Build the inversion test of an angle-encoding kernel

The feature map encodes $x=(x_0,x_1)$ on two qubits with one $R_X$ rotation per qubit,

$U(x)=R_X(x_0)\otimes R_X(x_1),\qquad R_X(\theta)=e^{-i\theta X/2}=\begin{pmatrix}\cos\frac\theta2&-i\sin\frac\theta2\\-i\sin\frac\theta2&\cos\frac\theta2\end{pmatrix},$

with feature $x_0$ on qubit $0$ (the most significant bit) and $x_1$ on qubit $1$. Starting from $\lvert00\rangle$, build the inversion test circuit

$V=U(x')^\dagger\,U(x)\qquad\text{for}\qquad x=(0.3,\,1.2),\quad x'=(1.0,\,0.4),$

so that the probability of the outcome $00$ is the fidelity kernel $K(x,x')=\lvert\langle\phi(x')\vert\phi(x)\rangle\rvert^2$. First prepare $\lvert\phi(x)\rangle=U(x)\lvert00\rangle$, then apply the inverse of the circuit that prepares $\lvert\phi(x')\rangle$.

How it is graded. Only $R_X$ gates are available, at most $4$ of them, on $2$ qubits. The grader simulates your circuit and compares, with absolute tolerance $10^{-6}$:

• check probabilities: the distribution $\big(P(00),P(01),P(10),P(11)\big)=(0.748605,\ 0.133816,\ 0.099748,\ 0.017830)$ of measuring both qubits;
• check p00: the expectation value of the projector $\lvert00\rangle\langle00\rvert=\tfrac14(II+ZI+IZ+ZZ)$, which equals $P(00)=K(x,x')=0.748605$.