1
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ 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 Jan 16 '18 at 16:37
0
$\begingroup$

Yes.

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.
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.