Order of Operations

In a quantum circuit, gates are applied in time order from left to right — the leftmost gate acts first. This seems obvious, but it has a critical consequence: quantum gates generally do not commute, so swapping two gates changes the answer.

Why order matters

Consider the $X$ gate (bit-flip) and the $H$ gate (Hadamard). Starting from $|0\rangle$:

• Applying $X$ first, then $H$: $H(X|0\rangle) = H|1\rangle = \tfrac{1}{\sqrt{2}}(|0\rangle - |1\rangle) = |-\rangle$.
• Applying $H$ first, then $X$: $X(H|0\rangle) = X|+\rangle = \tfrac{1}{\sqrt{2}}(|1\rangle + |0\rangle) = |+\rangle$.

Both end in an equal superposition, but the relative phase flips sign — they are physically distinguishable states. In matrix terms:

$HX \neq XH$, confirming that the two orderings produce different unitaries.

Reading and writing the order

In mathematical notation, operators are written right to left — $HX$ means $X$ acts first, then $H$. The circuit diagram convention goes the opposite way: gates are drawn left-to-right in the order they fire. The programming API follows the circuit diagram:

c.x(0);   // X fires first
c.h(0);   // H fires second  →  implements the unitary HX

Each method call appends one gate to the end of the timeline. Keep this left-to-right reading in mind whenever you transcribe a sequence of gates from a diagram into code.

Try it

Implement the operation "$X$ then $H$" (i.e. the unitary $HX$) on a single qubit. The grader checks the full unitary matrix, so the order of your gate calls must be exactly right.

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