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I know, that for a compound system $ |\psi \rangle_{AB} $ we can find the Schmidt basis, which is an unique one. Is it at the same time the basis, in which the two subsystems are minimally entangled?

If so, how this can be proved / disproved?

I think it would make sense to say, that when the Schmidt rank is equal to 1, the system is separable, because in the basis minimizing the entanglement we can represent the state $|\psi \rangle_{AB} $ as a product of two substates $ |\psi \rangle_A \otimes |\psi \rangle_B $.

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The Schmidt basis is just a special basis in which you can write a given bipartite state $|\psi\rangle_{AB}$. The state is always the same, regardless of which representation you choose (i.e., in which basis you express it). Thus, the entanglement - which is a property of the state, not of the chosen representation - does not change.

On the other hand, the Schmidt decomposition allows you to read off the entanglement of the state easily.

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  • $\begingroup$ I've formulated my question badly (of course entanglement is always the same, regardless of the representation...), but your answer clarifies everything. Thank you! $\endgroup$ – brzepkowski May 19 at 17:33

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