Past exam of the mathematics course of the University of Cambridge / 2007 / iii / Paper 34 / 4

Fix paired orthonormal bases in the two $d$-dimensional Hilbert spaces. A maximally entangled state isIt is normalized, and the partial traces of its rank-one density operator are ${\rho }_A={\rho }_B=I/d$. Thus all its Schmidt coefficients are $1/\sqrt{d}$ and its entanglement entropy is ${\log }_{2}d$. No pure bipartite state of these dimensions can have larger entanglement entropy: nonnegativity of quantum relative entropy applied to a reduced state and $I/d$ gives $0\le D({\rho }_A\Vert I/d)={\log }_{2}d-S({\rho }_A)$. Hence it is maximally entangled.

Write $|{\varphi }_A⟩={\sum }_j{c}_j|j{⟩}_A$ and choose the conjugate index vectorWith $|{\tilde{\Psi }}_{AB}⟩=\sqrt{d}|{\Psi }_{AB}⟩={\sum }_j|j{⟩}_A|j{⟩}_B$, the partial inner product givesThis is the relative state construction. The coefficient conjugation is defined relative to the fixed paired bases; it is essential, because the bra conjugates those coefficients again. Using the normalized entangled vector instead would give $|{\varphi }_A⟩/\sqrt{d}$.

The linear operator $M_A$ acts on the first factor, while the partial inner product acts on the second. They therefore commute in the contraction:Thus the operated vector is exactly the relative state obtained from the operated bipartite vector using the same conjugate index vector. For a physical normalized pure state, divide by $\Vert M_A{\varphi }_A\Vert$ when that norm is nonzero. If $M_A{\varphi }_A=0$, the vector identity still holds, but there is no normalized output state for that zero-probability branch. An arbitrary operator is not assumed unitary or norm preserving.

Use the unnormalized Choi matrixIt is positive because ${\Phi }_A$ is a completely positive map. Take its spectral decomposition, absorbing square roots of nonzero eigenvalues into the vectors:Expand $|v_k⟩={\sum }_{i,j}{v}_{k,ij}|i{⟩}_A|j{⟩}_B$ and define the Kraus operator $A_k$ by $(A_k{)}_{ij}=v_{k,ij}$. Equivalently,The preceding relative state identity then gives $(I\otimes ⟨{\varphi }_B^{\ast }|)v_k=A_k{\varphi }_A$. Inserting the spectral decomposition into the stated reconstruction identity yieldsThe conjugated index vector is required here; it is present in the PDF but missing in the converted TeX version of this identity.

Rank-one projectors span all operators by polarization. Explicitly, putting $P(w)=|w⟩⟨w|$ givesThus linearity extends the equality to every $X\in \mathcal{B}({\mathcal{H}}_A)$:Finally trace preservation and cyclicity of the trace giveFor pure-projector inputs this says the Hermitian operator ${\sum }_k{A}_k^{†}{A}_k-I$ has zero expectation on every vector, so it vanishes. ThereforeThis is the spectral Kraus decomposition. Using a normalized Choi state $J_{\Phi }/d$ instead would require the extra factor $\sqrt{d}$ when reshaping its eigenvectors into Kraus operators.