This exercise can be done in two steps.

1. Convert the state from $\ket{0}$ to $\alpha\ket{0} + \beta\ket{1}$.
   This can be done in a similar manner to the "Preparing a Rotated State" exercise, we first prepare an $\alpha\ket{0} -i\beta\ket{1}$ state using the $R_x$ gate, and then removing the relative phase of $-i$ by applying the $S$ gate, which would turn $\alpha\ket{0} -i\beta\ket{1}$ to $\alpha\ket{0} + \beta\ket{1}$.
   An alternative, simpler approach is to use the $R_y$ gate, which allows you to get the necessary state right away without introducing a relative phase:
$$
R_y(2\arctan\frac{\beta}{\alpha}) = \begin{bmatrix} \alpha & -\beta \\ \beta & \alpha \end{bmatrix}
$$
2. Add a phase of $e^{i\theta}$ to the $\ket{1}$ basis state using the $R_1(\theta)$ gate. This would turn $\alpha\ket{0} +\beta\ket{1}$ to $\alpha\ket{0} + e^{i\theta}\beta\ket{1}$.

The solution can be represented as $R_1(\theta)R_y(2\arctan\frac{\beta}{\alpha})$ or in matrix form as:
$$
\begin{bmatrix} 1 & 0 \\ 0 & e^{i\theta} \end{bmatrix}\begin{bmatrix} \alpha & -\beta \\ \beta & \alpha \end{bmatrix} = 
\begin{bmatrix} \alpha & -\beta \\ e^{i\theta}\beta & e^{i\theta}\alpha \end{bmatrix}
$$

This turns $\ket{0} = \begin{bmatrix} 1 \\ 0\end{bmatrix}$ into $\begin{bmatrix} \alpha & -\beta \\ e^{i\theta}\beta & e^{i\theta}\alpha \end{bmatrix} \begin{bmatrix} 1 \\ 0\end{bmatrix} = \begin{bmatrix} \alpha \\ e^{i\theta}\beta \end{bmatrix} = \alpha\ket{0} +e^{i\theta}\beta\ket{1}$.

@[solution]({
    "id": "single_qubit_gates__prepare_arbitrary_state_solution",
    "codePath": "./Solution.qs"
})
