Theory¶
In the previous notebooks we encoded the data, each having two features, with a simple $R_y(\theta)$ rotation. The angle for this rotation was chosen such that it rotated the state $\ket{0}$ to the desired state $\ket{\bf{x}}$ corresponding to the data. If we want to include more features for our data points we need to generalize this rotation. Instead of a simple rotation, we now need a combination of gates such that it maps $\ket{0\ldots 0} \mapsto \ket{\bf x} = \sum_i a_i \ket{i}$, where $\ket{i}$ is the $i^{th}$ entry of the computational basis $\left\{\ket{0\ldots0},\ldots,\ket{1\ldots1}\right\}$. Again we only work with normalised data, meaning $\sum_i \lvert a_i \rvert^2=1$. The general procedure how to initialize a state to an arbitrary superposed state can be found in the article by Long and Sun. In this notebook we will consider how to implement their scheme for 2 qubits, that is up to 4 features: \begin{equation} \ket{00} \mapsto a_{00} \ket{00} + a_{01} \ket{01} + a_{10} \ket{10} + a_{11} \ket{11} \end{equation}
For the implementation we closely follow the reference and use single bit rotation gates $U(\theta)$ defined by \begin{equation} U(\theta) = \begin{pmatrix} \cos(\theta) & \sin(\theta) \\ \sin(\theta) & -\cos(\theta) \end{pmatrix} = R_y(2\theta) \cdot Z \end{equation} and controlled versions of it. Because we will only act with these gates on $\ket{0}$ we can even drop the $Z$ gate.
For two qubits the scheme consists of three steps:
- Apply a bit rotation $U(\alpha_1)$ on the first qubit: \begin{equation} U(\alpha_1)\ket{0}\otimes\ket{0}= \sqrt{\abs{a_{00}}^2 + \abs{a_{01}}^2} \ket{00} + \sqrt{\abs{a_{10}}^2 + \abs{a_{11}}^2} \ket{10}, \end{equation} where $\alpha_1$ is given by \begin{equation} \alpha_1 = \arctan\left(\sqrt{\frac{\abs{a_{10}}^2 + \abs{a_{11}}^2}{\abs{a_{00}}^2 + \abs{a_{01}}^2}}\right) \end{equation}
- Next, apply a controlled-rotation $U(\alpha_2)$ on the second qubit, with as control the first qubit being 0. Choose $\alpha_2$ such that: \begin{equation} \cos(\alpha_2) = \frac{a_{00}}{\sqrt{\abs{a_{00}}^2+\abs{a_{01}}^2}}, \hspace{1cm} \sin(\alpha_2) = \frac{a_{01}}{\sqrt{\abs{a_{00}}^2+\abs{a_{01}}^2}} \end{equation}
- Lastly, apply a controlled-rotation $U(\alpha_3)$ on the second qubit, with the first qubit being 1 as control. Choose $\alpha_3$ such that: \begin{equation} \cos(\alpha_3) = \frac{a_{10}}{\sqrt{\abs{a_{10}}^2+\abs{a_{11}}^2}}, \hspace{1cm} \sin(\alpha_3) = \frac{a_{11}}{\sqrt{\abs{a_{10}}^2+\abs{a_{11}}^2}} \end{equation}
The angles $\alpha_2$ and $\alpha_3$ are chosen such that the root terms cancel out, leaving us with the desired result: \begin{equation} \begin{split} \sqrt{\abs{a_{00}}^2 + \abs{a_{01}}^2}& \ket{0}\otimes U(\alpha_2)\ket{0} + \sqrt{\abs{a_{10}}^2 + \abs{a_{11}}^2} \ket{1}\otimes U(\alpha_3)\ket{0}\\ &=a_{11} \ket{00} + a_{01} \ket{01} + a_{10} \ket{10} + a_{11} \ket{11} \end{split} \end{equation}
Note: this circuit makes it also possible to encode 3 features by simply setting $a_{11}=0$.
The circuit looks something like this:
