# How to prepare all the computational basis states by running the same quantum ansatz with distinct $\theta$ values?

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?

• This is very similar to a problem posted in the QHack 2023. I think that the policy for this type of questions is to wait for the end of the event to answer it Feb 19 at 20:18

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

$$$$U\left(\frac{\pi}{2}\right) = e^{-i \frac{\pi}{2} \mathcal{H}} = \begin{pmatrix} 0 & 0 & 0 & 1 \\ 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \end{pmatrix},$$$$

in which case our Hamiltonian is given by

$$$$\mathcal{H} = \frac{2i}{\pi} \ln \left[ \begin{pmatrix} 0 & 0 & 0 & 1 \\ 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \end{pmatrix} \right] = \frac{1}{2} \begin{pmatrix} -1 & 1 - i & -1 & 1 + i \\ 1 + i & -1 & 1 - i & -1 \\ -1 & 1 + i & -1 & 1 - i \\ 1 - i & -1 & 1 + i & -1 \end{pmatrix}.$$$$

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.

• Thank you! This of course works but I would like my parameter $\theta$ to have real continuous values (i.e. rotation angles). I just edited my question accordingly Feb 18 at 9:31