I was going through the qiskit textbook and in this chapter I came across a statement under the topic "Kickback with the T-gate" related to the Controlled-Z gate that

the controlled-Z rotation gates are symmetrical in fashion (two controls instead of a control and a target). There is no clear control or target qubit for all cases.

What does it imply exactly?

enter image description here

  • $\begingroup$ A releted answer. $\endgroup$ Aug 31, 2020 at 17:39
  • 2
    $\begingroup$ CZ(control=i,target=j)=CZ(control=j,target=i)...maybe a bit surprising at first...so you can pick either bit for control/target $\endgroup$
    – unknown
    Aug 31, 2020 at 17:50

2 Answers 2


For the mathematical explanation, check here: Why is the action of controlled-Z unaltered by exchanging target control qubits?

Maybe it would help you to see CZ in a different (symmetric) notation, like its current representation in Qiskit:

from qiskit import *
circuit = QuantumCircuit(2)

a CZ gate represented as a symmetric gate


Exchanging the two qubits swaps the basis states $|01\rangle \leftrightarrow |10\rangle$, but keeps $|00\rangle$ and $|11\rangle$ unchanged. Suppose you have a gate whose action on the computational basis is

$$ |00\rangle \to a|00\rangle \\ |01\rangle \to b|01\rangle \\ |10\rangle \to c|10\rangle \\ |11\rangle \to d|11\rangle. $$

If you swap the inputs you obtain the gate whose action on the computational basis is

$$ |00\rangle \to a|00\rangle \\ |01\rangle \to \color{red}{c}|01\rangle \\ |10\rangle \to \color{red}{b}|10\rangle \\ |11\rangle \to d|11\rangle. $$

Thus, all such gates are unchanged under exchange of qubits if and only if $b=c$.

Controlled-$Z$ is just such a gate with $a=b=c=1$ and $d=-1$. In fact, all controlled rotations around the $Z$ axis such as the controlled-$S$ gate have $b=c=1$ and are therefore symmetric under qubit exchange and so we do not generally label their inputs as control and target.


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.