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The form $$ {\bf r} \cdot {\bf r} = \left|{\bf r}\right|^2 = \sum_{i,j} r_i r_j {\bf e}_i \cdot {\bf e}_j = \sum_{i,j} r_i r_j \delta_{ij} = \sum_i r_i r_i $$

is defined via a bilinear form that takes two vectors and outputs a real number with certain conditions. This quantity is usually referred to as the square of a vector. I have encountered here forms such as ${\bf r}^3, {\bf r}^4, ...$. Now ${\bf r} \cdot {\bf r} \equiv {\bf r}^2$ and is $SO(3)$ invariant. How does one show ${\bf r}^3$ is not $SO(3)$ invariant and consequently only even powers are invariant under $SO(3)$.

The question is tantamount to: How is ${\bf r}^j$ ($j>2$) defined?

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    $\begingroup$ The linked file is 27 pages long; can you give a more precise location than "here"? $\endgroup$ Feb 15, 2022 at 17:36
  • $\begingroup$ Are you asking about "rotational symmetry of the Hamiltonian" on page 12? It might be better to ask on physics stack exchange then. $\endgroup$ Feb 15, 2022 at 17:39

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You want $\mathbf{r}^{j+k}=\mathbf{r}^j\otimes\mathbf{r}^k$ for a suitable definition of $\otimes$, right? The definition$$\mathbf{r}^{2k}=(\mathbf{r}\cdot\mathbf{r})^k,\,\mathbf{r}^{2k+1}=\mathbf{r}^{2k}r$$works (if $\otimes$ multiplies scalars by scalars and vectors in the usual ways, and takes the dot product of two vectors). This is the standard definition of $\mathbf{r}^n$. However, since $\mathbf{r}^{2k}=r^{2k}$ with $r:=|\mathbf{r}|$, $\mathbf{r}^{2k}$ is often simply written as $r^{2k}$.

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  • $\begingroup$ So basically for $n$ even this is a scalar, for $n$ odd a vector, correct? Then rotational invariance for even $n$ follows from being a power of the scalar length, which is rotationally invariant, whereas non-invariance for odd $n$ follow from being a vector? $\endgroup$ Feb 15, 2022 at 17:40
  • $\begingroup$ @Chill2Macht That's right, yes. $\endgroup$
    – J.G.
    Feb 15, 2022 at 17:40
  • $\begingroup$ @J.G. So an odd power is defined to be a vector? $\endgroup$
    – TheDawg
    Feb 15, 2022 at 18:14
  • $\begingroup$ @TheDawg That's right, yes. $\endgroup$
    – J.G.
    Feb 15, 2022 at 18:31
  • $\begingroup$ @J.G. Right, but then you cant even define the Hamiltonian for odd powers by construction, so the $SO(3)$ argument made is not a necessary condition for odd powers to vanish as they cant appear by construction regardless of rotational invariance. So why even talk about rotational invariance and implying odd powers are not allowed? $\endgroup$
    – TheDawg
    Feb 15, 2022 at 18:56

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