Changing Basis

The computational basis $\{|0\rangle, |1\rangle\}$ is not the only valid basis for a qubit. Any pair of orthonormal states spans the same space, so a state that looks simple in one basis can look like an equal superposition in another. Choosing the right basis often makes a calculation — or an algorithm — far more transparent.

The X basis (Hadamard basis)

A second natural basis is built from the two equal-superposition states introduced in the previous lesson:

|+\rangle = \frac{1}{\sqrt{2}}\big(|0\rangle + |1\rangle\big), \qquad |-\rangle = \frac{1}{\sqrt{2}}\big(|0\rangle - |1\rangle\big).

These are orthonormal — you can check directly that $\langle +|-\rangle = 0$ and $\langle +|+\rangle = \langle -|-\rangle = 1$ — so they form a valid basis called the X basis (or Hadamard basis). The Pauli- $X$ operator has $|+\rangle$ and $|-\rangle$ as its eigenvectors with eigenvalues $+1$ and $-1$ , respectively, which is why the name sticks.

Rewriting states across bases

Every state $|\psi\rangle$ can be expanded in either basis. To find the X-basis coefficients of a computational-basis state, express $|0\rangle$ and $|1\rangle$ in terms of $|+\rangle$ and $|-\rangle$ . Adding and subtracting the definitions above gives

|0\rangle = \frac{1}{\sqrt{2}}\big(|+\rangle + |-\rangle\big), \qquad |1\rangle = \frac{1}{\sqrt{2}}\big(|+\rangle - |-\rangle\big).

|\psi\rangle = \frac{\alpha + \beta}{\sqrt{2}}|+\rangle + \frac{\alpha - \beta}{\sqrt{2}}|-\rangle.

Preparing an X-basis state in a circuit

To prepare $|+\rangle$ or $|-\rangle$ starting from $|0\rangle$ , you use $H$ (and an optional $X$ flip beforehand). Concretely:

H|0\rangle = |+\rangle, \qquad H|1\rangle = H\bigl(X|0\rangle\bigr) = |-\rangle.

The state $|-\rangle$ lives entirely in the X basis: it is the $-1$ eigenstate of $X$ , so measuring in the X basis would give a definite outcome. Measuring in the Z basis instead gives outcome 0 or 1 with equal probability $\frac{1}{2}$ , because the Z-basis amplitudes both have magnitude $\frac{1}{\sqrt{2}}$ .

Try it

Starting from $|0\rangle$ , prepare the X-basis state $|-\rangle$ . You will need two gates. The grader checks the full state vector, so leaving the qubit in $|0\rangle$ will not pass.

Run your code to see the quantum state.

After running, the State Vector panel should show amplitudes $\approx +0.707$ on $|0\rangle$ and $\approx -0.707$ on $|1\rangle$ — that is $\frac{1}{\sqrt{2}}|0\rangle - \frac{1}{\sqrt{2}}|1\rangle = |-\rangle$ .