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Given a 2-qubits system, the 4 computational basis states are $|00\rangle$, $|01\rangle$, $|10\rangle$, $|11\rangle$.
Is it possible to prepare these states by a one-parameter quantum circuit "ansatz" $U(\theta)$ passing 4 distinct continuous values of $\theta \in \mathbb{R}$?
If yes, how does this parametric circuit look like and what is the process to build it?
You can do it by performing a rotation of $\theta$ in the Fourier basis. I'll give the reasoning for an arbitrary number of qubits.
Let us denote by $\mathsf{QFT}$ the unitary matrix corresponding to the QFT transformation. Note that: $$\mathsf{QFT}|x\rangle=\sum_k\mathrm{e}^{\frac{2\mathrm{i}\pi kx}{2^n}}|k\rangle$$ In particular, this means that, for $x$ and $y$ being two integers: $$\mathsf{QFT}^\dagger\sum_k\mathrm{e}^{\frac{2\mathrm{i}\pi kx}{2^n}}\mathrm{e}^{\frac{2\mathrm{i}\pi ky}{2^n}}|k\rangle=|x+y\pmod{2^n}\rangle$$ So if we perform a rotation in the Fourier basis, we end up with an addition in the computational basis.
Thus, our idea is that we want to apply a rotation of $\theta\in\mathbb{N}$ in the Fourier basis on $|0\rangle$ so that we end up with $|\theta\rangle$.
Suppose you start from the $|0\rangle$ state. You then apply a layer of Hadamard gates: $$\frac{1}{\sqrt{2^n}}\sum_x|x\rangle$$ You now apply a Phase gate with parameter $2^{n-i+1}2\theta\frac\pi{2^{n}}=\theta\frac\pi{2^{i-2}}$ on the $i$-th qubit. For an arbitrary basis state $|x\rangle$, let us denote: $$x=\sum_{i=0}^{n-1}b_i2^{n-i+1}$$ This is simply the binary decomposition of $x$. Note that $b_i$ represents the state of the $i$-th qubit. Because of how the Phase gate works, the rotation on the $i$-th wire will have an effect only if $b_i=1$. Thus, if the state is $|x\rangle$, the angle of the rotation that we apply is: $$\sum_{i=0}^{n-1}b_i2^{n-i+1}2\theta\frac{\pi}{2^n}=2x\theta\frac\pi{2^n}$$ Which means that the state is now: $$\frac{1}{\sqrt{2^n}}\sum_x\mathrm{e}^{\frac{2\mathrm{i}\pi x\theta}{2^n}}|x\rangle$$ Which means that applying $\mathsf{QFT}^\dagger$ on this state yields $|\theta\rangle$.
The answer to the question is affirmative. Here is an example of a single-parameter two-qubit circuit $U(\theta)$ that allows to prepare all four computational basis states starting from the fiducial state $|00\rangle$:
Specifically, the parameter values to obtain each computational basis state (up to a negligible global phase factor) are the following:
$U(0) |00\rangle = |00\rangle$.
$U(\frac{\pi}{2}) |00\rangle = |01\rangle$.
$U(\pi) |00\rangle = |10\rangle$.
$U(\frac{3\pi}{2}) |00\rangle = |11\rangle$.
The way I arrived at this circuit goes as follows. Let the parameterized $4 \times 4$ unitary matrix we are seeking, $U(\theta)$, be the time-evolution operator of some two-qubit Hamiltonian $\mathcal{H}$, $U(\theta) = e^{-i \theta \mathcal{H}}$. Assuming a priori the relation between the values of the parameter $\theta$ and the computational basis states that are enumerated above, we want
\begin{equation} U\left(\frac{\pi}{2}\right) = e^{-i \frac{\pi}{2} \mathcal{H}} = \begin{pmatrix} 0 & 0 & 0 & 1 \\ 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \end{pmatrix}, \end{equation}
in which case our Hamiltonian is given by
\begin{equation} \mathcal{H} = \frac{2i}{\pi} \ln \left[ \begin{pmatrix} 0 & 0 & 0 & 1 \\ 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \end{pmatrix} \right] = \frac{1}{2} \begin{pmatrix} -1 & 1 - i & -1 & 1 + i \\ 1 + i & -1 & 1 - i & -1 \\ -1 & 1 + i & -1 & 1 - i \\ 1 - i & -1 & 1 + i & -1 \end{pmatrix}. \end{equation}
Now that we have the two-qubit Hamiltonian $\mathcal{H}$, the desired single-parameter two-qubit unitary $U(\theta) = e^{-i \theta \mathcal{H}}$ is unequivocally defined. The only missing step is decomposing such two-qubit circuit in terms of basis gates for an arbitrary value of the parameter $\theta$. There are multiple ways of doing this; I opted for diagonalizing $\mathcal{H}$, as in $\mathcal{H} = V D V^{\dagger}$, which allows to express $U(\theta)$ as $U(\theta) = V e^{-i \theta D} V^{\dagger}$, thus leaving the $\theta$-dependence entirely to the diagonal part, which is easier to decompose.
How about $$ U(\theta) = X^{\text{bin}(\theta)_0} \otimes X^{\text{bin}(\theta)_1} $$ for $\theta\in \{0,1,2,3\}$ and $\text{bin}: \{0,1,2,3\} \rightarrow \{0,1\}^2$ being the binary representation, $\text{bin}(0) = (0,0)$, $\text{bin}(1) = (0,1)$, etc.
