Purification of quantum state

In quantum mechanics, especially quantum information, purification refers to the fact that every mixed state acting on finite-dimensional Hilbert spaces can be viewed as the reduced state of some pure state.

In purely linear algebraic terms, it can be viewed as a statement about positive-semidefinite matrices.

Statement

Let $\rho$ be a density matrix acting on a Hilbert space $H_{A}$ of finite dimension n. Then it is possible to construct a second Hilbert space $H_{B}$ and a pure state $|\psi \rangle \in H_{A}\otimes H_{B}$ such that $\rho$ is the partial trace of $|\psi \rangle \langle \psi |$ with respect to $H_{B}$ . While the initial Hilbert space $H_{A}$ might correspond to physically meaningful quantities, the second Hilbert space $H_{B}$ 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 $H_{B}$ should also correspond to a physical space, such as the environment. The exact form of $H_{B}$ in such cases will depend on the problem. Here is a proof of principle, showing that at very least $H_{B}$ has to have dimensions greater than or equal to $H_{A}$ .

With these statements in mind, if,

\operatorname {tr_{B}} \left(|\psi \rangle \langle \psi |\right)=\rho ,

we say that $|\psi \rangle$ purifies $\rho$ .

Proof

A density matrix is by definition positive semidefinite. So ρ can be diagonalized and written as $\rho =\sum _{i=1}^{n}p_{i}|i\rangle \langle i|$ for some basis $\{|i\rangle \}$ . Let $H_{B}$ be another copy of the n-dimensional Hilbert space with an orthonormal basis $\{|i'\rangle \}$ . Define $|\psi \rangle \in H_{A}\otimes H_{B}$ by

|\psi \rangle =\sum _{i}{\sqrt {p_{i}}}|i\rangle \otimes |i'\rangle .

Direct calculation gives

{\begin{aligned}\operatorname {tr_{B}} \left(|\psi \rangle \langle \psi |\right)&=\operatorname {tr_{B}} \left[\left(\sum _{i}{\sqrt {p_{i}}}|i\rangle \otimes |i'\rangle \right)\left(\sum _{j}{\sqrt {p_{j}}}\langle j|\otimes \langle j'|\right)\right]\\&=\operatorname {tr_{B}} \left(\sum _{i,j}{\sqrt {p_{i}p_{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{aligned}}

This proves the claim.

Note

The purification is not unique, but if during the construction of $|\psi \rangle$ in the proof above $H_{B}$ is generated by only the $\{|i'\rangle \}$ for which $p_{i}$ is non-zero, any other purification $|\varphi \rangle$ on $H_{A}\otimes H_{C}$ induces an isometry $V:H_{B}\to H_{C}$ such that $|\varphi \rangle =(I\otimes V)|\psi \rangle$ .
The vectorial pure state $|\psi \rangle$ is in the form specified by the Schmidt decomposition.
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 partial trace of a positive map remains a positive map.