Misplaced Pages

Revision as of 20:35, 16 June 2006

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.

Statement

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

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

Proof

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

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

Direct calculation gives

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

This proves the claim.

Note

• The vectorial pure state ${\displaystyle |\psi \rangle }$ is in the form specified by the Schmidt decomposition.

• Since square root decompositions of a positive semidefinite matrix are not unique, neither are purifications.

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.

• Quantum information science