# Unitary equivalence of purifications

Purifications

Let $\rho$ be a mixed quantum state on system $\mathcal{A}$. We say that a pure quantum state $|\psi\rangle$ on a composite system $\mathcal{A} \otimes \mathcal{B}$ is a purification of $\rho$ if

$\rho = \mathrm{Tr}_B (|\psi\rangle \langle\psi|)$

The following theorem is a basic result in quantum information theory.

Theorem 1 (unitary equivalence of purifications). If $|\psi\rangle$ and $|\phi\rangle$ are two purifications of $\rho$ then

$|\phi\rangle = (I_A \otimes U) |\psi\rangle$

for some unitary $U$ that acts only on system $\mathcal{B}$.

Schmidt decomposition

The above result is usually proved using the following theorem.

Theorem 2 (Schmidt decomposition). Any bipartite quantum state $|\psi\rangle$ on $\mathcal{A} \otimes \mathcal{B}$ can be written as

$|\psi\rangle = \sum_i \sqrt{p_i} |\alpha_i\rangle |\beta_i\rangle$

for some orthonormal bases $\{|\alpha_i\rangle\}$ and $\{|\beta_i\rangle\}$ on systems $A$ and $B$, respectively, and a probability distribution $\{p_i\}$.

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

$|\psi\rangle = \sum_i \sqrt{p_i} |\alpha_i\rangle |\beta_i\rangle$

Let us expand the first register of the second state $|\phi\rangle$ in the basis $\{|\alpha_i\rangle\}$:

$|\phi\rangle = \sum_i |\alpha_i\rangle |\tilde{\gamma}_i\rangle$

where $\{|\tilde{\gamma}_i\rangle\}$ are some arbitrary (non-normalized) vectors. Then we have

$\mathrm{Tr}_B (|\psi\rangle \langle\psi|) = \sum_{i,j} \sqrt{p_i p_j} |\alpha_i\rangle \langle\alpha_j| \mathrm{Tr} (|\beta_i\rangle \langle\beta_j|)$

$= \sum_{i,j} \sqrt{p_i p_j} |\alpha_i\rangle \langle\alpha_j| \delta_{ij}$

and

$\mathrm{Tr}_B (|\phi\rangle \langle\phi|) = \sum_{i,j} |\alpha_i\rangle \langle\alpha_j| \mathrm{Tr} (|\tilde{\gamma}_i\rangle \langle\tilde{\gamma}_j|)$

$= \sum_{i,j} |\alpha_i\rangle \langle\alpha_j| \langle\tilde{\gamma}_j|\tilde{\gamma}_i\rangle$

By assumption, both partial traces are equal. Since $\{|\alpha_i\rangle\}$ is an orthonormal basis, the coefficients in both expressions must be the same:

$\sqrt{p_i p_j} \delta_{ij} = \langle\tilde{\gamma}_j|\tilde{\gamma}_i\rangle$

This means that $|\tilde{\gamma}_i\rangle$ are pairwise orthogonal and $\||\tilde{\gamma}_i\rangle\| = \sqrt{p_i}$ (it could happen that $p_i = 0$ for some $i$, but this is not a problem). Equivalently, this equation says that we can find an orthonormal basis $\{|\gamma_i\rangle\}$ such that $|\tilde{\gamma}_i\rangle = \sqrt{p_i} |\gamma_i\rangle$. Then we see that $|\phi\rangle = (I \otimes U) |\psi\rangle$, where $U = \sum_i |\gamma_i\rangle \langle\beta_i|$ is the unitary change of basis from $\{|\beta_i\rangle\}$ to $\{|\gamma_i\rangle\}$.