Choose the <qubit> <computational basis> $|0\rangle,|1\rangle$, also called the $Z$ basis, and the <Hadamard basis> $|\pm\rangle=(|0\rangle\pm|1\rangle)/\sqrt2$, also called the $X$ basis. Compare a coherent preparation of $|+\rangle$ with choosing $|0\rangle$ or $|1\rangle$ at random with equal <probabilities> and discarding that label:
$$
\rho_+=\frac12\begin{pmatrix}1&1\\1&1\end{pmatrix},\qquad
\rho_{\mathrm{mix}}=\frac12|0\rangle\langle0|+\frac12|1\rangle\langle1|=\frac12\begin{pmatrix}1&0\\0&1\end{pmatrix}.
$$
These are original ideal preparation models. Their $Z$-basis <probabilities> are both $(1/2,1/2)$, because their <quantum state populations> agree. Their $X$-basis <probabilities> are respectively $(1,0)$ and $(1/2,1/2)$. Indeed $\langle+|\rho_+|+\rangle=1$, whereas $\langle+|I/2|+\rangle=1/2$. The missing off-diagonal <quantum coherence> therefore changes a prediction that the common populations cannot determine.

\Image[/quantum-state-preparation-branches.png]
{title=Coherent preparation and random mixing agree in the Z basis but differ in the X basis}
{description=Original ideal <qubit> preparation and detection branches. The random-choice branches are mutually exclusive trials; forgetting the label averages <probabilities> without recombining <probability amplitudes>. Each measurement branch uses a fresh preparation. The displayed <Born rule> probabilities assume ideal detection and are not measured counts or the apparatus of the experiment below.}

For a relative <quantum phase> $\phi$, the coherent family instead has
$$
|\psi_\phi\rangle=\frac{|0\rangle+e^{i\phi}|1\rangle}{\sqrt2},\qquad
\rho_\phi=\frac12\begin{pmatrix}1&e^{-i\phi}\\e^{i\phi}&1\end{pmatrix},\qquad
P_X(\pm)=\frac{1\pm\cos\phi}{2}.
$$
The last formula follows by adding or subtracting the two <probability amplitudes> before taking the squared <modulus>. At $\phi=\pi$, the state is $|-\rangle$ and the $X$ outcomes switch to $(0,1)$, while the $Z$ outcomes remain $(1/2,1/2)$. In contrast, $-|+\rangle$ changes both components by the same <global phase> and retains the $X$ outcomes $(1,0)$. In <photon polarization>, the identification $|0\rangle=|H\rangle$, $|1\rangle=|V\rangle$ makes $|+\rangle$ and $|-\rangle$ the two orthogonal diagonal <linear polarizations>; these labels describe measurement bases, not spatial axes for photon <spin>.

\Image[/quantum-state-global-relative-phase.png]
{title=A circle of global-phase representatives gives one state, while a relative sign changes X outcomes}
{description=Original ideal model. The top circle is the <complex number> $e^{i\gamma}$ multiplying the entire vector; it is not a <Bloch sphere> or a path through distinct physical states. All its representatives map to the same <density operator>. Below, changing the relative <quantum phase> from zero to $\pi$ gives different <density operators> and complementary $X$-basis <probabilities>. No experimental observations are plotted.}

Even perfect statistics in one <basis> cannot certify a preparation. The above $Z$ data fit both <density matrices>. Conversely, distinct preparation histories can yield the same <quantum state>:
$$
\frac12|0\rangle\langle0|+\frac12|1\rangle\langle1|
=\frac12|+\rangle\langle+|+\frac12|-\rangle\langle-|=\frac I2.
$$
The off-diagonal terms cancel in the right-hand ensemble. Every measurement on the <qubit> alone then has identical <probabilities> for those two histories. A retained classical label would allow conditioning on different members of the <quantum state ensemble>; it is extra information beyond the unlabelled <density operator>.


 Coherent preparation and incoherent mixing (source code)