Purification of quantum state

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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.

1 Statement
- 1.1 Proof
  - 1.1.1 Note
2 An application: Stinespring's theorem

[edit] Statement

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

$\operatorname{Tr}_B | \psi \rangle \langle \psi | = \rho.$

[edit] Proof

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

$| \psi \rangle = \sum_{i} |i \rangle \otimes | i' \rangle.$

Direct calculation gives

$\operatorname{Tr}_B | \psi \rangle \langle \psi | = \operatorname{Tr}_B \sum_{i, j} |i \rangle \langle j | \otimes | i' \rangle \langle j'| = \rho.$

This proves the claim.

[edit] Note

The vectorial pure state $| \psi \rangle$ 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.