Controlled Phase Gates

The controlled-phase gate $CP(\lambda)$ is a two-qubit gate that leaves every basis state unchanged except $|11\rangle$, which it multiplies by the complex number $e^{i\lambda}$:

In matrix form (rows and columns ordered $|00\rangle, |01\rangle, |10\rangle, |11\rangle$):

You can verify it is unitary: the four columns are orthonormal and the determinant has magnitude 1.

Why only $|11\rangle$ acquires a phase

The gate is controlled: it acts on the target only when the control is $|1\rangle$. The single-qubit $P(\lambda)$ gate adds a phase $e^{i\lambda}$ to $|1\rangle$ and leaves $|0\rangle$ alone. $CP(\lambda)$ makes that action conditional on the control being $|1\rangle$, so the phase appears exactly on $|11\rangle$ — the only basis state where both qubits are $|1\rangle$.

Common special cases

Three frequently used controlled-phase angles have their own shorthand names:

| Angle $\lambda$ | Gate name | $e^{i\lambda}$ | |---|---|---| | $\pi/4$ | $CT$ (controlled-$T$) | $e^{i\pi/4}$ | | $\pi/2$ | $CS$ (controlled-$S$) | $i$ | | $\pi$ | $CZ$ (controlled-$Z$) | $-1$ |

For example, $CZ$ is a special case of $CP(\pi)$ and is symmetric: it does not matter which qubit you call "control" and which you call "target," because $|11\rangle$ is the only state it changes.

Try it

Apply a controlled-phase of $\pi/2$ with qubit 0 as the control and qubit 1 as the target. The grader checks the full two-qubit unitary matrix, so doing nothing will not pass.

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