Dyade Dirac Notation: Why Last Equation?

[LagrangeEuler]

\{\vec{A},\vec{B}\}\cdot \vec{C}=\vec{A}(\vec{B}\cdot \vec{C})
\vec{C} \cdot \{\vec{A},\vec{B}\}=(\vec{C}\cdot \vec{A}) \vec{B}

I want to write dyade in Dirac notation.

(|\vec{A}\rangle\langle\vec{B}|)|\vec{C}\rangle= |\vec{A}\rangle\langle\vec{B}|\vec{C}\rangle
\langle\vec{C}|(|\vec{A}\rangle\langle\vec{B}|)=< \vec{C} |\vec{A}\rangle|\vec{B}\rangle

Why not

\langle \vec{C} |\vec{A}\rangle\langle \vec{B}|

in last equation?

 

[dextercioby]

It should be, it should have the same 'bra-keting' as the C in the LHS.

 

[LagrangeEuler]

I'm confused. In QM |\psi \rangle and \langle \psi | are vectors from some vector space and his dual respectively. But in some tensor analyses I don't see difference between |\vec{A}\rangle and \langle \vec{A}|.
In this case precisely. I have some number \langle|\rangle which multiply vector.

 

[Fredrik]

I'm confused. In QM |\psi \rangle and \langle \psi | are vectors from some vector space and his dual respectively. But in some tensor analyses I don't see difference between |\vec{A}\rangle and \langle \vec{A}|.
In this case precisely. I have some number \langle|\rangle which multiply vector.

I don't follow what you're saying here. Yes, $\langle C|A\rangle$ is a number. Why is that a problem?

There's no need to use the arrow notation for vectors here (unless what you have in mind is that $\vec A$ is a vector in some other vector space $\mathbb R^3$), since kets are always vectors.