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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.

Statement

Let ρ be a density matrix acting on a Hilbert space ${\displaystyle H_{A}}$ of finite dimension n. Then it is possible to construct a second Hilbert space ${\displaystyle H_{B}}$ and a pure state ${\displaystyle |\psi \rangle \in H_{A}\otimes H_{B}}$ such that ρ is the partial trace of ${\displaystyle |\psi \rangle \langle \psi |}$ with respect to ${\displaystyle H_{B}}$. While the initial Hilbert space ${\displaystyle H_{A}}$ might correspond to physically meaningful quantities, the second Hilbert space ${\displaystyle H_{B}}$ needn't have any physical interpretation whatsoever. However, in physics the process of state purification is assumed to be physical, and so the second Hilbert space ${\displaystyle H_{B}}$ should also correspond to a physical space, such as the environment. The exact form of ${\displaystyle H_{B}}$ in such cases will depend on the problem, here is simply a proof of principle, showing that at very least ${\displaystyle H_{B}}$ has to have dimensions greater than or equal to ${\displaystyle H_{A}}$.

With these statements in mind, if,

${\displaystyle \operatorname {tr_{B}} \left(|\psi \rangle \langle \psi |\right)=\rho ,}$

we say that ${\displaystyle |\psi \rangle }$ purifies ${\displaystyle \rho }$. Notice that we do not say it is *the* purification of ${\displaystyle \rho }$ because there may be many states that solve this equation.

Proof

A density matrix is by definition positive semidefinite. So ρ can be diagonalized and written as ${\displaystyle \rho =\sum _{i=1}^{n}p_{i}|i\rangle \langle i|}$ for some basis ${\displaystyle \{|i\rangle \}}$. Let ${\displaystyle H_{B}}$ be another copy of the n-dimensional Hilbert space with an orthonormal basis ${\displaystyle \{|i'\rangle \}}$. Define ${\displaystyle |\psi \rangle \in H_{A}\otimes H_{B}}$ by

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

Direct calculation gives

${\displaystyle \operatorname {tr_{B}} \left(|\psi \rangle \langle \psi |\right)=\operatorname {tr_{B}} \left}$

${\displaystyle =\operatorname {tr_{B}} \left(\sum _{i,j}{\sqrt {p_{i}p_{j}}}|i\rangle \langle j|\otimes |i'\rangle \langle j'|\right)=\sum _{i,j}\delta _{ij}{\sqrt {p_{i}p_{j}}}|i\rangle \langle j|=\rho .}$

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.
• 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 of a positive map remains a positive map.

An application: Stinespring's theorem

This section needs expansion. You can help by adding to it. (June 2008)

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.

• Linear algebra
• Quantum information science