Tensor Structure of Operations

When you have a register of $n$ qubits, applying a single-qubit gate $G$ to qubit $k$ means tensoring $G$ with identity on every other qubit. For a three-qubit register the full operator is

Writing out those $8 \times 8$ matrices by hand would be tedious — so quantum programming frameworks handle the tensor product automatically. You simply call the gate and name the qubit.

The $n$-qubit state space

An $n$-qubit register lives in the tensor-product space $(\mathbb{C}^2)^{\otimes n}$, which has dimension $2^n$. The computational basis consists of all $n$-bit strings $|b_{n-1} \cdots b_1 b_0\rangle$. For three qubits the eight basis states are $|000\rangle, |001\rangle, \ldots, |111\rangle$.

The state vector tracks one complex amplitude for each basis state, so a three-qubit register needs $2^3 = 8$ amplitudes.

Applying a gate to one qubit

Suppose the register starts in $|000\rangle$ and you apply $H$ only to qubit 0:

Expanding into the eight-dimensional basis (qubit 0 is most significant in this platform):

Now add an $X$ gate on qubit 2 at the same time (independent gates commute when they act on different qubits):

What the simulator does

Every time you call c.h(0), the simulator constructs the operator $H \otimes I^{\otimes (n-1)}$ and multiplies it into the state vector. It then does the same for c.x(2) with $I^{\otimes 2} \otimes X$. You never write the big matrix — the library builds it from the qubit index and the known $2 \times 2$ gate.

This is the core idea behind all quantum programming frameworks: gate calls become tensor-product operators applied to the full state vector.

Try it

Start from $|000\rangle$ (a three-qubit register). Apply $H$ to qubit 0 and $X$ to qubit 2, leaving qubit 1 untouched. The grader checks the final state vector.

After running, you should see nonzero amplitudes only on $|001\rangle$ and $|101\rangle$, each with magnitude $\tfrac{1}{\sqrt{2}} \approx 0.707$.

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