The Controlled-Z Gate

The controlled-Z gate (CZ) multiplies the amplitude of $|11\rangle$ by $-1$ and leaves all other basis states unchanged:

In matrix form, with the two-qubit computational basis ordered $|00\rangle, |01\rangle, |10\rangle, |11\rangle$:

Symmetry: no control, no target

Look at the matrix — it is diagonal. Because a diagonal matrix is symmetric, swapping the role of the two qubits gives exactly the same gate. There is no physical difference between calling qubit 0 the "control" and qubit 1 the "target": CZ($q_0$, $q_1$) = CZ($q_1$, $q_0$). This symmetry sets CZ apart from CNOT, which has a definite control and a definite target.

Formally, CZ is the tensor product of $Z$ conditioned on the other qubit being $|1\rangle$:

but because $Z$ also leaves $|0\rangle$ unchanged and flips the sign of $|1\rangle$, you could equally write $I \otimes |0\rangle\langle 0| + Z \otimes |1\rangle\langle 1|$ — the same matrix.

Phase kicks and relative phases

When CZ acts on a product state $|{+}\rangle|1\rangle$, the phase it introduces is relative, not global, and therefore physically observable:

The $|01\rangle$ amplitude is untouched (one qubit is $|0\rangle$ so CZ does nothing), while the $|11\rangle$ amplitude picks up the $-1$ phase. The resulting state is a superposition with a minus sign — a different quantum state from the input that a later Hadamard or interference experiment could distinguish.

Try it

Prepare qubit 0 in $|{+}\rangle$ and qubit 1 in $|1\rangle$, then apply CZ. Observe how the sign of the $|11\rangle$ component flips while the $|01\rangle$ component is unchanged.

After running, check the amplitude of $|11\rangle$ in the statevector panel: it should be $-\tfrac{1}{\sqrt{2}} \approx -0.707$, confirming the phase kick.

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