Question
Let $Q$ be an operator. Under what circumstances is the complex number $\langle a|Q| b\rangle$ equal to the complex number $(\langle b|Q| a\rangle)$ ' for any states $|a\rangle$ and $|b\rangle ?$
Step 1
First, we know that $\langle a|Q| b\rangle$ is a complex number, which we can denote as $c$. So, $c = \langle a|Q| b\rangle$. Show more…
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If $a, b, c, p, q, r$ are three non-zero complex numbers such that $\frac{p}{a}+\frac{q}{b}+\frac{r}{c}=1+i$ and $\frac{a}{p}+\frac{b}{q}+\frac{c}{r}=0$, then value of $\frac{p^{2}}{a^{2}}+\frac{q^{2}}{b^{2}}+\frac{r^{2}}{c^{2}}$ is (A) 0 (B) $-1$ (C) $2 i$ (D) $-2 i$
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