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For two operators $A, B$ defined on Hibert space $H_n$, the state is $\rho$, then there is

$$\langle AB \rangle +\sqrt{\langle A \rangle - (\langle A \rangle)^2}\sqrt{\langle B \rangle - (\langle B \rangle)^2} \geq \langle A \rangle \langle B \rangle$$

In the derivation of (23) in the following paper (https://arxiv.org/abs/0705.2024), it was claimed to be one form of Cauchy-Schwarz inequality on expectation values of operators, but I failed to see why it is true.

Any insights would be appreciated.

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    $\begingroup$ I'm guessing you're missing a squared inside the square roots? if you have $\langle A^2\rangle-\langle A\rangle^2$ in there (and same for $B$) this is essentially Heisenberg's uncertainty principle, see en.wikipedia.org/wiki/…. Also, what is "it" here? Can you edit the post to link to the source? $\endgroup$
    – glS
    Commented Dec 3, 2022 at 11:03
  • $\begingroup$ @glS Thanks for the reply. In the original derivation, it is indeed $\langle A \rangle$. I attach the source link in the edited question. $\endgroup$
    – QubitTy
    Commented Dec 3, 2022 at 20:26
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    $\begingroup$ it's not super obvious from that equation in the paper. Maybe $A,B$ are projections, and therefore $A=A^2$? $\endgroup$
    – glS
    Commented Dec 4, 2022 at 10:57
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    $\begingroup$ @glS A and B are projectors. $\endgroup$
    – QubitTy
    Commented Dec 5, 2022 at 4:24

1 Answer 1

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I think the inequality you gave $$\tag{1} \langle AB \rangle +\sqrt{\langle A \rangle - (\langle A \rangle)^2}\sqrt{\langle B \rangle - (\langle B \rangle)^2} \geq \langle A \rangle \langle B \rangle$$ makes no sense in general.

Take $n=2$, $A=2I$, $B=\frac{1}{2}I$ and let $\rho$ be a maximally mixed state, i.e. $\rho = \frac{1}{2}I$.
Then $\sqrt{\langle A \rangle - \langle A \rangle^2} = \sqrt{-2}$ is a complex number, and $\sqrt{\langle B \rangle - \langle B \rangle^2}$ is a real number. Hence, the LHS of Eq (1) is a complex number. Essentially, the inequality attempts to compare a complex number on the LHS with a real number on RHS.

For this to make sense, much stronger assumptions may be needed.

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    $\begingroup$ while true, the closely related inequality $\sqrt{\langle A^2\rangle-\langle A\rangle^2} \sqrt{\langle B^2\rangle-\langle B\rangle^2} \ge |\langle A B\rangle - \langle A\rangle \langle B \rangle|$ does indeed hold, and is nothing but Heisenberg's uncertainty principle in a slightly more general form. Well, it holds for pure states at least, and it's a direct application of CS there. Essentially $\langle \psi,AB \psi\rangle|^2\le \|A \psi\|^2 \|B\psi\|^2$ with suitably chosen operators $A,B$ and with $\psi$ the pure state. $\endgroup$
    – glS
    Commented Dec 4, 2022 at 17:12
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    $\begingroup$ the given equation might reduce to this if, say, $A,B$ are projectors, and the state pure, and for some reason $\langle AB\rangle \le \langle A\rangle \langle B\rangle$. Note that in such scenarios $\langle A\rangle\ge \langle A\rangle^2$ so the square roots can be safely assumed to be real $\endgroup$
    – glS
    Commented Dec 4, 2022 at 17:15
  • $\begingroup$ @glS ah interesting! Thanks! What do I google to see the proof of that? $\endgroup$
    – MonteNero
    Commented Dec 5, 2022 at 3:40

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