In a nutshell my question is: "In the bracket notation, we can operate in the bra and in the ket with the same operator like $$\langle a|(A|b\rangle)=(\langle a|A)|b\rangle~?$$ So this operations are correct? $$A|b_i\rangle=\sqrt{b_i}|b_i-1\rangle\leftrightarrow \langle b_i| A^\dagger=(\sqrt{b_i})^*\langle b_i-1|;$$

$$A^\dagger|b_i\rangle=\sqrt{b_i+1}|b_i+1\rangle\leftrightarrow \langle b_i|A=(\sqrt{b_i+1})^*\langle b_i+1|; $$ then applying the first equation above in the Ket $$ \langle b_i|A|b_i+1\rangle=\langle b_1|\big(A|b_i+1\rangle\big)=\langle b_1|\big(\sqrt{b_i+1}|b_i\rangle \big)=\sqrt{b_i+1} $$ Now applying the second relation in the Bra $$ \langle b_i|A|b_i+1\rangle=\big(\langle b_i|A\big)|b_i+1\rangle= (\sqrt{b_i+1})^*\langle b_i+1|b_i+1\rangle=(\sqrt{b_i+1})^* $$ For that to be true it's required that the values $\in \mathcal{R}$.


1 Answer 1


We have to be careful when we use bra-ket notation to deal with non-hermitian operators. In these cases, it is more convinient to use $(\cdot,\cdot)$ notation.

Being $a$ and $b$ two states, and $A$ an operator,

$$(a,Ab)=(A^\dagger a,b),$$

which in bra-ket notation would be $$\langle a|A|b\rangle=\langle a|\Big(A|b\rangle\Big)=\Big(A^\dagger|a\rangle\Big)^\dagger|b\rangle.$$

Then, you can see that your last equation is right only if your $A$ was hermitian.

  • $\begingroup$ I used the second equation and took its hermitian as you said. I understood then. Thanks :) $\endgroup$ Mar 4, 2021 at 1:01

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