# Commutive property of the Bra-ket notation

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 $$\langle a\vert \hat A\vert b\rangle= \langle a\vert\left[\alpha \vert c\rangle +\beta \vert f\rangle\right] = \langle a\vert\alpha \vert c\rangle +\langle a\vert \beta \vert f\rangle =\alpha \langle a\vert c\rangle +\beta \langle a\vert f\rangle.$$
• 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 $\langle A|\boldsymbol{A}|B\rangle = \langle A| \cdot \boldsymbol{A} \cdot |B\rangle$ Dec 1, 2018 at 17:48
• not sure why you need the $"\cdot"$ in there since $3$ is a scalar. Dec 1, 2018 at 17:50
• Oh, so more like $\langle A|3|B\rangle = \langle A| \cdot 3 |B\rangle$? Dec 1, 2018 at 17:50
• 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 $\langle a\vert b\rangle=\vec a\cdot \vec b$. Dec 1, 2018 at 17:51
• 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 $|\Psi \rangle$. Dec 1, 2018 at 17:56