Schmidt decomposition for tripartite system $ABC$ with vanishing mutual information between $A$ and $C$

Question

Suppose I have a tripartite system $ABC$ in a pure state $|\psi_{ABC}\rangle$ with mutual information $I(A:C)=0$. This implies that the reduced density matrix $\rho_{AC}$ factorizes as $\rho_{AC} = \rho_A \otimes \rho_C$.

How do I show that this implies the existence of a Schmidt decomposition of $|\psi_{ABC}\rangle$ of the form \begin{align} |\psi_{ABC}\rangle = \sum_{kl} \sqrt{\lambda_k p_l} |\psi_k\rangle_A \otimes |\phi_{kl}\rangle_B \otimes |\varphi_l\rangle_C \end{align} where $|\psi_k\rangle_A$, $|\phi_{kl}\rangle_B$, $|\varphi_l\rangle_C$ are orthonormal states on Hilbert spaces $A$,$B$, and $C$ respectively?

Answer 1

TL;DR: The key observation is that Schmidt basis on a subsystem consists of eigenvectors of the reduced state of that subsystem. Consequently, if the reduced state is a product state then its Schmidt basis can be chosen to consist of pure product states.

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We begin by writing down the Schmidt decomposition for $|\psi\rangle_{ABC}$ with respect to the partitioning of $ABC$ into subsystems $B$ and $AC$

$$ |\psi\rangle_{ABC}=\sum_i s_i|\beta_i\rangle_B|\alpha_i\rangle_{AC}\tag1 $$

which allows us to write an eigendecomposition for $\rho_{AC}=\rho_A\otimes\rho_C$ in terms of the Schmidt basis

$$ \rho_A\otimes\rho_C=\sum_i s_i^2|\alpha_i\rangle_{AC}\langle \alpha_i|_{AC}.\tag2 $$

However, if

$$ \rho_A=\sum_k\lambda_k|\psi_k\rangle_A\langle\psi_k|_A\quad \rho_C=\sum_l p_l|\varphi_l\rangle_C\langle\varphi_l|_C\tag3 $$

are eigendecompositions of $\rho_A$ and $\rho_C$, then

$$ \rho_A\otimes\rho_C=\sum_{kl}\lambda_k p_l|\psi_k\rangle_A|\varphi_l\rangle_C\langle\psi_k|_A\langle\varphi_l|_C\tag4 $$

is also an eigendecomposition of $\rho_A\otimes\rho_C$. Comparing $(2)$ and $(4)$, we see that both $s_i^2$ and $\lambda_k p_l$ are eigenvalues of $\rho_A\otimes\rho_C$. Therefore, the ranges of the index $i$ and the pair $k,l$ coincide, there is a bijection $f$ such that $f(k, l)=i$ and

$$ s_{f(k,l)}^2=\lambda_k p_l.\tag5 $$

Moreover, if the eigenvalues $s_i^2$ are all distinct, then

$$ |\alpha_{f(k,l)}\rangle_{AC}\equiv|\psi_k\rangle_A|\varphi_l\rangle_C\tag6 $$

where $\equiv$ denotes equality up to global phase. If there are repeated eigenvalues then $(6)$ does not necessarily hold. However, in this case, the unitary freedom$^1$ in the subspaces corresponding to equal Schmidt coefficients in $(1)$ allows us to choose $|\alpha_i\rangle_{AC}$ as in $(6)$.

Now, we can upgrade equality up to global phase in $(6)$ to proper equality by absorbing the phase factor into $|\beta_i\rangle_B$. Denoting the new basis, with phase factors absorbed, as $|\phi_{kl}\rangle_B$, we have

$$ \begin{align} |\beta_{f(k,l)}\rangle_B|\alpha_{f(k,l)}\rangle_{AC}&=|\phi_{kl}\rangle_B|\psi_k\rangle_A|\varphi_l\rangle_C\\ &=|\psi_k\rangle_A|\phi_{kl}\rangle_B|\varphi_l\rangle_C. \end{align}\tag7 $$

Finally, substituting $(5)$ and $(7)$ into $(1)$, we obtain

$$ \begin{align} |\psi\rangle_{ABC}&=\sum_i s_i|\beta_i\rangle_B|\alpha_i\rangle_{AC}\\ &=\sum_{kl} s_{f(k,l)}|\beta_{f(k,l)}\rangle_B|\alpha_{f(k,l)}\rangle_{AC}\\ &=\sum_{kl}\sqrt{\lambda_k p_l}|\psi_k\rangle_A|\phi_{kl}\rangle_B|\varphi_l\rangle_C \end{align}\tag8 $$

as desired.

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^{$^1$ This is analogous to unitary freedom in singular value decomposition from which it derives.}