Measuring in the X Basis

Every measurement we have seen so far has been a computational-basis (or $Z$-basis) measurement: the qubit collapses to $|0\rangle$ or $|1\rangle$. But qubits can also be measured in other bases. The $X$ basis uses the eigenstates of the Pauli $X$ operator,

A qubit in $|+\rangle$ gives outcome $0$ in the $X$ basis with certainty, and a qubit in $|−\rangle$ gives outcome $1$ with certainty — exactly the same kind of deterministic readout that $|0\rangle$ and $|1\rangle$ give in the $Z$ basis.

Why rotating first works

Most quantum hardware only provides $Z$-basis measurement. The standard trick is to rotate into the $Z$ basis before measuring: apply a Hadamard gate, then measure. Because $H$ swaps the two bases,

a $Z$-basis measurement on the rotated state returns the correct $X$-basis outcome. Schematically:

This is the general strategy: to measure in any basis, first apply the unitary that rotates that basis onto the computational basis, then perform a standard measurement.

Example: reading out $|−\rangle$

Starting from $|0\rangle$, we can prepare $|−\rangle$ in two steps:

1. Apply $X$ to flip to $|1\rangle$.
2. Apply $H$ to get $H|1\rangle = |−\rangle$.

To read out the $X$-basis value, we then apply another $H$ (the rotation step) and measure:

The double application of $H$ around $|−\rangle$ is not a coincidence: $H$ is its own inverse ($H^2 = I$), so the second $H$ exactly undoes the first while simultaneously performing the basis-change needed for measurement.

Try it

Build the circuit that prepares $|−\rangle$ and measures it in the $X$ basis. The grader checks the probability distribution: outcome 1 should appear with probability 1 and outcome 0 with probability 0.

After running, the probability bar for 1 should be full and the bar for 0 empty — the signature of a deterministic $X$-basis measurement on $|−\rangle$.

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