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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.

Callum
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user1587566
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    Actually, you should use $Ry$ for preparing the state you mentioned. – Martin Vesely Jul 20 '23 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>? – user1587566 Jul 21 '23 at 16:46

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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 $\cos\frac\theta2 \vert 0 \rangle - i \sin\frac\theta2 \vert 1 \rangle$.

slimmerikko
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  • 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>? – user1587566 Jul 21 '23 at 16:45
  • 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 $R_Y(\theta)$. – slimmerikko Jul 22 '23 at 19:26