Physics Asked by Vinicius Fuckner Linhares on May 2, 2021
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 $$langle a|(A|brangle)=(langle a|A)|brangle~?$$ So this operations are correct?
$$A|b_irangle=sqrt{b_i}|b_i-1rangleleftrightarrow langle b_i| A^dagger=(sqrt{b_i})^*langle b_i-1|;$$
$$A^dagger|b_irangle=sqrt{b_i+1}|b_i+1rangleleftrightarrow langle b_i|A=(sqrt{b_i+1})^*langle b_i+1|; $$
then applying the first equation above in the Ket $$ langle b_i|A|b_i+1rangle=langle b_1|big(A|b_i+1ranglebig)=langle b_1|big(sqrt{b_i+1}|b_irangle big)=sqrt{b_i+1} $$
Now applying the second relation in the Bra
$$ langle b_i|A|b_i+1rangle=big(langle b_i|Abig)|b_i+1rangle= (sqrt{b_i+1})^*langle b_i+1|b_i+1rangle=(sqrt{b_i+1})^* $$
For that to be true it’s required that the values $in mathcal{R}$.
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 $$langle a|A|brangle=langle a|Big(A|brangleBig)=Big(A^dagger|arangleBig)^dagger|brangle.$$
Then, you can see that your last equation is right only if your $A$ was hermitian.
Answered by AFG on May 2, 2021
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