Past exam of the mathematics course of the University of Cambridge / 2005 / iii / Paper 37 / 2

Diagonalize the reduced density matrix on Alice's Hilbert space:For every positive eigenvalue, define a vector on Bob's Hilbert space byThe definition of the partial trace givesThus the $|b_j⟩$ are orthonormal. Extend the $|a_j⟩$ to an orthonormal basis of Alice's Hilbert space. If $|a⟩$ lies in the zero-eigenvalue subspace of ${\rho }_A$, thenThere is therefore no component of $|{\Psi }_{AB}⟩$ in those additional basis directions. Expansion in Alice's orthonormal basis yields the Schmidt decompositionIts positive Schmidt coefficients are $s_j=\sqrt{{\lambda }_j}$ and its Schmidt rank is $r$. Taking the two partial traces and using the orthonormality of both families givesHence the two reduced density matrices have exactly the same nonzero eigenvalues, including multiplicities. Their numbers of zero eigenvalues may differ when the local Hilbert spaces have different dimensions. In particular their Von Neumann entropies agree.

The coefficient matrix in the computational orthonormal bases, with Alice's index labelling rows, isSince the entries are real, the reduced density matrices areBoth have trace one and determinant $1/9$. Their characteristic polynomial is ${\lambda }^2-\lambda +1/9$, so the nonzero eigenvalues areBoth are positive. Thus the Schmidt rank, or the Schmidt number of this pure state, isIf “Schmidt numbers” denotes the individual positive Schmidt coefficients instead, those areTheir squares are ${\lambda }_{\pm }$; giving both the Schmidt rank and the Schmidt coefficients removes this terminology ambiguity. The pure state is entangled because its Schmidt rank exceeds one.

Factor the coefficients into a tensor product:This is already a Schmidt decomposition with one term. ConsequentlyEach reduced density matrix is the rank-one orthogonal projector onto $(|0⟩-|1⟩)/\sqrt{2}$, so its other eigenvalue is zero. The joint pure state is a product state, with no entanglement.