# Cauchy-Schwarz inequality of expectation values of operators

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.

• 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?
– glS
Dec 3, 2022 at 11:03
• @glS Thanks for the reply. In the original derivation, it is indeed $\langle A \rangle$. I attach the source link in the edited question. Dec 3, 2022 at 20:26
• it's not super obvious from that equation in the paper. Maybe $A,B$ are projections, and therefore $A=A^2$?
– glS
Dec 4, 2022 at 10:57
• @glS A and B are projectors. Dec 5, 2022 at 4:24

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.

• 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.
– glS
Dec 4, 2022 at 17:12
• 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
– glS
Dec 4, 2022 at 17:15
• @glS ah interesting! Thanks! What do I google to see the proof of that? Dec 5, 2022 at 3:40