Inverse Gates

Every quantum gate must be reversible — that is guaranteed by the requirement that gates are unitary ($U^\dagger U = I$). For a gate $U$, its inverse is $U^{-1} = U^\dagger$ (the conjugate transpose). For the Pauli gates this turns out to be especially simple.

The Pauli matrices are Hermitian and unitary

The three Pauli gates are

Each matrix is equal to its own conjugate transpose: $X^\dagger = X$, $Y^\dagger = Y$, $Z^\dagger = Z$. A matrix that equals its own conjugate transpose is called Hermitian. When a matrix is both Hermitian and unitary, the two conditions $U^\dagger = U$ and $U^\dagger U = I$ combine to give

You can verify this directly for each Pauli by squaring the matrix. For $X$:

The same calculation for $Y$ and $Z$ gives the same result. A matrix that squares to the identity satisfies $U^{-1} = U$, so:

Each Pauli gate is its own inverse.

Concretely:

• $X$ applied twice returns any state to itself: $X|0\rangle = |1\rangle$, then $X|1\rangle = |0\rangle$.
• $Y$ applied twice returns any state exactly to itself: $Y^2 = I$, so no phase of any kind is introduced.
• $Z$ applied twice leaves both amplitudes unchanged: $Z^2|0\rangle = |0\rangle$, $Z^2|1\rangle = (-1)^2|1\rangle = |1\rangle$.

Try it

The circuit below starts with the qubit in $|0\rangle$ and immediately applies $X$ to produce $|1\rangle$. Your task is to undo that flip by applying the inverse of $X$. The grader checks the full state vector, so the qubit must end in $|0\rangle$.

After running: the Bloch vector returns to the north pole, and the amplitude on $|0\rangle$ is exactly $1$. Two $X$ gates cancel completely because $X^2 = I$.

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