Two-state ensembles of a maximally mixed qubit ($I/2=p|u⟩⟨u|+(1-p)|v⟩⟨v|⟺p=1/2,⟨u|v⟩=0$)

For normalized pure states $|u⟩,|v⟩$, 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: $1/2=p+(1-p)|⟨u|v⟩{|}^2\ge p$ and $1/2=1-p+p|⟨u|v⟩{|}^2\ge 1-p$. Hence $p=1/2$ and the overlap vanishes. Sufficiency is the resolution of the identity in an orthonormal basis.