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In 3D, the rank two tensorial physical quantities, for example, the electric susceptibility, the conductivity, the stress tensor etc, are in general, not irreducible representations i.e. neither traceless symmetric tensors nor antisymmetric tensors but an arbitrary tensor of rank two.

From the point of view of physics why is it then important to find out the irreducible representations of $SO(3)$ and their transformation properties? I mean, suppose I determine the transformations of the irreps (e.g. traceless symmetric tensors or antisymmetric tensors of rank two), where and how do I use it? What do I use it for? How is it useful in doing or understanding nature? If possible, please provide some insight with examples.

Note I know that under the action of a most general rotation, the five (traceless) symmetric combinations of the components of a rank two tensor $T$ linearly mix among themselves. Ditto is the case with the three antisymmetric combinations of the components of $T$. The trace is also invariant under rotation (i.e. scalar). In particular, I know that $${\bf 9}={\bf 5}\oplus{\bf 3}\oplus{\bf 1}.$$ But the thing I am asking above is that what do we understand physically by knowing this or how is this useful?

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In a nutshell: If we know the tensor/spinor/vector components in one coordinate system, and we know how it transforms, then we know it in every coordinate system, which is pretty useful. This sort of reasoning carries over to many group representations.

Useful group-theoretic information is provided by knowledge of e.g. Casimirs, fusion rules, Clebsch-Gordan coefficients, etc.

Example with the isospin group $G=SU(2)$ in the standard model: One can use group & representation theory to find the branching ratio

$$ \frac{\Gamma(\Delta^0\rightarrow n+\pi^0 )}{\Gamma( \Delta^0\rightarrow p+\pi^- )} ~=~ 2, \qquad \frac{\Gamma(\Delta^+\rightarrow p+\pi^0 )}{\Gamma( \Delta^+\rightarrow n+\pi^+ )} ~=~ 2, $$

cf. e.g. this & this Phys.SE post; or Chapter 8, eqs. (8.11) & (8.14), of these 't Hooft's lecture notes. The pdf file is available here.

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  • $\begingroup$ -1: Actually, if you know the invariant definition of a tensor then you know what the tensor looks like in every frame as well as knowing it's transformational law - just like vectors. $\endgroup$ Commented May 9, 2022 at 9:16
  • $\begingroup$ I updated the answer. $\endgroup$
    – Qmechanic
    Commented May 9, 2022 at 10:09
  • $\begingroup$ It doesn't change my opinion - unfortunately. $\endgroup$ Commented May 9, 2022 at 10:10

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