Multiple qubits and tensor products

Combine the components

For $\lvert a\rangle=(a_0,a_1)^T$ and $\lvert b\rangle=(b_0,b_1)^T$,

$$\lvert a\rangle\otimes\lvert b\rangle= \begin{pmatrix}a_0b_0\\a_0b_1\\a_1b_0\\a_1b_1\end{pmatrix}.$$

Our basis order is $\lvert00\rangle,\lvert01\rangle,\lvert10\rangle,\lvert11\rangle$. The leftmost qubit, q0, is the most significant bit. Other frameworks may display bitstrings differently; check before comparing results.

A complete example

Take $\lvert a\rangle=\lvert+\rangle$ and $\lvert b\rangle=\lvert1\rangle$. Their joint vector is

$$\frac1{\sqrt2}\begin{pmatrix}0\\1\\0\\1\end{pmatrix} =\frac{\lvert01\rangle+\lvert11\rangle}{\sqrt2}.$$

Measuring gives 01 or 11 equally often. The second qubit is always 1; the first is equally likely to be 0 or 1.

More states, more memory

An n-qubit pure state has $2^n$ complex amplitudes. A general classical statevector simulation therefore needs exponential memory. This describes the cost of storing an arbitrary statevector, not a claim that a quantum device lets you read all those amplitudes in one shot.

Some structured states and circuits admit much more efficient classical simulation. Exponential state-space dimension alone is not evidence of useful quantum advantage.

Practise

In the gate playground, apply H to q0 and X to q1. Verify the expected 01/11 distribution. Then change X to q0 instead: explain the result by tracing the ordering rather than guessing from the bitstring labels.

Reference: IBM Quantum — Multiple systems.