| Revision as of 19:54, 14 April 2007 editMct mht (talk | contribs)4,336 edits complete positivity, which is true for the partial trace, is not needed here (although i would agree that the redirect probably doesn't make this very clear).← Previous edit | Latest revision as of 09:41, 28 February 2022 edit undoChristian75 (talk | contribs)Extended confirmed users, New page reviewers, Pending changes reviewers, Rollbackers114,675 edits {{R with history}} | ||
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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 there exist a 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> | |||
| :<math>\operatorname{Tr}_B | \psi \rangle \langle \psi | = \rho.</math> | |||
| === Proof === | |||
| A density matrix is by definition positive semidefinite. So ρ has square root factorization <math>\rho = A A^* = \sum_{i =1} ^n | i \rangle \langle i |</math>. Let <math>H_B</math> be another copy of the ''n''-dimensional Hilbert space with any orthonormal basis <math>\{ | i' \rangle \}</math>. Define <math>| \psi \rangle \in H_A \otimes H_B</math> by | |||
| :<math>| \psi \rangle = \sum_{i} |i \rangle \otimes | i' \rangle.</math> | |||
| Direct calculation gives | |||
| :<math> | |||
| \operatorname{Tr}_B | \psi \rangle \langle \psi | = | |||
| \operatorname{Tr}_B \sum_{i, j} |i \rangle \langle j | \otimes | 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 ] is a ]. | |||
| == An application: Stinespring's theorem == | |||
| {{sect-stub}} | |||
| By combining ] and purification of a mixed state, we can recover the ] for the finite dimensional case. | |||
| ] | |||
| ] | |||
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