= Qubit state
{title2=$\rho=\tfrac12(I+\mathbf r\cdot\boldsymbol\sigma)$}
A <qubit state> specifies measurement <probabilities>, rather than a pre-existing answer to every possible measurement. For a <pure state>, normalized vectors differing by a <global phase> represent the same physical state:
$$
|\psi\rangle\sim e^{i\gamma}|\psi\rangle,\qquad \gamma\in\mathbb R.
$$
Equivalently, the <pure state> is a ray: the one-dimensional complex <linear subspace> spanned by any nonzero representative. Dividing a nonzero vector by its <norm> fixes its length, but leaves the <global phase> free. The <density operator> is unchanged because $(e^{i\gamma}|\psi\rangle)(e^{i\gamma}|\psi\rangle)^\dagger=|\psi\rangle\langle\psi|$. In contrast, multiplying only one nonzero component by $e^{i\phi}$ changes a relative <quantum phase> and can change measurement <probabilities>. <Foundations I: States and Ensembles by John Preskill>[Preskill, Physics 229 notes, sections 2.1 and 2.2.1, state rays and qubit parametrization]{ref}.
After choosing a <global phase>, every <pure state> of a <qubit> can be written
$$
|\psi(\theta,\phi)\rangle=\cos(\theta/2)|0\rangle+e^{i\phi}\sin(\theta/2)|1\rangle,
\qquad 0\leq\theta\leq\pi,\quad 0\leq\phi<2\pi.
$$
At $\theta=0$ or $\pi$, the parameter $\phi$ does not distinguish physical states: only one component survives. Elsewhere it is the relative <quantum phase>. The corresponding unit <Bloch vector> is $\mathbf r=(\sin\theta\cos\phi,\sin\theta\sin\phi,\cos\theta)$ on the existing <Bloch sphere>. Its angles describe a space of states; for a <photon polarization> <qubit> they are not the photon's position or direction of travel. <Foundations I: States and Ensembles by John Preskill>[Preskill, Physics 229 notes, section 2.3.2, equations (2.62)–(2.68)]{ref}.
The <Born rule> in the <computational basis> gives $P(0)=\cos^2(\theta/2)$ and $P(1)=\sin^2(\theta/2)$. These <probabilities> discard $\phi$. In the <Hadamard basis> $|\pm\rangle=(|0\rangle\pm|1\rangle)/\sqrt2$, adding or subtracting the <probability amplitudes> instead gives
$$
P_X(\pm)=\frac12|\alpha\pm\beta|^2
=\frac12\pm\operatorname{Re}(\alpha^*\beta)
=\frac12\bigl(1\pm\sin\theta\cos\phi\bigr).
$$
This last equality follows by substituting the angular representative. In the basis $|\pm i\rangle=(|0\rangle\pm i|1\rangle)/\sqrt2$, the same calculation gives
$$
P_Y(\pm)=\frac12|\alpha\mp i\beta|^2
=\frac12\pm\operatorname{Im}(\alpha^*\beta)
=\frac12\bigl(1\pm\sin\theta\sin\phi\bigr).
$$
Thus both the real and imaginary parts of <quantum coherence> can be observed, using different measurement settings. For example, $(|0\rangle+|1\rangle)/\sqrt2$ and $(|0\rangle-|1\rangle)/\sqrt2$ both give $P(0)=P(1)=1/2$, but the first gives $P_X(+)=1$ and the second $P_X(+)=0$. A <global phase> could never cause this difference.
\Image[/qubit-states-phase-interference.png]
{title=Identical computational-basis probabilities can hide different relative phases and coherence}
{description=Exact model probabilities for $(|0\rangle+e^{i\phi}|1\rangle)/\sqrt2$, compared with the incoherent mixture $I/2$. Each curve requires repeated preparations measured in the indicated basis; these are not experimental data.}
For a general <qubit state>, write its <density matrix> in the <computational basis> as
$$
\rho=\begin{pmatrix}p&c\\c^*&1-p\end{pmatrix},\qquad
p\in\mathbb R,\quad c\in\mathbb C,\quad
0\leq p\leq1,\quad |c|^2\leq p(1-p).
$$
These are exactly the <Hermitian>, unit-<trace>, <positive semidefinite> constraints. Positivity means every <eigenvalue> is nonnegative, not that every matrix entry is nonnegative. The general definition is given in <density matrix>. https://quantum.cloud.ibm.com/learning/en/courses/general-formulation-of-quantum-information/density-matrices/density-matrix-basics[Watrous, IBM Quantum Learning, Density matrix basics, Definition and Connection to quantum state vectors]{ref}.
For this two-by-two <matrix>, the explicit constraint follows from its <determinant>, $\det\rho=p(1-p)-|c|^2$: with <trace> one, a nonnegative <determinant> and real <eigenvalues> rule out a negative <eigenvalue>. Its <eigenvalues> are
$$
\lambda_\pm=\frac12\left(1\pm\sqrt{(2p-1)^2+4|c|^2}\right).
$$
The <pure states> have $|c|^2=p(1-p)$ and <eigenvalues> $1,0$; the <mixed states> have strict inequality and two positive <eigenvalues>. For a normalized representative, $p=|\alpha|^2$ and $c=\alpha\beta^*$. A statistical preparation $\rho=\sum_j w_j|\psi_j\rangle\langle\psi_j|$, with $w_j\geq0$ and $\sum_jw_j=1$, averages <probabilities>, rather than adding <probability amplitudes>. Such a preparation is mixed if the positive-weight vectors do not all represent the same ray. A <reduced density matrix> of an entangled pair can also be mixed even when the pair has a <pure state>. https://quantum.cloud.ibm.com/learning/en/courses/general-formulation-of-quantum-information/density-matrices/density-matrix-basics[Watrous, IBM Quantum Learning, Density matrix basics, Interpretation of density matrices]{ref}.
For a concrete check, the <pure state> $(\sqrt3|0\rangle+|1\rangle)/2$ and the <mixed state> obtained by preparing $|0\rangle$ with <probability> $3/4$ and $|1\rangle$ with <probability> $1/4$ have
$$
\rho_{\mathrm{pure}}=\begin{pmatrix}3/4&\sqrt3/4\\\sqrt3/4&1/4\end{pmatrix},\qquad
\rho_{\mathrm{mix}}=\begin{pmatrix}3/4&0\\0&1/4\end{pmatrix}.
$$
Both have <trace> one and the same <computational-basis measurement> <probabilities>. Their <eigenvalues> are respectively $(1,0)$ and $(3/4,1/4)$; the respective values of their <purity of a density operator> $\operatorname{Tr}(\rho^2)$ are $1$ and $5/8$. Their <Hadamard basis> plus <probabilities> are respectively $1/2+\sqrt3/4\approx0.9330$ and $1/2$. By contrast, setting $p=3/4,c=1/2$ would give <determinant> $-1/16$ and <eigenvalue> $(2-\sqrt5)/4<0$: normalized diagonal entries alone do not make a valid <qubit state>.
The same constraints become especially beautiful in the <Bloch ball> representation. With the <Pauli matrices> $\boldsymbol\sigma=(X,Y,Z)$ and the <identity operator> $I$,
$$
\rho=\tfrac12(I+r_xX+r_yY+r_zZ),\qquad
\mathbf r=(2\operatorname{Re}c,-2\operatorname{Im}c,2p-1),\qquad |\mathbf r|\leq1.
$$
The <eigenvalues> are $(1\pm|\mathbf r|)/2$, so the <Bloch sphere> is precisely the <pure state> boundary, and $I/2$ is the maximally mixed centre. For each axis $j$, the <Born rule> gives $P_j(\pm)=(1\pm r_j)/2$. This unifies a geometric picture, matrix positivity, and experimental frequencies in one description. For the equatorial family in the phase diagram, $p=1/2,c=e^{-i\phi}/2$ and $\mathbf r=(\cos\phi,\sin\phi,0)$; replacing it by $I/2$ erases the transverse components and every fringe. See <quantum state tomography> for how independent ensembles measured along the three axes estimate these components. <Foundations I: States and Ensembles by John Preskill>[Preskill, Physics 229 notes, section 2.3.2, equations (2.62)–(2.70)]{ref}.
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