Gisin–Hughston–Jozsa–Wootters theorem

Consider a mixed state ${\displaystyle \rho =\sum _{i}p_{i}|\phi _{i}\rangle \langle \phi _{i}|}$ of the system ${\displaystyle S}$, where the states ${\displaystyle |\phi _{i}\rangle }$ are not assumed to be mutually orthogonal. We can add an auxiliary space ${\displaystyle {\mathcal {H}}_{A}}$ with an orthonormal basis ${\displaystyle \{|a_{i}\rangle \}}$, then the mixed state can be obtained as reduced density operator from the pure bipartite state

${\displaystyle |\Psi _{SA}\rangle =\sum _{i}{\sqrt {p_{i}}}|\phi _{i}\rangle |a_{i}\rangle .}$

More precisely, ${\displaystyle \rho =\mathrm {Tr} _{A}|\Psi _{SA}\rangle \langle \Psi _{SA}|}$. The state ${\displaystyle |\Psi _{SA}\rangle }$ is thus called the purification of ${\displaystyle \rho }$. Since the auxiliary space and the basis can be chosen arbitrarily, the purification of a mixed state is not unique; in fact, there are infinitely many purifications of a given mixed state.

Consider a mixed quantum state ${\displaystyle \rho }$ with two different realizations as ensemble of pure states as ${\displaystyle \rho =\sum _{i}p_{i}|\phi _{i}\rangle \langle \phi _{i}|}$ and ${\displaystyle \rho =\sum _{j}q_{j}|\varphi _{j}\rangle \langle \varphi _{j}|}$. Here both ${\displaystyle |\phi _{i}\rangle }$and ${\displaystyle |\varphi _{j}\rangle }$ are not assumed to be mutually orthogonal. There will be two corresponding purifications of the mixed state ${\displaystyle \rho }$ reading as follows:

Purification 1: ${\displaystyle |\Psi _{SA}^{1}\rangle =\sum _{i}{\sqrt {p_{i}}}|\phi _{i}\rangle |a_{i}\rangle }$;

Purification 2: ${\displaystyle |\Psi _{SA}^{2}\rangle =\sum _{j}{\sqrt {q_{j}}}|\varphi _{j}\rangle |b_{j}\rangle }$.

The sets ${\displaystyle \{|a_{i}\rangle \}}$and ${\displaystyle \{|b_{j}\rangle \}}$ are two collections of orthonormal bases of the respective auxiliary spaces. These two purifications only differ by a unitary transformation acting on the auxiliary space, viz., there exists a unitary matrix ${\displaystyle U_{A}}$such that ${\displaystyle |\Psi _{SA}^{1}\rangle =I\otimes U_{A}|\Psi _{SA}^{2}\rangle }$.[7] Therefore, ${\displaystyle |\Psi _{SA}^{1}\rangle =\sum _{j}{\sqrt {q_{j}}}|\varphi _{i}\rangle \otimes U_{A}|b_{j}\rangle }$, which means that we can realize the different ensembles of a mixed state just by choosing to measure different observables of one given purification.