1. May 4, 2012

### 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?

2. May 4, 2012

### dextercioby

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

3. May 4, 2012

### 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.

4. May 4, 2012

### Fredrik

Staff Emeritus
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.