Schrödinger–HJW theorem

Let ${\displaystyle {\mathcal {H}}_{S}}$ be a finite-dimensional Hilbert space, and consider a generic (possibly mixed) quantum state ${\displaystyle \rho }$ defined on ${\displaystyle {\mathcal {H}}_{S}}$, and admitting a decomposition of the form

$${\displaystyle \rho =\sum _{i}p_{i}|\phi _{i}\rangle \!\langle \phi _{i}|,}$$

for a collection of (not necessarily mutually orthogonal) states ${\displaystyle |\phi _{i}\rangle \in {\mathcal {H}}_{S}}$, and coefficients ${\displaystyle p_{i}\geq 0}$ such that ${\textstyle \sum _{i}p_{i}=1}$. Note that any quantum state can be written in such a way for some ${\displaystyle \{|\phi _{i}\rangle \}_{i}}$ and ${\displaystyle \{p_{i}\}_{i}}$.[8] Any such ${\displaystyle \rho }$ can be purified, that is, represented as the partial trace of a pure state defined in a larger Hilbert space. More precisely, it is always possible to find a (finite-dimensional) Hilbert space ${\displaystyle {\mathcal {H}}_{A}}$ and a pure state ${\displaystyle |\Psi _{SA}\rangle \in {\mathcal {H}}_{S}\otimes {\mathcal {H}}_{A}}$ such that ${\displaystyle \rho =\operatorname {Tr} _{A}(|\Psi _{SA}\rangle \!\langle \Psi _{SA}|)}$. Furthermore, the states ${\displaystyle |\Psi _{SA}\rangle }$ satisfying this are all and only those of the form

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

for some orthonormal basis ${\displaystyle \{|a_{i}\rangle \}_{i}\subset {\mathcal {H}}_{A}}$. The state ${\displaystyle |\Psi _{SA}\rangle }$ is then referred to as 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.[9] Because all of them admit a decomposition in the form given above, given any pair of purifications ${\displaystyle |\Psi \rangle ,|\Psi '\rangle \in {\mathcal {H}}_{S}\otimes {\mathcal {H}}_{A}}$, there is always some unitary operation ${\displaystyle U:{\mathcal {H}}_{A}\to {\mathcal {H}}_{A}}$ such that

$${\displaystyle |\Psi '\rangle =(I\otimes U)|\Psi \rangle .}$$

Consider a mixed quantum state ${\displaystyle \rho }$ with two different realizations as ensemble of pure states as ${\textstyle \rho =\sum _{i}p_{i}|\phi _{i}\rangle \langle \phi _{i}|}$ and ${\textstyle \rho =\sum _{j}q_{j}|\varphi _{j}\rangle \langle \varphi _{j}|}$. 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: ${\displaystyle |\Psi _{SA}^{1}\rangle =\sum _{i}{\sqrt {p_{i}}}|\phi _{i}\rangle \otimes |a_{i}\rangle }$;
• Purification 2: ${\displaystyle |\Psi _{SA}^{2}\rangle =\sum _{j}{\sqrt {q_{j}}}|\varphi _{j}\rangle \otimes |b_{j}\rangle }$.

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 ${\displaystyle |\Psi _{SA}^{1}\rangle =(I\otimes U_{A})|\Psi _{SA}^{2}\rangle }$.[10] Therefore, ${\textstyle |\Psi _{SA}^{1}\rangle =\sum _{j}{\sqrt {q_{j}}}|\varphi _{j}\rangle \otimes U_{A}|b_{j}\rangle }$, which means that we can realize the different ensembles of a mixed state just by making different measurements on the purifying system.