# When does this weaker version of this inequality become an equality?

Let $A$ and $B$ be Hermitian operators and $\langle \rangle$ be the usual quantum mechanics inner product. We know from the inequality that:

$$|\langle A B\rangle| \leq \left\langle A \right\rangle \left\langle B \right\rangle$$

But a weaker inequality would be:

$$\implies \Re(\langle A B \rangle) \leq |\langle A B\rangle| \leq \left\langle A\right\rangle \left\langle B \right\rangle$$

$$\implies \Re(\langle A B \rangle) \leq \left\langle A\right\rangle \left\langle B \right\rangle$$

But when does this weaker version of this inequality become an equality?

$$\Re(\langle A B \rangle) = \left\langle A\right\rangle \left\langle B \right\rangle$$

• Does $A= k B$ where $k>0$ work? – M. T Sep 8 '16 at 8:41
• what's $⟨A⟩$ ? you mean $|A|$ ? – marmouset Sep 8 '16 at 8:44
• Sorry, I seem to have confused somethings and have now retyped the question. – drewdles Sep 8 '16 at 9:11

In fact, it is common in the proof of the Cauchy-Schwarz inequality to note that we can only have $$|\langle A \mid B \rangle| = \langle A\rangle \langle B \rangle$$ if $A$ and $B$ are multiples of each other. That is, either $A = 0$ or $B = kA$ for some $k \in \Bbb C$.
Note however that we can only have $\Re(\langle A \mid B \rangle) = \langle A \rangle \langle B \rangle$ if we also have $|\langle A \mid B \rangle| = \langle A\rangle \langle B \rangle$, which is easy to see from your $3$-part inequality. So, if $\Re(\langle A \mid B \rangle) = \langle A \rangle \langle B \rangle$, we may conclude that $A$ and $B$ are multiples of each other.
Use the above to conclude that $\Re(\langle A \mid B \rangle) = \langle A \rangle \langle B \rangle$ if and only if $A = 0$ or $B = kA$ for some $k \in \Bbb R$.