Let be a mixed quantum state on system . We say that a pure quantum state on a composite system is a purification of if
The following theorem is a basic result in quantum information theory.
Theorem 1 (unitary equivalence of purifications). If and are two purifications of then
for some unitary that acts only on system .
The above result is usually proved using the following theorem.
Theorem 2 (Schmidt decomposition). Any bipartite quantum state on can be written as
for some orthonormal bases and on systems and , respectively, and a probability distribution .
However, standard proofs of Theorem 1 are often based on the following (wrong) argument: given two spectral decompositions of the same matrix, the corresponding eigenvalues and eigenvectors in both decompositions must be equal. This is true only if the given matrix has a non-degenerate spectrum. Otherwise the eigenvectors corresponding to a repeated eigenvalue are not uniquely determined (any orthonormal basis of the corresponding subspace is a valid set of eigenvectors).
A simple proof of Theorem 1
Here is an elementary proof of Theorem 1 which does not run into problems in case of a degenerate spectrum. By Theorem 2 we can write the first state as
Let us expand the first register of the second state in the basis :
where are some arbitrary (non-normalized) vectors. Then we have
By assumption, both partial traces are equal. Since is an orthonormal basis, the coefficients in both expressions must be the same:
This means that are pairwise orthogonal and (it could happen that for some , but this is not a problem). Equivalently, this equation says that we can find an orthonormal basis such that . Then we see that , where is the unitary change of basis from to .