# Does Rx(θ) applied on pure states create superposition?

I'm struggling to find if $$\text{Rx}(\theta)$$ gate would convert a pure state qubit $$|0\rangle$$ to a superposition $$\cos( \theta) |0\rangle + \sin(\theta) |1\rangle$$.

A definitive answer with reference will be appreciated.

• Actually, you should use $Ry$ for preparing the state you mentioned. Jul 20, 2023 at 21:08
• Thanks, but I made a mistake. It should be Rz(θ) = [[cos θ,− sin θ],[sin θ,cos θ]] instead. Applied to pure |0>, would create a superposition cos θ |0> + sin θ |1>? Jul 21, 2023 at 16:46

In Nielsen and Chuang it is given that the matrix represntation of $$R_X(\theta)$$ is $$R_X(\theta) = \begin{pmatrix} \cos\frac\theta2 & -i\sin\frac\theta2 \\ -i\sin\frac\theta2 & \cos\frac\theta2 \end{pmatrix}.$$ From this you can see that $$\vert 0 \rangle$$ is sent to a superposition of $$\vert 0 \rangle$$ and $$\vert 1 \rangle$$ unless $$\sin\frac\theta2 = 0$$ or $$\cos\frac\theta2=0$$ so $$\theta = 0$$ or $$\theta = \pi$$.
So for $$\theta=0$$, we have $$R_X(\theta) \vert 0 \rangle = \vert 0 \rangle$$, and for $$\theta=\pi$$, we have $$R_X(\theta)\vert 0 \rangle = \vert 1 \rangle$$, but for all other $$\theta$$ you will get a superposition, namely $$\cos\frac\theta2 \vert 0 \rangle - i \sin\frac\theta2 \vert 1 \rangle$$.
• The matrix that you mention is not $R_z(\theta)$. We have $R_z(\theta) = e^{-i\theta Z/2} = \begin{pmatrix} e^{-i\theta/2} & 0 \\ 0 & e^{i\theta / 2} \end{pmatrix}$. Therefore, applying $R_z(\theta)$ to a pure state will always give you the same pure state back (up to global phase). As mentioned in a different comment, to get the pure state you mention, you will need $R_Y(\theta)$. Jul 22, 2023 at 19:26