}}\left(\begin{array}{c} 1\pm\sqrt{2} \\ 1 \end{array}\right).$$ If you equate these values to $$\cos\frac{\theta}{2}$$ and $$\sin\frac{\theta}{2}$$, then you can calculate $$\sin\theta=2\sin\frac{\theta}{2}\cos\frac{\theta}{2}=\pm\frac{1}{\sqrt{2}}$$ So, $$\theta=\frac{\pi}{4},\frac{5\pi}{4}$$ and $$\phi=0$$. Note that since the angles are $$\pi$$ apart (this is always the case), the two eigenvectors represent the same line through the centre of the Bloch sphere, just pointing in opposite directions.
Just to add to the answers above: the Hadamard gate is simply a 180 degree rotation around the $$(\hat{x}+\hat{z})/\sqrt{2}$$ axis.
$$y$$-rotation is defined as $$Ry(\theta)= \begin{pmatrix} \cos(\theta/2) & -\sin(\theta/2) \\ \sin(\theta/2) & \cos(\theta/2) \\ \end{pmatrix}.$$ Setting $$\theta = \pi/2$$ you get $$Ry(\pi/2)=\frac{1}{\sqrt{2}} \begin{pmatrix} 1 & -1 \\ 1 & 1 \end{pmatrix},$$ which is "almost" Hadamard gate (up to phase if the gate is applied on state $$|1\rangle$$). To resolve this, you can apply $$X$$ gate to the result above: $$XRy(\pi/2)=\frac{1}{\sqrt{2}} \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} \begin{pmatrix} 1 & -1 \\ 1 & 1 \end{pmatrix}= \frac{1}{\sqrt{2}} \begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix} = H.$$ Now we have to represent $$X$$ as a rotation on a Bloch sphere. You did it yourself but since $$Rx(\theta)= \begin{pmatrix} \cos(\theta/2) & -i\sin(\theta/2) \\ -i\sin(\theta/2) & \cos(\theta/2) \\ \end{pmatrix}$$ we have $$Rx(\pi)= \begin{pmatrix} 0 & -i \\ -i & 0 \\ \end{pmatrix} =-iX.$$
This means that $$Rx(\pi)Ry(\pi/2)= -iH$$, i.e. the Hadamard gate up to global phase $$-i$$ which can be in case of single-qubit gate neglected.