Undergrad Multiplying three vector operators

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The discussion centers on the multiplication of three vector operators, specifically how to express the product of vector operators A, B, and C in component form. It is clarified that the type of product—inner, cross, or tensor—must be specified, as they yield different results. The inner product results in a scalar that can be multiplied by another vector, while the cross product is not associative and requires careful ordering. Additionally, the tensor product does not mix components but combines them in a straightforward manner. The conversation also touches on the quantum operators corresponding to classical quantities, indicating a transition from classical to quantum mechanics.
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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?
 
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Kashmir said:
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
 
Kashmir said:
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}##
 
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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##?
 
Insights auto threads is broken atm, so I'm manually creating these for new Insight articles. Towards the end of the first lecture for the Qiskit Global Summer School 2025, Foundations of Quantum Mechanics, Olivia Lanes (Global Lead, Content and Education IBM) stated... Source: https://www.physicsforums.com/insights/quantum-entanglement-is-a-kinematic-fact-not-a-dynamical-effect/ by @RUTA

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