Let's say we have two quantum many body states $|\psi_1\rangle$ and $|\psi_2\rangle$(or equivalently, two quantum circuit $U_1$ and $U_2$ ), also an ancilla qubit $\alpha|0\rangle+\beta|1\rangle$. The goal is to prepare a state $\alpha|\psi_1\rangle+\beta|\psi_2\rangle$.

Can this be done efficiently using quantum circuits?

If $|\psi_1\rangle$ and $|\psi_2\rangle$ are logical state in quantum error correction. This is called encoding circuit. My question is, in general, could this be done?

  • $\begingroup$ I'm not sure if this is exactly what you mean, but there is some recent work about the impossibility of having a perfect "quantum adder", that is, a circuit that sends arbitrary $\lvert\psi\rangle$ and $\lvert\phi\rangle$ into $\lvert\psi\rangle+\lvert\phi\rangle$. See Alvarez-Rodriguez et al. (1411.4534) and Oszmaniec et al. (1505.04955). Could you instead give a reference for the "encoding circuit" you mention? $\endgroup$
    – glS
    Commented Jan 16, 2018 at 16:37

1 Answer 1



Given a circuit for a unitary $U$ on $\mathbb C^d$, we can always build a circuit for a controlled-$U$ gate acting on $\mathbb C^2\otimes C^d$, this is, a gate acting as \begin{align} |0\rangle|x\rangle&\mapsto |0\rangle |x\rangle\\ |1\rangle|x\rangle&\mapsto |1\rangle (U| x\rangle) \end{align} (i.e., $U$ acts on $\mathbb C^d$ if the control qubit is $|1\rangle$). Such a gate is obtained by replacing any elementary gate $G$ in the circuit for $U$ by a controlled-$G$ gate. (E.g., if the gate set is single-qubit rotations and CNOTs, replace them by controlled unitaries and Toffolis, for which efficient circuits are known.)

Now you can build your state by starting as follows:

  1. Start with $(\alpha|0\rangle + \beta |1\rangle)|x\rangle$.
  2. Apply $U_1$ to the second register.
  3. Apply a controlled-$(U_2U_1^{-1})$ gate.
  • $\begingroup$ I don't this is correct, you'd still have an entangled qubit, which is not present in the stated goal. $\endgroup$
    – Pablo
    Commented Oct 4, 2023 at 11:37
  • $\begingroup$ @Pablo Very true. Guess I did not read the question properly back then. Since I cannot delete an accepted answer, I guess I'll fix it when I find the time. Thanks for noting this! $\endgroup$ Commented Oct 4, 2023 at 17:16

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