# For tetrapartite state, and another way of decomposition, is the Schmidt basis separable?

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?

## 1 Answer

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.