When applying a Hadamard Gate Wikipedia defines it as $XR_y(\frac{\pi}{2}) = H$. The effect of a Pauli Gate $X$ is defined as a Rotation of $\pi$ radians about the x-Axis on the Bloch sphere. The effect of the Rotation Operator Gate $R_x(\theta)$ is defined as a rotation about the x-Axis by an angle $\theta$ on the Bloch sphere.
Why is it, that it's $R_x(\pi) \neq X$? If I apply $R_x(\pi)R_y(\frac{\pi}{2})$ to the state $|0\rangle{}$ I end up with a superposition state that differs by $-i$ from what it is supposed to be according to the definitions of the Hadamard Gate.
I get that it's $R_x(\pi) = -iX$ according to the definition $R_x(\theta)=e^{-i\theta X/2}$. My question is what the physical background of it is.
Why can't I use the rotation operator with $\theta = \pi$ in the Hadamard Gate, but have to use $X$ instead?
Best Answer
Recall the matrix exponential of a vector of Pauli matrices $$\exp(-i \theta\mathbf{n}\cdot\boldsymbol{\sigma}/2)=\mathbb{I}\cos\frac{\theta}{2}+i \mathbf{n}\cdot\boldsymbol{\sigma}\sin\frac{\theta}{2}.$$ By this token, we find $$R_x(\pi)=\exp(-i \sigma_x \pi/2)=-i \sigma_x.$$ Physically, there is no difference between this operator $-i\sigma_x$ and $\sigma_x$, because one can never distinguish between two states that differ by a global phase: $$|\psi\rangle\sim e^{i\varphi}|\psi\rangle.$$ Similarly, this means that there is no physical difference between the gate $H$ and the gate $-iH$. It so happens that the standard definition of $H$ has a particular global phase, so the Wikipedia page is probably trying to find operators that match this global phase. Moreover, when more than one qubit is involved, the relative phases become important, so it is good to find a standard definition of operators like $H$ that do not have ambiguous phases. To be rigorous, you can always say something like $$H=UR_x(\pi)R_y(\pi/2),\qquad U=i=e^{i\pi/2}.$$
Here's an example of how relative phases can matter when you have multiple qubits. Let's say you have a gate that applies $H$ one qubit $\#2$ when qubit $\#1$ is in state $|1\rangle$ and does nothing when qubit $\#1$ is in state $|0\rangle$. As an operator, this would be written as $$cH\equiv| 0\rangle_1\langle 0|\otimes\mathbb{I}_2+| 1\rangle_1\langle 1|\otimes H_2\neq \equiv| 0\rangle_1\langle 0|\otimes\mathbb{I}_2+| 1\rangle_1\langle 1|\otimes iH_2 \equiv ciH,$$ where the second form uses $iH$ instead of $H$ as the controlled operation. Acting on a state like $$|\Psi\rangle=\frac{|0\rangle_1+|1\rangle_1}{\sqrt{2}}\otimes |0\rangle_2=\frac{|0\rangle_1\otimes |0\rangle_2+|1\rangle_1\otimes |0\rangle_2}{\sqrt{2}},$$ where we know that $$H_2|0\rangle_2=\frac{|0\rangle_2+|1\rangle_2}{\sqrt{2}},$$ leads to the pair: \begin{equation} \begin{aligned} cH|\Psi\rangle&=\frac{|0\rangle_1\otimes|0\rangle_2+|1\rangle_1\otimes \frac{|0\rangle_2+|1\rangle_2}{\sqrt{2}}}{\sqrt{2}} \\ &=\frac{\sqrt{2}|0\rangle_1\otimes|0\rangle_2+|1\rangle_1\otimes |0\rangle_2+|1\rangle_1\otimes|1\rangle_2}{2} \\ ciH|\Psi\rangle&=\frac{|0\rangle_1\otimes|0\rangle_2+|1\rangle_1\otimes i\frac{|0\rangle_2+|1\rangle_2}{\sqrt{2}}}{\sqrt{2}} \\ &=\frac{\sqrt{2}|0\rangle_1\otimes|0\rangle_2+i|1\rangle_1\otimes |0\rangle_2+i|1\rangle_1\otimes|1\rangle_2}{2}\neq cH|\Psi\rangle. \end{aligned} \end{equation} So which global phase you use does not matter when there is only one qubit involved, but you need to make everything consistent when you have more than one qubit, which is why it is useful to stick with one convention.