1
$\begingroup$

I am trying to understand this distillation circuit (related to this question), but I am confused by the meaning of the "parity control Z" operations represented with this symbol: enter image description here

The explanation in the simulator is "Include a qubit's Z observable in the column parity control. Gates in the same column only apply if an odd number of parity controls are satisfied".

Does it mean that if the the first $k$ qubits are affected with this operator, and some unitary $U$ is applied on the last $n-k$ qubit, the resulting operation is a controlled unitary operation only applied if the first $k$ qubits have an odd number of $1$ once the state is expanded in the computational basis?

It seems to be the case with simple examples but I don't know if I just found out particular cases.

$\endgroup$

1 Answer 1

1
$\begingroup$

Does it mean that if the the first k qubits are affected with this operator [...] a controlled unitary operation only applied if the first k qubits have an odd number of 1 once the state is expanded in the computational basis?

That's right. You can think of the parity controls as building up a multi-qubit Pauli operator, and other operations in the same column only apply in the -1 eigenspace of that operator. If all the parity controls are Z basis, this is equivalent to conditioning operations on an odd number of the parity controls being ON (in the computational basis).

$\endgroup$
2
  • 1
    $\begingroup$ Oh ok: so in different terms it corresponds to the unitary is $(1+P)/2 \otimes I + (1-P)/2 \otimes U$, where $P$ is the Pauli operator corresponding to the $X_{\text{par}}$, $Y_{\text{par}}$, $Z_{\text{par}}$ that have been put on the "control" side. $\endgroup$ Aug 29, 2023 at 16:27
  • 1
    $\begingroup$ @MarcoFellous-Asiani Right. $\endgroup$ Aug 29, 2023 at 16:35

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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