Misplaced Pages

Purification of quantum state

Article snapshot taken from Wikipedia with creative commons attribution-sharealike license. Give it a read and then ask your questions in the chat. We can research this topic together.

This is an old revision of this page, as edited by Mct mht (talk | contribs) at 19:54, 14 April 2007 (complete positivity, which is true for the partial trace, is not needed here (although i would agree that the redirect probably doesn't make this very clear).). The present address (URL) is a permanent link to this revision, which may differ significantly from the current revision.

Revision as of 19:54, 14 April 2007 by Mct mht (talk | contribs) (complete positivity, which is true for the partial trace, is not needed here (although i would agree that the redirect probably doesn't make this very clear).)(diff) ← Previous revision | Latest revision (diff) | Newer revision → (diff)

In quantum mechanics, especially quantum information, purification refers to the fact that every mixed state acting on finite dimensional Hilbert spaces can be viewed as the reduced state of some pure state.

In purely linear algebraic terms, it can be viewed as a statement about positive-semidefinite matrices.

Statement

Let ρ be a density matrix acting on a Hilbert space H A {\displaystyle H_{A}} of finite dimension n, then there exist a Hilbert space H B {\displaystyle H_{B}} and a pure state | ψ H A H B {\displaystyle |\psi \rangle \in H_{A}\otimes H_{B}} such that the partial trace of | ψ ψ | {\displaystyle |\psi \rangle \langle \psi |} with respect to H B {\displaystyle H_{B}}

Tr B | ψ ψ | = ρ . {\displaystyle \operatorname {Tr} _{B}|\psi \rangle \langle \psi |=\rho .}

Proof

A density matrix is by definition positive semidefinite. So ρ has square root factorization ρ = A A = i = 1 n | i i | {\displaystyle \rho =AA^{*}=\sum _{i=1}^{n}|i\rangle \langle i|} . Let H B {\displaystyle H_{B}} be another copy of the n-dimensional Hilbert space with any orthonormal basis { | i } {\displaystyle \{|i'\rangle \}} . Define | ψ H A H B {\displaystyle |\psi \rangle \in H_{A}\otimes H_{B}} by

| ψ = i | i | i . {\displaystyle |\psi \rangle =\sum _{i}|i\rangle \otimes |i'\rangle .}

Direct calculation gives

Tr B | ψ ψ | = Tr B i , j | i j | | i j | = ρ . {\displaystyle \operatorname {Tr} _{B}|\psi \rangle \langle \psi |=\operatorname {Tr} _{B}\sum _{i,j}|i\rangle \langle j|\otimes |i'\rangle \langle j'|=\rho .}

This proves the claim.

Note

  • The vectorial pure state | ψ {\displaystyle |\psi \rangle } is in the form specified by the Schmidt decomposition.
  • Since square root decompositions of a positive semidefinite matrix are not unique, neither are purifications.
  • In linear algebraic terms, a square matrix is positive semidefinite if and only if it can be purified in the above sense. The if part of the implication follows immediately from the fact that the partial trace is a positive map.

An application: Stinespring's theorem

This section needs expansion. You can help by adding to it.

By combining Choi's theorem on completely positive maps and purification of a mixed state, we can recover the Stinespring dilation theorem for the finite dimensional case.

Categories: