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#REDIRECT ]
In ], especially ], '''purification''' refers to the fact that every ] acting on ] can be viewed as the ] of some pure state.


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In purely linear algebraic terms, it can be viewed as a statement about ].

== Statement ==

Let <math>\rho</math> be a ] acting on a ] <math>H_A</math> of finite dimension ''n''. Then it is possible to construct a second Hilbert space <math>H_B</math> and a pure state <math>| \psi \rangle \in H_A \otimes H_B</math> such that <math>\rho</math> is the partial trace of <math>| \psi \rangle \langle \psi |</math> with respect to <math>H_B</math>. While the initial Hilbert space <math>H_A</math> might correspond to physically meaningful quantities, the second Hilbert space <math>H_B</math> 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 <math>H_B</math> should also correspond to a physical space, such as the environment. The exact form of <math>H_B</math> in such cases will depend on the problem. Here is a ], showing that at very least <math>H_B</math> has to have dimensions greater than or equal to <math>H_A</math>.

With these statements in mind, if,
:<math>\operatorname{tr_B} \left( | \psi \rangle \langle \psi | \right )= \rho,</math>

we say that <math>| \psi \rangle</math> purifies <math>\rho</math>.

=== Proof ===

A density matrix is by definition positive semidefinite. So ρ can be ] and written as <math>\rho = \sum_{i =1} ^n p_i | i \rangle \langle i |</math> for some ] <math>\{ | i \rangle \}</math>. Let <math>H_B</math> be another copy of the ''n''-dimensional Hilbert space with an ] <math>\{ | i' \rangle \}</math>. Define <math>| \psi \rangle \in H_A \otimes H_B</math> by

:<math>| \psi \rangle = \sum_{i} \sqrt{p_i} |i \rangle \otimes | i' \rangle.</math>

Direct calculation gives

:<math>
\begin{align}
\operatorname{tr_B} \left( | \psi \rangle \langle \psi | \right ) &=
\operatorname{tr_B} \left \\
&=\operatorname{tr_B} \left( \sum_{i, j} \sqrt{p_ip_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 | \\
&= \sum_{i} \sqrt{p_i}^2 |i\rangle \langle i| \\
&= \rho.
\end{align} </math>

This proves the claim.

==== Note ====

* The purification is not unique, but if during the construction of <math> | \psi \rangle </math> in the proof above <math>H_B</math> is generated by only the <math>\{ | i' \rangle \}</math> for which <math> p_i </math> is non-zero, any other purification <math> | \varphi \rangle </math> on <math> H_A \otimes H_C </math> induces an ] <math> V: H_B \to H_C </math> such that <math> | \varphi \rangle = (I \otimes V ) | \psi \rangle </math>.
* The vectorial pure state <math>| \psi \rangle</math> is in the form specified by the ].
* Since ] ] of a positive semidefinite matrix are not unique, neither are purifications.
* In linear algebraic terms, a square matrix is positive semidefinite ] it can be purified in the above sense. The ''if'' part of the implication follows immediately from the fact that the ] of a positive map remains a ].

== An application: Stinespring's theorem ==
{{Expand section|date=June 2008}}
By combining ] and purification of a mixed state, we can recover the ] for the finite-dimensional case.

{{DEFAULTSORT:Purification Of Quantum State}}
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