Solution
ID: past-exam-of-the-quantum-theory-and-quantum-computers-course-of-the-university-of-oxford/2010/3/solution
Past exam of the quantum theory and quantum computers course of the University of Oxford 2010 3 Solution by
Codex 0 2026-10-31
A density operator is a positive self-adjoint trace-class operator with trace one. In finite dimensions this is a positive semidefinite Hermitian matrix with . Its expectation of an observable is whenever the expectation is defined.
In the source's square-root notation, take the unique positive square root . It has Hilbert-Schmidt norm one and satisfies , soThe standard vector purification is the associated vector on a system and a chosen copy. Tracing out the copy gives ; measuring gives the same trace formula. Thus the operator square root and the purified vector are two related objects, distinguished by this vectorization.
For a general coefficient matrix , expansion of the expectation givesThus . Positivity is immediate from , and its trace is one by normalization of .
If is nonsingular, it is square and is positive definite. DefineThen , which for a square matrix also implies . Thus is unitary, , and .
In an eigenbasis of , its square root is . Hence . DefineUnitarity gives , and substitution gives the Schmidt decompositionFor singular or rectangular , the same conclusion follows on its support by a singular value decomposition; the stated nonsingular case needs only the unitary calculation above.
For the supplied Bell state, the requested matrices areThe Schmidt vectors can be chosen as , , and the two coefficients are . The square root's canonical vector purification is ; the original Bell vector is a different purification of the same reduced state, related by a unitary on the second factor. This illustrates the unitary freedom of purification.
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