= Solution
A <density operator> represents a <pure state> exactly when it has rank one, equivalently $\rho^2=\rho$ or $\operatorname{Tr}\rho^2=1$. The spectra just found give
$$
\boxed{\rho_1=|\psi_1\rangle\langle\psi_1|,\quad |\psi_1\rangle=\frac{|0\rangle-|1\rangle}{\sqrt2};\qquad
\rho_3=|\psi_3\rangle\langle\psi_3|,\quad |\psi_3\rangle=|1\rangle.}
$$
Their <wavefunctions> are defined up to a <global phase>. The other two are <mixed states>, with purity $(3/4)^2+(1/4)^2=5/8$, and have no single-wavefunction representation as rank-one projectors.
Back to article page
