1
$\begingroup$

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?).

$\endgroup$
1
  • 1
    $\begingroup$ 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. $\endgroup$ Dec 3, 2020 at 15:34

1 Answer 1

1
$\begingroup$

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:

$$ N_{ijk} = S_{ijk} - \frac 1 {30}[(\delta_{ij}+\delta_{ji})(T_{llk}+T_{lkl}+T_{kll})+ (\delta_{ik}+\delta_{ki})(T_{llj}+T_{ljl}+T_{jll})+(\delta_{kj}+\delta_{jk})(T_{lli}+T_{lil}+T_{ill})]$$

where

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:

enter image description here

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.