Past exam of the mathematics course of the University of Cambridge / 2008 / iii / Paper 58 / 3

The Schmidt decomposition theorem says that a normalized vector in the tensor product of two finite-dimensional Hilbert spaces can be writtenwhere both families are orthonormal. The numbers $\sqrt{{\lambda }_j}$ are its Schmidt coefficients. The Schmidt rank $r$ and coefficients, including multiplicities, are uniquely determined by the state; the local vectors may be changed within degenerate eigenspaces and by compensating phases. Zero coefficients may be appended to make $n_1$ terms, since $n_2\ge n_1$.

For a proof, form the reduced density matrix ${\rho }_1={\mathrm{Tr}}_2|\psi ⟩⟨\psi |$. It is a Hermitian matrix that is positive semidefinite, with trace one. The spectral theorem supplies an orthonormal basis $u_1,\dots ,u_{n_1}$ of eigenvectors with nonnegative eigenvalues ${\lambda }_j$. Define the unnormalized relative vectors in the second Hilbert space byCompleteness of the first orthonormal basis gives $|\psi ⟩={\sum }_j|u_j⟩|w_j⟩$. In any orthonormal basis $e_b$ of the second factor, the definition of the partial trace givesThus a zero eigenvalue gives $w_j=0$, and for each positive eigenvalue the vector $v_j=w_j/\sqrt{{\lambda }_j}$ has unit norm. Distinct such vectors are orthogonal by the same identity. Substituting them gives the required Schmidt decomposition, with at most $n_1$ nonzero terms.

Taking the other partial trace now givesTherefore the two reduced density matrices have identical nonzero eigenvalues, namely the squared Schmidt coefficients. This also proves uniqueness of the coefficient multiset and shows that the pure state is a product state exactly when $r=1$.

Let $Q$ be Alice's unknown qubit, with pure state $|\varphi {⟩}_Q=\alpha |0⟩+\beta |1⟩$. Alice and Bob initially share the Bell state $|{\Phi }^{+}{⟩}_{AB}=(|00⟩+|11⟩)/\sqrt{2}$. Alice holds $Q,A$ and Bob holds $B$. Define the Bell basisExpanding in this orthonormal basis gives the Bell-basis teleportation identityFor example, the last branch on $B$ is $\alpha |1⟩-\beta |0⟩$, which is $XZ|\varphi ⟩$; the ordering of the Pauli operators therefore matters.

Alice performs a projective measurement in the Bell basis on her two local qubits. Each outcome has probability $1/4$, independent of the unknown amplitudes. She encodes the four possible outcomes as two classical bits and sends them to Bob. For outcomes ${\Phi }^{+},{\Phi }^{-},{\Psi }^{+},{\Psi }^{-}$, respectively, Bob applies the local unitary operationsThese are inverses of the corresponding branch Pauli operators, so Bob obtains exactly the unknown input qubit in every branch. Using $XZ$ instead of $ZX$ as the last correction also succeeds up to an irrelevant overall phase. Alice need not know $\alpha$ or $\beta$. Every quantum operation is local to one of the parties, and the only transmitted information is the two-bit classical outcome. The shared Bell pair is consumed.

Before the message arrives, Bob's density operator isso no unknown-state information is available without the classical communication. Alice's original qubit has been included in the destructive Bell-basis measurement; the protocol creates no extra copy. Since the corrected transformation is the identity on every input vector and is linear, it also preserves entanglement with an external reference and transfers mixed inputs. Thus this is deterministic quantum teleportation, using precisely one shared maximally entangled Bell pair, two classical bits, and local operations.