Past exam of the introduction to quantum information course of the University of Oxford / 2015 / 2

For any normalized vector $|u⟩$, the Born rule assigns the probability $⟨u|\rho |u⟩$ to the corresponding rank-one orthogonal projection. This must be nonnegative for every $u$, which is exactly positive semidefiniteness. For any complete orthonormal basis $\{|j⟩\}$, the outcome probabilities must sum to one:Thus a Hermitian matrix represents a physical density matrix only ifThese conditions are also sufficient in finite dimension. The spectral theorem writes $\rho ={\sum }_j{p}_j|u_j⟩⟨u_j|$ with $p_j\ge 0$ and ${\sum }_j{p}_j=1$, describing a quantum state ensemble of pure states with valid mixing probabilities. A Hermitian matrix alone could have negative eigenvalues, hence negative measurement probabilities.

Expand the given pure state in the ordered computational basis $00,01,10,11$:The coefficients have squared magnitudes $1/3,1/6,1/3,1/6$, summing to one. The density matrix is the outer product $v{v}^{†}$, not the diagonal matrix of these four probabilities: its off-diagonal entries retain quantum coherence. ThereforeThe outer product is a Hermitian matrix and is positive semidefinite: $u^{†}\rho u=|v^{†}u{|}^2\ge 0$. Its trace is one, and ${\rho }^2=\rho$, as required for a pure state.

Let $a_{ij}$ be the coefficient of $|ij⟩$. The partial trace contracts the repeated index of the discarded qubit:For the first qubit, each diagonal entry is $1/3+1/6=1/2$ and the upper off-diagonal entry is $1/3-1/6=1/6$. For the second qubit, the diagonal entries are $2/3$ and $1/3$, while the upper off-diagonal entry is $-1/\sqrt{18}+1/\sqrt{18}=0$. Thus the two reduced density matrices, each in computational basis order $0,1$, areBoth have eigenvalues $2/3,1/3$ and trace one, so both are mixed states even though the joint quantum state is pure. Their different eigenvectors are visible in the useful rewriteAbsorbing the minus sign into one local basis vector makes this a Schmidt decomposition with nonnegative coefficients. It also checks both partial traces and exhibits the entanglement responsible for the mixed local descriptions.

For a Hermitian matrix $A$, the spectral theorem gives $A=U\mathrm{diag}({\lambda }_1,\dots ,{\lambda }_d)U^{†}$, with real eigenvalues and a unitary matrix $U$. HenceThe square root here is the unique positive square root of an operator. Taking the trace, which is unchanged by unitary similarity, provesFor a density matrix, all eigenvalues are nonnegative and sum to one. Its trace norm is thereforeThe absolute values cannot be omitted for a general Hermitian matrix, since positive and negative eigenvalues can cancel in its trace but never in its trace norm.

With no access to the discarded qubit, the two preparations are exactly the reduced density matrices in c, with equal prior probabilities. Their difference isSince $\mathrm{Tr}\Delta =0$, its eigenvalues are $\pm \sqrt{2}/6$. The trace norm and trace distance are consequentlyThe Holevo–Helstrom theorem gives the optimumTo establish attainability, rather than only an upper bound, measure the Hermitian operatorwhere $X,Z$ are Pauli matrices. Since $\Delta =(\sqrt{2}/6)B$, report first qubit for the positive outcome; second qubit for the negative outcome. The corresponding orthogonal projections are $E_{\pm }=(I\pm B)/2$. Their success probability iswhich reaches the bound. Identical eigenvalues of the two density matrices do not make them indistinguishable: their eigenvectors differ, and this quantum measurement detects that difference.