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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?

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  • $\begingroup$ A releted answer. $\endgroup$ Aug 31 '20 at 17:39
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    $\begingroup$ 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 '20 at 17:50
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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)
circuit.cz(0,1)
circuit.draw('mpl')

a CZ gate represented as a symmetric gate

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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.

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