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

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?

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.

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.

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