I am new to quantum gates but do not understand the connection between the $RX$ and $X$ gates. I know that

$$R X(\theta)=\exp \left(-i \frac{\theta}{2} X\right)=\left(\begin{array}{cc} \cos \frac{\theta}{2} & -i \sin \frac{\theta}{2} \\ -i \sin \frac{\theta}{2} & \cos \frac{\theta}{2} \end{array}\right)$$

Meanwhile the $X$ gate is given by

$$X = \left(\begin{array}{cc} 0 & 1 \\ 1 & 0 \end{array}\right)$$

Is there a value of $\theta$ such that the two are the same? I see that choosing $\theta = \pi/2$ gives the result upto an overall factor of $-i$. Is that it or is there a deeper connection between the two gates? Is there a similar connection between the $Y$ and $RY$ gates and the $Z$ and $RZ$ gates such that the rotated gates are more general than the $X, Y$ and $Z$ gates?


1 Answer 1


You're almost correct - choosing $ \theta = \pi$ does yield $$ \begin{bmatrix} 0 & -i \\ -i & 0 \end{bmatrix} $$

Because this differs from the $X$ gate by a constant factor global phase ($ -i$), the gates are equivalent. (See here to learn more about the global phase).

This connection holds similarly for $ RY$ and $Y$, and $RZ$ and $Z$. A way to visualize this is the Bloch sphere: in essence, these gates are rotations about the $X, Y, Z$ axes (respectively):

Bloch sphere

So essentially our Pauli primitives are $\pi$ rotations over the respective axis.


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