controlled-Z rotation gates in symmetrical fashion

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?

• A releted answer. Aug 31 '20 at 17:39
• 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 Aug 31 '20 at 17:50

2 Answers

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')


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.