Big-Endian vs Little-Endian Qubits

Every quantum toolkit stores the state vector as an ordered list of $2^n$ amplitudes. The question is: which basis state does index $k$ refer to? The answer depends on whether the framework uses a big-endian or little-endian qubit convention — and mixing them up is one of the most common sources of subtle bugs in quantum programming.

The two conventions

Consider a 2-qubit system with qubits labeled $q_0$ and $q_1$ . There are exactly four computational-basis states:

|00\rangle,\quad |01\rangle,\quad |10\rangle,\quad |11\rangle.

To turn a bit-string $q_0 q_1$ into an integer index $k$ we need an ordering rule.

Big-endian (MSB first): qubit 0 is the most significant bit (MSB). The index is

k = q_0 \cdot 2^{n-1} + q_1 \cdot 2^{n-2} + \cdots + q_{n-1} \cdot 2^{0}.

State $|10\rangle$ (qubit 0 = 1, qubit 1 = 0) maps to index $1 \cdot 2 + 0 \cdot 1 = 2$ .

Little-endian (LSB first): qubit 0 is the least significant bit (LSB). The index is

k = q_0 \cdot 2^{0} + q_1 \cdot 2^{1} + \cdots + q_{n-1} \cdot 2^{n-1}.

State $|10\rangle$ (qubit 0 = 1, qubit 1 = 0) now maps to index $1 \cdot 1 + 0 \cdot 2 = 1$ .

The same physical state, a different index. This platform uses big-endian: qubit 0 is always the most significant bit.

Encoding an integer

To encode a non-negative integer $v$ into an $n$ -qubit big-endian register, write $v$ in binary with exactly $n$ bits (padding with leading zeros if needed), then flip qubit $i$ with an $X$ gate whenever bit $n - 1 - i$ equals 1.

For $n = 3$ and $v = 5$ (binary $101$ ):

| Bit position | Value | Qubit | Gate | |---|---|---|---| | bit 2 (MSB) | 1 | qubit 0 | X | | bit 1 | 0 | qubit 1 | — | | bit 0 (LSB) | 1 | qubit 2 | X |

Applying $X$ to qubits 0 and 2 produces $|101\rangle$ , index 5 in big-endian order.

Why it matters

A controlled gate applies to a specific (control, target) qubit pair. If you believe qubit 0 is the MSB but your tool treats it as the LSB, you will write the control and target on the wrong physical lines, computing a completely different unitary. The state vector will look plausible — it will be normalized — but the algorithm will silently produce wrong answers. Keeping the convention explicit is therefore not a stylistic choice: it is correctness.

Try it

Encode the integer 6 (binary $110$ ) in a 3-qubit big-endian register. Remember: qubit 0 is the MSB. The grader checks the full state vector, so you must target the right qubits.

Run your code to see the quantum state.

After running, look at the state-vector tab: all amplitude should sit at index 6, confirming that $|110\rangle$ is the only populated basis state.