# What is the connection between $RX$ gates and $X$ gates (similar for $Y$ and $Z$)?

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?

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):

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

• I'd clarify this a bit: equivalent, unless you want to use a controlled version of them. The habit of ignoring the global phase of the gate can come back and bite painfully: quantumcomputing.stackexchange.com/questions/5973/… Commented Sep 11, 2020 at 16:24
• Hi, may you explain why I can't reproduce such a property within wolframalpha.com, inputing e^{i*{{0,1},{1,0}}*(-pi/2)}. Commented Sep 19, 2022 at 10:12
• Hi @DanieleCuomo, it might be valuable to create a new question with the full details of what you're attempting! Commented Oct 2, 2022 at 14:35
• I actually did it! quantumcomputing.stackexchange.com/questions/28224/… Commented Oct 2, 2022 at 15:07