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There exists a nice way of preparing any superposition (with real amplitudes — this is the case I'm interested in) of states $\{\ldots0001\rangle,\,|\ldots0010\rangle,\,|\ldots0100\rangle,\ldots\}$, etc. This can be achieved with an $O(\log_2 n)$-depth circuit having $O(n)$ gates. To do so, one takes the circuit from Fig. 5 here, and replaces the $G(1/2)$ gate with $R_y$ rotations.

I'm wondering if this idea can be somehow generalized in order to prepare any superposition (with real coefficients) of constant Hamming weight states. Clearly, such a circuit will contain at least $O\left({n}\choose{m}\right)$ parametric gates. Would be cool if the depth could be made, similarly to the case above, significantly smaller than the number of gates.

Please do not suggest the UCC ansatz :D

UPDATE

In the comments below, Mark S pointed out a paper in which the preparation of Dicke states is discussed. Those are the equal-weight superpositions of constant Hamming weight states. The circuit contains only $O(kn)$ gates, and cannot be generalized to case of interest in an obvious way, as it can be done in the first paper cited above. Still, may be useful.

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  • $\begingroup$ Have you reviewed this question and answer? $\endgroup$ – Mark S Aug 11 at 22:44
  • $\begingroup$ This question and answer also appear relevant. $\endgroup$ – Mark S Aug 12 at 0:00
  • $\begingroup$ Thanks for the links! I really like the algorithm from "Deterministic Preparation of Dicke States". Do you think it can be adapted to my case? It only contains $O(nk) $ gates, and it's unclear to me how I can introduce $n\choose k$ free parameters. $\endgroup$ – mavzolej Aug 12 at 10:12
  • $\begingroup$ What is meant by "free parameters"? Are you asking for a distribution other than the uniform distribution (which is what i think the Dicke states will give you)? You could add another ancilla qubit, conditionally rotate the ancilla from $\vert 0\rangle$ to $\vert 1\rangle$ for each basis vector in your Dicke state by an amount given by (an appropriately normalized version of) your parameters, and post-select the ancilla being $\vert 1\rangle$ maybe? $\endgroup$ – Mark S Aug 12 at 14:44
  • $\begingroup$ I need to have $({n\choose k} - 1)$ real parameters in the circuit so that for any possible superposition of the $n \choose k$ states with real amplitudes there would exist a set of parameters preparing such a superposition. I provide an example for $k=1$ in the first paragraph of my question. $\endgroup$ – mavzolej Aug 12 at 14:56
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Here's an idea for somewhere to get started... (I have not worked through this in any more detail than presented here, but it looks plausible.) Let $$ U_{\Lambda}(\alpha_1,\alpha_2,\ldots,\alpha_{|\Lambda|}) $$ be a unitary that acts on a set of qubits $\Lambda$ such that, if those qubits start in the state $|0\rangle^{\otimes|\Lambda|}$, they are output in the state $$ \sum_i\alpha_i|0\rangle^{\otimes(i-1)}|1\rangle|0\rangle^{\otimes|\Lambda|-i}. $$ You say you already know how to do this.

Now consider wanting to make a state $$ \sum\beta_{i<j}|ij\rangle $$ (where $|ij\rangle$ is a shorthand to say 1s on qubits $i,j$ and 0 elsewhere) on $N$ qubits. Presumably we could do this by

  1. let $\alpha_i=\sqrt{\sum_{j:j>i}|\beta_{ij}|^2}$.
  2. Implement $U_{1,2,\ldots,N}(\alpha_1,\alpha_2,\ldots,\alpha_N)$.
  3. Apply controlled-$U_{2,3,\ldots ,N}(\tilde\beta_{12},\tilde\beta_{13},\ldots,\tilde\beta_{1N})$ controlled off qubit 1, where $\tilde\beta_{ij}=\beta_{ij}/\alpha_i$. This gets all the terms correct where the first qubit is in the 1 state.
  4. Then apply controlled-$U_{3,4,\ldots,N}(\tilde\beta_{23},\tilde\beta_{24},\ldots,\tilde\beta_{2N})$ where qubit 1 is controlled off $|0\rangle$, and qubit 2 is controlled off being in state $|1\rangle$.
  5. Repeat in order until all are done. On the $n^{th}$ round, you control off $n$ qubits such that the first $n-1$ are in the $|0\rangle$ state and the $n^{th}$ is in the $|1\rangle$ state.

This should get you the correct 2-excitation state. But this is a procedure that you can nest to get higher excitation states: for the $k$-excitation case, you first perform the division in steps 1 & 2, then replace the $U$s in the controlled-$U$ with the $k-1$ excitation version.

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