The Y Gate

The three Pauli gates — $X$, $Y$, and $Z$ — form a natural family. You have already seen $X$ (the bit-flip). Now meet $Y$, the gate that flips both the amplitude values and introduces an imaginary factor.

What Y does

Acting on the computational basis states, the Pauli-Y gate gives

Its matrix is

The imaginary unit $i$ in the off-diagonal entries is what sets $Y$ apart from $X$. You can verify the first rule directly: multiply $Y$ by the column vector $\binom{1}{0}$ and you get $\binom{0}{i}$, which represents the state $i|1\rangle$.

Relating Y to X and Z

The Pauli matrices satisfy a compact product identity:

To check it, multiply the matrices step by step. First compute

then multiply by $i$:

On the Bloch sphere, $Y$ is a 180° rotation about the $y$-axis. By comparison, $X$ rotates about $x$ and $Z$ rotates about $z$. All three Pauli gates are their own inverses: $Y^2 = I$. You can confirm this by direct multiplication, or by noting that $Y$ is both unitary ($Y^\dagger Y = I$, so $Y^{-1} = Y^\dagger$) and Hermitian ($Y^\dagger = Y$); combining these gives $Y^{-1} = Y^\dagger = Y$, hence $Y \cdot Y = I$.

Try it

Apply $Y$ to a qubit starting in $|0\rangle$. The simulator will compare the full state vector, so a do-nothing circuit will not pass.

After running, the amplitude on $|1\rangle$ should be $i$ and the amplitude on $|0\rangle$ should be $0$. The measurement probability is entirely on outcome 1, just as with $X|0\rangle$ — but the amplitude carries an extra factor of $i$.

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