I'm struggling when it comes to understanding the commutive properties of the Bra-ket notation in quantum mechanics. I understand how to work with constants, bra and kets. However, the second I start introducing eigen-equations such as $$\hat{x}|x'\rangle = x'|x'\rangle$$ to solve problems like $$\langle\psi_p|\hat{x}|x'\rangle = x'\psi_p^*(x')$$ I instantly lose grip on the rules for the Bra-ket notation. Am I allowed to simply pull the $x'$ out of the bracket like this: $\langle\psi_p|x'|x'\rangle=x'\langle\psi_p|x'\rangle$?

My book only state the rules for operators between a bra and a ket, $\langle a|\boldsymbol{A}|b\rangle$, but not how to handle problem including eigen-equations such as this one.


When writing $\hat x\vert x'\rangle=x'\vert x'\rangle$, the $x'$ is actually a number (aka a scalar) and so can be moved about like a regular number, so that $$ \langle \psi_p\vert\hat x\vert x'\rangle = \langle \psi_p \vert x'\vert x'\rangle = x'\langle\psi_p\vert x'\rangle $$ because $x'\in \mathbb{R}$, much in the way that $\langle b\vert 3\vert b\rangle=3\langle a\vert b\rangle$.

Indeed if $\hat A\vert b\rangle = \alpha \vert c\rangle +\beta \vert f\rangle$ then $$ \langle a\vert \hat A\vert b\rangle= \langle a\vert\left[\alpha \vert c\rangle +\beta \vert f\rangle\right] = \langle a\vert\alpha \vert c\rangle +\langle a\vert \beta \vert f\rangle =\alpha \langle a\vert c\rangle +\beta \langle a\vert f\rangle. $$

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  • $\begingroup$ Okai, great, but for clarity, is $\langle A|3|B\rangle = \langle A| \cdot 3 \cdot |B\rangle$? And if there is an operator instead $\langle A|\boldsymbol{A}|B\rangle = \langle A| \cdot \boldsymbol{A} \cdot |B\rangle$ $\endgroup$ – Gjert Dec 1 '18 at 17:48
  • $\begingroup$ not sure why you need the $"\cdot" $ in there since $3$ is a scalar. $\endgroup$ – ZeroTheHero Dec 1 '18 at 17:50
  • $\begingroup$ Oh, so more like $\langle A|3|B\rangle = \langle A| \cdot 3 |B\rangle$? $\endgroup$ – Gjert Dec 1 '18 at 17:50
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    $\begingroup$ The bra vector already implicitly contains a dot product so there is no need for "$\cdot$" in there. In terms of usual vector one would have, for instance, $\langle a\vert \to \vec a\cdot$ so that $\langle a\vert b\rangle=\vec a\cdot \vec b$. $\endgroup$ – ZeroTheHero Dec 1 '18 at 17:51
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    $\begingroup$ Extra note: It could also be viewed as follows. It is hidden in the fact that one can write $\hat{x}= x\hat{1}$ in the position representation. Therefore the $x$ is a scalar and can be pulled out of the bracket and the identity maps $|\Psi \rangle$ to $|\Psi \rangle$. $\endgroup$ – Mathphys meister Dec 1 '18 at 17:56

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