I Multiplying three vector operators

[Kashmir]

Given vector operators as
$$\mathbf{A} = (A_{1}, A_{2} ,A_{3}) $$

$$\mathbf{B} = (B_{1}, B_{2} ,B_{3}) $$

$$\mathbf{C} = (C_{1}, C_{2} ,C_{3}) $$

I know that for two vector operators $$\begin{equation}
\mathbf{Q} \mathbf{P} = \sum_{\alpha = 1}^{3} Q_{\alpha} P_{\alpha}
\end{equation}$$

What is $$\mathbf{A}\mathbf{B}\mathbf{C}$$ in component form?

 

[topsquark]

Given vector operators as
$$\mathbf{A} = (A_{1}, A_{2} ,A_{3}) $$

$$\mathbf{B} = (B_{1}, B_{2} ,B_{3}) $$

$$\mathbf{C} = (C_{1}, C_{2} ,C_{3}) $$

I know that for two vector operators $$\begin{equation}
\mathbf{Q} \mathbf{P} = \sum_{\alpha = 1}^{3} Q_{\alpha} P_{\alpha}
\end{equation}$$

What is $$\mathbf{A}\mathbf{B}\mathbf{C}$$ in component form?

You didn't specify what kind of product you are looking at. (There are several.) But from appearances you are doing an inner (or "dot") product.

So
$\displaystyle \textbf{A} \cdot \textbf{B} = \sum_{i = 1}^3 A_i B_i = \lambda$, which is a scalar, so

$\left ( \textbf{A} \cdot \textbf{B} \right ) \textbf{C} = \left ( \begin{matrix} \lambda C_1 \\ \lambda C_2 \\ \lambda C_3 \end{matrix} \right )$

There is also the cross product, but you have to specify which pair you are multiplying first as the cross product is not associative. And finally, there's the tensor product, but that one's easy as there isn't any mixing: $\textbf{A} \otimes \textbf{B} \otimes \textbf{C}$. There isn't really much of a component form for this, just
$\left ( \begin{matrix} A_1 \\ A_2 \\ A_3 \end{matrix} \right ) \otimes \left ( \begin{matrix} B_1 \\ B_2 \\ B_3 \end{matrix} \right ) \otimes \left ( \begin{matrix} C_1 \\ C_2 \\ C_3 \end{matrix} \right )$

-Dan

 

[Dale]

Given vector operators as
$$\mathbf{A} = (A_{1}, A_{2} ,A_{3}) $$

$$\mathbf{B} = (B_{1}, B_{2} ,B_{3}) $$

$$\mathbf{C} = (C_{1}, C_{2} ,C_{3}) $$

I know that for two vector operators $$\begin{equation}
\mathbf{Q} \mathbf{P} = \sum_{\alpha = 1}^{3} Q_{\alpha} P_{\alpha}
\end{equation}$$

What is $$\mathbf{A}\mathbf{B}\mathbf{C}$$ in component form?

I wouldn't write it this way because it is ambiguous. I would write $$\begin{equation}
\mathbf{Q} \cdot \mathbf{P} = \sum_{\alpha = 1}^{3} Q_{\alpha} P_{\alpha}
\end{equation}$$ for clarity. This distinguishes between an operation $\mathbf{Q} \cdot \mathbf{P}$ which takes two vectors and produces a scalar and an operation $\lambda \mathbf{Q}$ which takes a vector and a scalar and produces a vector.

Note that $(\mathbf{A} \cdot \mathbf{B}) \mathbf{C} \ne \mathbf{A}(\mathbf{B} \cdot \mathbf{C})$ and also note that $\mathbf{A} \cdot \mathbf{B} \cdot \mathbf{C}$ does not exist nor does $\mathbf{A} \mathbf{B} \mathbf{C}$

 

[Kashmir]

Thank you.

I'm looking for the product which appears as the result of quantisation rule which takes the classical expression to the quantum one.

Suppose the classical quantites we have are $$\mathscr A_1(\mathbf{r,p,t})=\mathbf{r(p.r)}$$

$$\mathscr A_2(\mathbf{r,p,t})=\mathbf{(r.p)r}$$

What will be the quantum operators $\mathbf A_1$,$\mathbf A_2$ corresponding to $\mathscr A_1$ and $\mathscr A_2$?