Consider two tetrapartite quantum states $|\phi\rangle^{AA^\prime BB^\prime}$ and $|\psi_1\rangle^{AA^\prime}|\psi_2\rangle^{BB^\prime}$ in a finite dimentional Hilbert space $\mathcal{H}^A\otimes\mathcal{H}^{A^\prime}\otimes\mathcal{H}^B\otimes\mathcal{H}^{B^\prime}$, such that $$\| |\phi\rangle\langle\phi|^{AA^\prime BB^\prime} -|\psi_1\rangle\langle\psi_1|^{AA^\prime}\otimes|\psi_2\rangle\langle\psi_2|^{BB^\prime} \|_p \leq \varepsilon .$$

If we attempt to do the Schmidt decomposition for $|\phi\rangle^{AA^\prime BB^\prime}$ with respect to systems $AB$ and $A^\prime B^\prime$ (not $AA^\prime$ and $BB^\prime$), that Schmidt basis is close to separable for $AA^\prime$ and $BB^\prime$?

Since $|\psi_1\rangle^{AA^\prime}|\psi_2\rangle^{BB^\prime}$ is decomposed as $$|\psi_1\rangle^{AA^\prime}|\psi_2\rangle^{BB^\prime}=\sum_{i,j}\sqrt{q_i}\sqrt{q_j}|e_i\rangle^A|f_j\rangle^B|\tilde{e}_i\rangle^{A\prime}|\tilde{f}_j\rangle^{B^\prime},$$

if $|\phi\rangle^{AA^\prime BB^\prime}$ is decomposed as $$|\phi\rangle^{AA^\prime BB^\prime} = \sum_\mu\sqrt{p_\mu}|g_\mu\rangle^{AB}|\tilde{g}_\mu\rangle^{A^\prime B^\prime},$$

I wonder that $|g_\mu\rangle^{AB}$ (and $|g_\mu\rangle^{A^\prime B^\prime}$) are close to separable state (like $|e_i\rangle^A|f_j\rangle^B$ (and $|\tilde{e}_i\rangle^{A^\prime }|\tilde{f}_j\rangle^{B^\prime}$) ) or not?


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This is immediately disproved by the structure of the counterexamples to your previous question Does proximity of two bipartite states in a norm force high overlap between the elements of the Schmidt bases?: Schmidt decompositions need not be close to each other in any way. You can copy either of those examples one-by-one and will find a counterexample.

To give a concrete example, with the four Bell states $|\phi_i\rangle$, you have $$ \sum \tfrac12|\phi_i\rangle_{AB}\otimes |\phi_i\rangle_{A'B'} = \tfrac12(\lvert 00\rangle + \lvert 11)_{AA'}\otimes (\lvert 00\rangle + \lvert 11)_{BB'}\ , $$ i.e. separable in that partition. If you now perturb the Schmidt coefficients of the l.h.s., the Schmidt decomposition will become unique, with maximally entangled Schmidt vectors in the $AA'$ vs. $BB'$ partition.


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