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I am trying to implement the quantum k-means algorithm proposed in https://arxiv.org/pdf/1909.04226.pdf.

In the equation (8) of the manuscript we need to implement a state $|\psi\rangle = \frac{1}{\sqrt{2}}\left[|0\rangle \otimes |X_i\rangle + |1\rangle \otimes |X_j\rangle\right]$ like the one shown in the title, where $X_i$, $X_j$ are normalised vectors.

One solution proposed in the paper for implementing the $|\psi\rangle$ state is to use Hadamard and CSWAP gates. Assuming that $X_i$ and $X_j$ can be prepared with one qubit each, if one apply a Hadamard gate to the first qubit and afterwards a CSWAP gate, we would have the state $|\gamma\rangle = \frac{1}{\sqrt{2}}\left[|0\rangle \otimes |X_i\rangle \otimes |X_j\rangle + |1\rangle \otimes |X_j\rangle \otimes |X_i\rangle \right]$. So I need to somehow "get rid" of the leftmost qubit. Here are some of the things I have tried/thought about until now:

  • As the $|\gamma\rangle$ state is not separable, we can't use the partial trace or just reset the leftmost qubit.
  • We can apply a SWAP gate between the leftmost qubit and a new qubit q3, because then equation (9) of the manuscript wouldn't hold.

Any ideas on how to encode the $|\psi\rangle$ state with or without using the mentioned procedure ?

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    $\begingroup$ If you can prepare a state $X_i$ you can remove it by running the inverse algorithm. Just control it off your first qubit! $\endgroup$
    – DaftWullie
    Oct 13, 2023 at 15:18

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Suppose that $U_i$ prepares the state $\left|X_i\right\rangle$. Start by applying an Hadamard gate on the first qubit: $$\frac{1}{\sqrt{2}}\left(|00\rangle+|10\rangle\right)$$ Then apply $U_j$ controlled off the first qubit: $$\frac{1}{\sqrt{2}}\left(|00\rangle+|1\rangle\otimes\left|X_j\right\rangle\right)$$ Apply an $X$ gate on the first qubit: $$\frac{1}{\sqrt{2}}\left(|10\rangle+|0\rangle\otimes\left|X_j\right\rangle\right)$$ Apply $U_i$ controlled off the first qubit: $$\frac{1}{\sqrt{2}}\left(|1\rangle\otimes\left|X_i\right\rangle+|0\rangle\otimes\left|X_j\right\rangle\right)$$ And finally reapply an $X$ gate on the first qubit to obtain the state you're looking for.

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