The Z Gate

The Pauli-Z gate is the third of the three Pauli single-qubit gates. Unlike X, which swaps amplitudes, Z leaves $|0\rangle$ untouched and multiplies the $|1\rangle$ amplitude by $-1$:

$Z|0\rangle = |0\rangle, \qquad Z|1\rangle = -|1\rangle.$

Its matrix is

$Z = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}.$

Phase, not amplitude

When the qubit is in a basis state $|0\rangle$ or $|1\rangle$, Z has no effect you can detect by measuring: the probability of each outcome is $|\text{amplitude}|^2$, and multiplying by $-1$ does not change the absolute value. This minus sign is called a relative phase.

The effect of Z only becomes observable when the qubit is in a superposition. Apply Z to the equal superposition $|+\rangle = \tfrac{1}{\sqrt{2}}\big(|0\rangle + |1\rangle\big)$:

The state $|{-}\rangle$ has the same measurement probabilities as $|+\rangle$ — both give 0 or 1 with probability $\tfrac{1}{2}$ — but the relative phase between the two amplitudes has flipped. That difference matters enormously in interference-based algorithms: a Hadamard applied after Z will rotate $|{-}\rangle$ back to $|1\rangle$, not $|0\rangle$, demonstrating that the phase is physically real.

On the Bloch sphere, Z is a rotation by $\pi$ around the $z$-axis: it leaves the north and south poles fixed while flipping the equatorial states $|+\rangle \leftrightarrow |{-}\rangle$ and $|{+i}\rangle \leftrightarrow |{-i}\rangle$.

Try it

Prepare $|+\rangle$ with a Hadamard, then apply Z to produce $|{-}\rangle$. The grader checks the full state vector, so the sign of the $|1\rangle$ amplitude matters.

After running, you should see the amplitude on $|0\rangle$ is $+\tfrac{1}{\sqrt{2}} \approx 0.707$ and the amplitude on $|1\rangle$ is $-\tfrac{1}{\sqrt{2}} \approx -0.707$. Both probabilities are still $\tfrac{1}{2}$, confirming that Z changes phase without changing measurement outcomes — unless you look in a basis that is sensitive to the phase.

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