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As stated in the title, let $M$ be a linear operator on a finite bipartite Hilbert space. Suppose $0\leq M^{AB}\leq \mathbb{I}$ and $0\leq M^A,M^B\leq\mathbb{I}$, where $M^A=\mathrm{Tr}_B\left(M^{AB}\right)$ and $M^B=\mathrm{Tr}_A\left(M^{AB}\right)$. Is it always true that $$ M^{AB} \leq M^A\otimes \mathbb{I}_B? $$ It trivially holds for product operators, that is $M^{AB} = M^A\otimes M^B$, but the general statement is not clear to me.

Any help is appreciated.

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2 Answers 2

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Not true.
Just take Bell state $M^{AB} = |v\rangle\langle v|$, where $|v\rangle = \frac{1}{\sqrt{2}}(|00\rangle + |11\rangle)$. It has eigenvalue 1.
But $M^A \otimes I_2= \frac{1}{2}I_2 \otimes I_2 = \frac{1}{2}I_4$.

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  • $\begingroup$ $M^{AB}$ doesn't even have to be entangled, it can be the product of two pure states and still have eigenvalue 1. $\endgroup$ Jun 2, 2021 at 20:41
  • $\begingroup$ But in such a case the inequality will hold. $\endgroup$
    – Danylo Y
    Jun 2, 2021 at 21:02
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    $\begingroup$ Oh, you are assuming that $M^A=\mathrm{Tr}_B (M^{AB})$? That wasn't explicit in the question but it definitely helps motivate the question. $\endgroup$ Jun 2, 2021 at 22:14
  • $\begingroup$ @QuantumMechanic Yes, that is indeed what I meant. Thanks for pointing out! $\endgroup$
    – user16106
    Jun 3, 2021 at 7:55
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No, but it is true that $$M^{AB} \le d\, M^A \otimes \mathbb{I}_B,$$ where $d$ is the minimum of $d_A$ and $d_B$. This inequality is tight in the sense that entangled states saturate it, as the other answer shows. A proof can be found in Appendix A of this paper.

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