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For 1-site I was able to use an Rx gate followed by an Ry gate to generate every possible state. For two-sites, I initially tried to apply this to each qubit but realized this doesn't work because it does not entangle the qubits. I have seen many examples of ansatz but they all seem to be optimized to particular hamiltonians and some don't generate every state. Is there a very simple 2-site ansatz I can use to generate all states?

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Actually, we do not need to consider a generic two-qubit unitary since we start from the fiducial state $|00\rangle$, and therefore all that matters is the first column of the $4 \times 4$ matrix representation of the circuit that we apply onto the fiducial state. As @MicheleAmoretti correctly pointed out, an arbitrary SU(4) matrix can be parameterized with 15 free parameters. An arbitrary two-qubit state, in turn, takes only 6 free parameters (cf., e.g., @AdamZalcman's answer to this earlier question). The same reduction of the number of free parameters was already alluded to in the question itself: Two rotation gates suffice to prepare an arbitrary single-qubit state, but it takes 3 free parameters to encode an arbitrary SU(2) matrix.

As for the actual parameterized quantum circuit that realizes such a generic two-qubit state, we can consider a symmetric bipartition to reshape the 4-dimensional vector into a $2 \times 2$ matrix and then perform its singular value decomposition. This allows to derive a quantum circuit with a single CNOT and six rotation gates, as explained in Section VII of this paper.

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  • $\begingroup$ Thank you for the detailed response! $\endgroup$
    – user22395
    Feb 27, 2023 at 5:19
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The minimal circuit for a generic 2-qubit unitary $U$ is described in the following paper: https://arxiv.org/pdf/quant-ph/0308033.pdf (see Fig. 2). This amounts to 3 CNOT gates, 3 rotation gates, and 4 arbitrary single qubit gates which become 12 single rotation gates using ZYZ decomposition.

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  • $\begingroup$ Thank you for the resource. When you say 4 arbitrary single qubit gates can be used, does this mean the Rx, Ry or Rz gates can be used in these positions? $\endgroup$
    – user22395
    Feb 28, 2023 at 1:36
  • $\begingroup$ Arbitrary means, in general, a combination of Rx, Ry and Rz (e.g., ZYZ). $\endgroup$ Feb 28, 2023 at 6:43

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