Past exam of the mathematics course of the University of Cambridge / 2006 / iii / Paper 60 / 1

Each operator has trace one, so its characteristic polynomial is ${\lambda }^2-\lambda +\det \rho$. The first and last are diagonal; for the others, the symmetric and antisymmetric vectors diagonalize the matrix. Their eigenvalues areFor ${\rho }_1$, $(1,-1{)}^T/\sqrt{2}$ has eigenvalue one and $(1,1{)}^T/\sqrt{2}$ has eigenvalue zero. For ${\rho }_2$, those eigenvalues are $1/4$ and $3/4$, respectively. All four density operators are positive semidefinite, as these spectra confirm.

A density operator represents a pure state exactly when it has rank one, equivalently ${\rho }^2=\rho$ or $\mathrm{Tr}{\rho }^2=1$. The spectra just found giveTheir 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.

Two finite-dimensional Hermitian matrices are related by unitary conjugation if and only if they have the same multiset of eigenvalues. Necessity follows because conjugation preserves the characteristic polynomial; sufficiency follows by mapping orthonormal eigenbases in the spectral theorem. Consequentlyand no member of the first pair is equivalent to a member of the second. For an explicit example, the Hadamard gate $H$ gives $H{\rho }_0{H}^{†}={\rho }_2$ and $H{\rho }_3{H}^{†}={\rho }_1$. Multiplying $H$ by a scalar phase does not change either conjugation.

The Von Neumann equation integrates to $\rho (t)=U(t)\rho (0)U(t{)}^{†}$, so the spectrum and purity of a density operator are invariant under any allowed open-loop control. These are the kinematic restrictions.

There is no additional orbit restriction here: the generators $i{\sigma }_z$ and $i{\sigma }_x$, together with their matrix commutator, span the special unitary Lie algebra:For freely chosen durations and piecewise continuous real control values, this gives density operator controllability. The connected reachable group is $SU(2)$; its action by conjugation is the same as that of $U(2)$, because a global phase cancels. Thus exactly the pairs ${\rho }_0\leftrightarrow {\rho }_2$ and ${\rho }_1\leftrightarrow {\rho }_3$ can be interconverted. No coherent control can turn either mixed state into either pure state, even though every orientation within each spectral orbit is reachable.