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Irreducible decomposition of a Dyadic operator
Recently, I've been studying $SO(3)$ Group Theory for physics application and for a couple of days I'm struggling to understand how to get a 3D cartesian tensor, like: $T_{ij}=a_i\,b_j$ to be decomposed as $$\frac{\vec{a}\cdot\vec{b}}{3}\delta_{ij}+\frac{a_i\,b_j-a_j\,b_i}{2}+\left(\frac{a_i\,b_j+a_i\,b_j}{2}-\frac{\vec{a}\cdot\vec{b}}{3}\delta_{ij}\right),$$ i.e., this irreducible representation is invariant under rotation and it's commonly said that $3\otimes3=1+3+5$. I already know that a second order tensor can be decomposed as antisymmetric and symmetric parts, but I don't get it why in decomposition above there is a trace part (you may say that is because is a traceless tensor, but why it has to be?).
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$\begingroup$ The traceful part is proportional to I, and so invariant under rotations, $R^T I R=I$, whereas the symmetric traceless part S does transform nontrivially like a quintet, $R^T S R= S'$. You need the two decoupled irreps then to be trace-orthogonal, which fixes the normalizations. $\endgroup$
– Cosmas Zachos
Dec 3, 2020 at 15:34
The so-called natural form of a rank $N$ tensor is symmetric in all its indices and is traceless. It has $2N+1$ degrees of freedom that transform like $2N+1$ order-$N$ spherical harmonics: $Y_{l=N}^{m}(\theta, \phi)$.
So for rank-2, the natural form is:
$$ N_{ij} = \frac 1 2 (T_{ij}+T_{ji}) - \frac 1 3 \delta_{ij} T_{kk}$$
$$ N_{ij} = S_{ij} - \frac 1 3 \delta_{ij} T_{kk}$$
where $S_{ij}$ refers to the symmetric part (but not trace free).
The relation between spherical($T^{l,m}$) form and cartesian is:
$$ T^{2,\pm 2} = \frac 1 2 [S_{xx}-S_{yy}\pm 2iS_{xy}]$$
$$ T^{2,\pm 1} = \frac 1 2 [S_{zx}\pm iS_{yz}]$$
$$ T^{2,0} = \sqrt{\frac 2 3} S_{zz}$$
You will find that the $T^{2,m}$ are rotated just as the $Y_2^m(\theta,\phi)$ are.
At higher ranks, it gets involved. The symmetrized rank-3 tensor is:
$$ S_{ijk} = \frac 1 6 [T_{ijk}+T_{kij}+T_{jki}+T_{kji}+T_{jik}+T_{ikj}]$$
and the trace free version is:
$$ N_{ijk} = S_{ijk} - \frac 1 {30}[(\delta_{ij}+\delta_{ji})(T_{llk}+T_{lkl}+T_{kll})+ (\delta_{ik}+\delta_{ki})(T_{llj}+T_{ljl}+T_{jll})+(\delta_{kj}+\delta_{jk})(T_{lli}+T_{lil}+T_{ill})]$$
and the $(2\cdot 3+1)=7$ spherical tensors that transform like $Y_3^m$ are (according to my notes):
$$ T^{3,\pm3} = \frac 1 {\sqrt 8}[(-S_{xxx}+3S_{xyy}) \mp i(S_{yyy}-S_{xxy})]$$
$$ T^{3,\pm2} = \frac 1 2[-S_{xxz}-S_{yyz} \mp 2iS_{xyz}]$$
$$ T^{3,\pm 1} = \frac {\sqrt{15}} 3\big(\frac 1 {\sqrt 2}[\mp S_{zzz}-iS_{zzz}] + \frac 1 {\sqrt 8}[\mp(S_{xxx}-S_{xyy})+i(S_{yyy}\pm S_{xxy}]\big) $$
$$ T^{3,0} =\frac {\sqrt{10}} 3[\frac 1 {\sqrt 2}(S_{xzz}+iS_{yzz}) + S_{zzz}]$$
Rank-3 breaks down according to:
$${\bf 3} \otimes {\bf 3} \otimes {\bf 3} = {\bf 10} \oplus {\bf 8} \oplus {\bf 8} \oplus {\bf 1} $$
(where the ${\bf 1}$ is the familiar $\epsilon_{ijk})$. So what happened in creating the natural form tensor $S_{ijk}$ is that we subtracted off a vector trace: $S_{ijj}$ from the original 10 degrees-of-freedom:
$${\bf 10} \rightarrow {\bf 7} + {\bf 3} $$
Likewise, the octet is split into a rank-2 like object and a vector trace:
$${\bf 8} \rightarrow {\bf 5} + {\bf 3} $$
See Physical Review A, "Irreducible fourt-rank Cartesian tensors", Andrews and Ghoul, Volume 25, Number 5, Page 2647, [1992], for Rank-4, though it has typos in the indices, and looks something like this:
lurscher
JEBJEB
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