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?

Edit: 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 quantity 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 operator $\mathbf A_1$,$\mathbf A_2$ corresponding to $\mathscr A_1$ and $\mathscr A_2$?

  • $\begingroup$ What kind of product is it? $\endgroup$ Commented Jan 9, 2023 at 14:24
  • $\begingroup$ "I know that..." - who uses this convention? I would assume any convention of just putting vectors next to each other meant outer product $QP^{\alpha \beta}=Q^\alpha P^\beta$, (maybe symmetrised over the indices) $\endgroup$
    – jacob1729
    Commented Jan 9, 2023 at 14:31
  • $\begingroup$ Concerning your edit: do you want the operators to be Hermitian, or not? Have you specified your quantization ordering prescription? Where do these symbols come from? $\endgroup$ Commented Jan 9, 2023 at 17:17
  • $\begingroup$ @CosmasZachos, Yes I want them to be hermitian. I have not found ordering prescription in my text book. These symbols come from studying how we use classical expression to get the relevant quantum operators. $\endgroup$
    – Kashmiri
    Commented Jan 9, 2023 at 17:54
  • $\begingroup$ Isn't it obvious that $\mathbf A_1^\dagger =\mathbf A_2$, so their symmetrization, $\{ \mathbf A_1 ,\mathbf A_2\}$, is self-adjoint? $\endgroup$ Commented Jan 9, 2023 at 18:41

1 Answer 1


Your expression is not really defined.

Your formula your have stated is the definition of the dot product.

Either your expression is

$$(\vec{A} \cdot \vec{B}) \vec{C}$$

In which case it is the vector $$(A_{1}B_{1} + A_{2}B_{2} + A_{3}B_{3}) \begin{bmatrix} C_{1}\\C_{2}\\C_{3}\end{bmatrix} = \begin{bmatrix}(A_{1}B_{1}C_{1} + A_{2}B_{2}C_{1} + A_{3}B_{3}C_{1}\\ (A_{1}B_{1}C_{2} + A_{2}B_{2}C_{2} + A_{3}B_{3}C_{2}\\A_{1}B_{1}C_{3} + A_{2}B_{2}C_{3} + A_{3}B_{3}C_{3} \end{bmatrix} $$


$$\vec{A} (\vec{B} \cdot \vec{C})= (B_{1}C_{1} + B_{2}C_{2} + B_{3}C_{3}) \begin{bmatrix} A_{1}\\A_{2}\\A_{3}\end{bmatrix}$$

$$\begin{bmatrix}(B_{1}C_{1}A_{1} + B_{2}C_{2}A_{1} + B_{3}B_{3}A_{1}\\ B_{1}C_{1}A_{2} + B_{2}C_{2}A_{2} + B_{3}C_{3}A_{2}\\B_{1}C_{1}A_{3} + B_{2}C_{2}A_{3} + B_{3}C_{3}A_{3} \end{bmatrix} $$

Or as other answers have stated it is cross product


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