4
$\begingroup$

I need to use the following matrix gate in a quantum circuit:

$$\text{Sign Flip}=\left[\begin{matrix}0 & -1 \\ -1 & 0\end{matrix}\right]$$

$\text{Sign Flip}$ can be decomposed as (in terms of Pauli-$X$,$Y$,$Z$):

$$\begin{bmatrix} 0 & -1\\ -1 & 0\end{bmatrix} = \begin{bmatrix} 0 & 1\\ 1 & 0\end{bmatrix} \begin{bmatrix} 0 & -i\\ i & 0\end{bmatrix} \begin{bmatrix} 1 & 0\\ 0 & -1\end{bmatrix} \begin{bmatrix} 0 & -i\\ i & 0\end{bmatrix} \begin{bmatrix} 1 & 0\\ 0 & -1\end{bmatrix}$$

Is there any standard shorthand notation for the $\text{Sign Flip}$ gate? I don't really want to replace one simple custom gate by $5$ quantum gates in my circuit.

$\endgroup$
3
  • 1
    $\begingroup$ Is it physically different from Pauli-X gate? $\endgroup$
    – kludg
    Jul 28, 2018 at 9:34
  • $\begingroup$ Looks to me like -X but I must be seeing it totally wrong, what am I missing? $\endgroup$ Jul 28, 2018 at 18:32
  • 2
    $\begingroup$ @user1271772 Norbert and Blue discussed this in comments that have since been deleted. The conclusion was that this gate is physically equivalent to Pauli X. $\endgroup$ Jul 30, 2018 at 12:37

1 Answer 1

6
$\begingroup$

A unitary $U$ and $e^{i\phi}U$, which differs from it by a phase, act exact identically on any quantum state. Thus, they should really be considered the "same" unitary in terms of their action.

You can therefore use $X$ instead of your unitary, which is $-X$. This will have exactly the identical action in any circuit.

(Why is this? There are different ways to see this: Either since $|\psi\rangle$ and $e^{i\phi}|\psi\rangle$ describe the same quantum state, or by working with density operators on which $U$ acts as $\rho\mapsto U\rho U^\dagger$, such that phases cancel. Also, note that this is not true for controlled-unitaries -- but this is an entirely different question.)

$\endgroup$

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.