Past exam of the quantum theory and quantum computers course of the University of Oxford / 2008 / 3

A density operator is a positive self-adjoint trace-class operator with $\boxed{\mathrm{Tr}\rho =1}$. A finite expectation is $\boxed{⟨A{⟩}_{\rho }=\mathrm{Tr}(\rho A)}$. It represents a pure state exactly when $\rho =|v⟩⟨v|$ for a unit vector, equivalently ${\rho }^2=\rho$ or $\mathrm{Tr}{\rho }^2=1$.

For a finite-dimensional spectral decomposition $\rho ={\sum }_j{p}_j|v_j⟩⟨v_j|$, introduce an auxiliary space with orthonormal vectors $f_j$ and use the enlarged space $\mathcal{K}=\mathcal{H}\otimes \mathcal{E}$. DefineThen $\Vert {\Omega }_{\rho }{\Vert }^2=\sum p_j=1$ and $⟨{\Omega }_{\rho },\tilde{A}{\Omega }_{\rho }⟩=\sum p_j⟨v_j,A{v}_j⟩=\mathrm{Tr}(\rho A)$. This purification shows that subsystem mixedness can arise from entanglement with an unobserved environment even when the total state is pure. Subsystem measurements alone cannot distinguish these two descriptions.

A nonzero pure bipartite vector is entangled when it cannot be written as a tensor product of two vectors; equivalently its Schmidt rank exceeds one. For mixed bipartite states, entanglement means absence of a convex decomposition into product density operators.

For the nonorthogonal expansion in the question, the unnormalized reduced operator isIndeed $⟨\Psi ,(A\otimes I)\Psi ⟩={\sum }_{j,k}⟨{\xi }_k,A{\xi }_j⟩⟨{\eta }_k,{\eta }_j⟩=\mathrm{Tr}(A{S}_1)$. Positivity follows because $⟨v,S_{1}v⟩=\Vert {\sum }_j⟨v,{\xi }_j⟩{\eta }_j{\Vert }^2\ge 0$, and $\mathrm{Tr}{S}_1=\Vert \Psi {\Vert }^2=:Z$. Thus $\boxed{{\rho }_1=S_1/Z}$ is a density operator; if $\Psi$ is already normalized, $Z=1$. The Gram factors cannot be dropped merely because the vectors are called bases. For unnormalized $\Psi$, physical expectations likewise divide its raw quadratic form by $Z$.

For the Schmidt decomposition diagonalize $S_1$ in an orthonormal basis ${\xi }_j$ with its positive eigenvalues $c_j^2$. Expand $\Psi ={\sum }_j{\xi }_j\otimes w_j$ with $w_j=(⟨{\xi }_j|\otimes I)\Psi$. The reduced matrix gives $⟨w_k,w_j⟩=c_j^2{\delta }_{jk}$. Therefore ${\eta }_j=w_j/c_j$ are orthonormal andZero-eigenvalue components vanish because their squared norms are zero. This proves the decomposition for an arbitrary nonzero vector; the zero vector uses the empty sum. For a unit vector $\sum c_j^2=1$.

In the numerical example the coefficient matrix in the two product bases is $C=\left(\begin{array}{ccc}1& 2& 3\\ 0& 1& 2\end{array}\right)$. Taking the first-subsystem trace yieldsThe vector $(1,-2,1{)}^{\mathsf{T}}$ is in its nullspace, so rank is at most two; its leading $2\times 2$ minor has determinant one, so $\boxed{\mathrm{rank}{S}_2=2}$. This is the rank of $C$ and the Schmidt rank, proving entanglement. The displayed source ket has norm squared $19$, and the displayed reduced matrix has trace $19$. The physical density operator is $\boxed{{\rho }_2=S_2/19}$; its nonzero eigenvalues are $(19\pm \sqrt{337})/38$, since $C{C}^{\mathsf{T}}=\left(\begin{array}{cc}14& 8\\ 8& 5\end{array}\right)$. The raw matrix is correct for the unnormalized quadratic form actually printed.