Composing Gates

Quantum gates are linear maps, and like all linear maps they can be chained: the output of one gate becomes the input of the next. A sequence of gates is called a circuit, and reading it left to right tells you the order in which the operations are applied to the state.

Gate sequences as matrix products

Suppose you apply gate $A$ and then gate $B$ to a state $|\psi\rangle$ . The result is

$B\bigl(A|\psi\rangle\bigr) = (BA)|\psi\rangle.$

Notice the reversal: in matrix notation the rightmost factor is applied first. When you write code, however, you list the operations in the order they are applied — first $A$ , then $B$ — which matches the way you read a timeline.

A worked example: X then Z

Starting from $|0\rangle$ , apply $X$ and then $Z$ :

$X|0\rangle = |1\rangle$ — the NOT gate flips the bit.
$Z|1\rangle = -|1\rangle$ — Z multiplies the $|1\rangle$ amplitude by $-1$ .

So the final state is $-|1\rangle$ . The overall operation can be written as a single matrix product:

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

You can verify: $(ZX)|0\rangle = \binom{0}{-1} = -|1\rangle$ , matching the step-by-step calculation.

Order matters

Swapping the order to $Z$ first and then $X$ gives a different state:

|0\rangle \xrightarrow{Z} Z|0\rangle = |0\rangle \xrightarrow{X} X|0\rangle = |1\rangle.

Step by step:

$Z|0\rangle = |0\rangle$ — Z leaves $|0\rangle$ unchanged (it only introduces a sign on $|1\rangle$ ).
$X|0\rangle = |1\rangle$ .

The final state is $|1\rangle$ , which differs from $-|1\rangle$ only by a global phase of $-1$ . Because $X$ and $Z$ anti-commute ( $XZ = -ZX$ ), every pair of states produced by the two orderings is related by this global phase of $-1$ for any input — so the two orderings are never measurably distinguishable on their own. In contrast, gate pairs that neither commute nor anti-commute (such as $T$ and $X$ ) can produce states that differ by a relative phase, which does affect measurement statistics in a larger circuit.

Try it

Apply $X$ and then $Z$ to qubit 0, starting from $|0\rangle$ . The grader checks the full state vector — the sign of the amplitude matters here.

Run your code to see the quantum state.

After running, you should see that all the probability sits on $|1\rangle$ (as expected — the qubit is definitely $1$ ) but the amplitude is $-1$ , not $+1$ . That minus sign is the imprint of $Z$ .