Schmidt decomposition

Unitary equivalence of purifications

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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\}.

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