Gates Are Unitary

A quantum gate is more than just a transformation — it must be reversible and norm-preserving. These two properties are captured in a single word: unitary.

What unitarity means

A matrix $U$ is unitary if $U^\dagger U = I$ , where $U^\dagger$ is the conjugate transpose of $U$ . This condition has an immediate physical consequence: if $|\psi\rangle$ is any normalized quantum state, then $U|\psi\rangle$ is also normalized:

\langle \psi | U^\dagger U | \psi \rangle = \langle \psi | I | \psi \rangle = \langle \psi | \psi \rangle = 1.

In other words, the total probability of all outcomes is always exactly 1, no matter which gate you apply.

Verifying the X gate

The Pauli $X$ gate has the matrix

X = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}.

Its conjugate transpose is $X^\dagger = X$ (real and symmetric), so

X^\dagger X = X X = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = I.

Starting from $|0\rangle = \binom{1}{0}$ , applying $X$ gives $X|0\rangle = \binom{0}{1} = |1\rangle$ . The amplitude squared of the single non-zero component is $|1|^2 = 1$ — full probability, nothing lost.

The same is true for the $Y$ and $Z$ gates, and indeed for every gate you will encounter in this course. Unitarity is what ensures that quantum computations are always physically valid.

Try it

The $X$ gate maps $|+\rangle = \tfrac{1}{\sqrt{2}}(|0\rangle + |1\rangle)$ to $\tfrac{1}{\sqrt{2}}(|1\rangle + |0\rangle)$ , which still has amplitudes $\tfrac{1}{\sqrt{2}}$ for both basis states. Compute the sum of the squared amplitudes and return it. The grader checks that the result equals 1, confirming that $X$ preserves the norm.

Run your code to see the quantum state.