Basis Ordering and Indexing

A two-qubit state vector has four entries, one per basis state. To read or set the right entry you need to know which index corresponds to which label.

From labels to indices

Each computational basis state is a tensor product $|q_0\rangle \otimes |q_1\rangle$, written compactly as $|q_0 q_1\rangle$. The bits $q_0$ and $q_1$ each take the value 0 or 1.

In qbit-learning the convention is: qubit 0 is the most significant bit (MSB). So the index is computed exactly like reading a two-digit binary number with $q_0$ in the leading position:

Applying that formula to every two-qubit basis state gives the full table:

| Basis label | $q_0$ | $q_1$ | Index | |-------------|-------|-------|-------| | $|00\rangle$ | 0 | 0 | 0 | | $|01\rangle$ | 0 | 1 | 1 | | $|10\rangle$ | 1 | 0 | 2 | | $|11\rangle$ | 1 | 1 | 3 |

For example, $|10\rangle$ has $q_0 = 1$ and $q_1 = 0$, giving index $2 \cdot 1 + 0 = 2$.

Why MSB-first?

The choice to place qubit 0 at the most significant position is common in quantum computing toolkits and makes the index formula clean: the index of $|q_0 q_1 \cdots q_{n-1}\rangle$ in an $n$-qubit system is the ordinary binary integer whose bits are $q_0 q_1 \cdots q_{n-1}$.

Generalising to $n$ qubits

For $n$ qubits the state vector has $2^n$ entries. The index is still the binary integer formed by reading the qubit values from qubit 0 (MSB) to qubit $n-1$ (LSB):

Three qubits give eight basis states ($|000\rangle$ through $|111\rangle$ at indices 0–7), four qubits give sixteen, and so on.

Try it

Use the formula above to determine the state-vector index for the two-qubit basis state $|10\rangle$ and return that integer.

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