= Two-state ensembles of a maximally mixed qubit
{title2=$I/2=p|u\rangle\langle u|+(1-p)|v\rangle\langle v|\iff p=1/2,\ \langle u|v\rangle=0$}
For normalized <pure states> $|u\rangle,|v\rangle$, a two-member <quantum state ensemble> gives the <maximally mixed state> of a <qubit> exactly when its members are <orthogonal> and equally probable. To prove necessity, take <expectation values> of its <density matrix> in the two members: $1/2=p+(1-p)|\langle u|v\rangle|^2\geq p$ and $1/2=1-p+p|\langle u|v\rangle|^2\geq1-p$. Hence $p=1/2$ and the overlap vanishes. Sufficiency is the <resolution of the identity> in an <orthonormal basis>.
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