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| #REDIRECT ] | |||
| In ], especially ], '''purification''' refers to the fact that every ] acting on finite-dimensional Hilbert spaces can be viewed as the ] of some pure state. | |||
| {{R with history}} | |||
| In purely linear algebraic terms, it can be viewed as a statement about ]. | |||
| == Statement == | |||
| Let ρ be a density matrix acting on a Hilbert space <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 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 simply a proof of principle, 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>. Notice that we do not say it is *the* purification of <math>\rho</math> because there may be many states that solve this equation. | |||
| === 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 basis <math>\{ | i \rangle \}</math>. Let <math>H_B</math> be another copy of the ''n''-dimensional Hilbert space with an orthonormal basis <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> | |||
| \operatorname{tr_B} \left( | \psi \rangle \langle \psi | \right )= | |||
| \operatorname{tr_B} \left </math> | |||
| <math> | |||
| =\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 | = \rho. | |||
| </math> | |||
| This proves the claim. | |||
| ==== Note ==== | |||
| * The vectorial pure state <math>| \psi \rangle</math> is in the form specified by the ]. | |||
| * 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 ] 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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