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Lets say $\rho,\sigma$ satisfy $F(\rho,\sigma)=0$, i.e., they are quantum states living on orthogonal supports. What can we say about $F(\text{Tr}_A(\rho),\text{Tr}_B(\sigma))$?

I am looking for upper bounds here, so we clearly know that $$F(\text{Tr}_A(\rho),\text{Tr}_B(\sigma))\geq F(\rho,\sigma)=0$$ but this is trivial. Can we prove any upper bound? I don't mind if we also assume $\rho,\sigma$ are pure states, in which case can we give a non-trivial upper bound?

One example I tried was orthogonal EPR pairs which satisfied RHS=0 and LHS (i.e., after taking partial trace)=$\frac{1}{2}$. But is this the maximum one can achieve for orthogonal quantum states?

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    $\begingroup$ Do you really mean Tr_A in the first and Tr_B in the second state?? This is completely unreasonable, generally these are not even the same space! $\endgroup$ – Norbert Schuch Aug 29 '20 at 19:08
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The Bell states are orthogonal and their partial trace is equal - the maximally mixed states - and thus indistinguishable. So yes, this is the maximum you can achieve. Another extremal example could be a tensor product $|a\rangle\otimes|b_i\rangle$ with two orthogonal states $|b_i\rangle$.

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