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A rank 2 tensor $T_{ij}$ of 3D rotation group $SO(3)$ is a reducible representation. It has the decomposition $9=5+3+1$ where 5 is the symmetric traceless tensor, 3 is the vector and 1 is the scalar. When the rotation group acts on a vector, the representative matrix is a $3\times 3$ orthogonal matrix with unit determinant.

After we make this decomposition into irreducible tensors by block diagonalization of a $9\times 9$ matrix into a $5\times 5$, $3\times 3$ and $1\times 1$ blocks, I have some questions.

  • What kind of matrices are the $9\times 9$ square matrices which act on the $9\times 1$ general tensor $T_{ij}$?

  • What kind of matrices are the $5\times 5$ square matrices which act on the $5\times 1$ symmetric traceless tensor?

When I ask what kind I am asking if they are also orthogonal and having unit determinant. I guess not. In that case, what kind they are. I am quite sure the $3\times 3$ & $1\times 1$ matrices that act on $3$ and $1$ are respectively orthogonal with determinant one and the number 1 respectively?

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Take the $3\times 3$ rotation matrix $R(\Omega)$ and tensor it with itself: $R(\Omega)\otimes R(\Omega)$ will be a $9\times 9$ matrix acting on the components of the $9\times 1$ $T_{ij}$.

As you have guessed, $R(\Omega)\otimes R(\Omega)$ is reducible. The $5\times 5$ block contains the Wigner D-matrices for $L=2$, i.e the $D^2_{MM'}(\Omega)$ functions. The $3\times 3$ block will contain $D^{1}_{mm'}(\Omega)$ matrices and the $1\times 1$ block is the scalar rep with $D^0=1$. All these are representations of $SO(3)$ so are unitary with determinant $+1$.

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  • $\begingroup$ I did not understand. The block matrices are orthogonal or unitary? You said unitary with determinant +1. That confuses me. I understand why an orthogonal matrix with unit determinant should act on a vector representation. Because its magnitude has to remain unchanged. But why the $5\times 5$ block matrix also need to be orthogonal and unit determinant? Also, can you tell the nature of the $9\times9$ reducible matrix? Is it unitary/orthogonal and halso having determinant 1? What is $\Omega$ by the way? $\endgroup$ – mithusengupta123 Mar 12 '20 at 14:30
  • $\begingroup$ They are complex because one uses complex combinations of components, v.g $x+iy$, or because the $Y^L_M$ are complex: the eigenstates of $L_z$ are complex combinations of cartesian components. All reps of SO(3) are equivalent to unitary ones, and the complex matrices with elements $D^L_{mm'}$ are entries in a unitary matrix. The properties of the $5\times 5$ follow because the decomposition of a tensor product of two irreps with det=+1 contains irreps with det=+1. Basically the tensor product of SO(3) irreps decomposes into a sum of SO(3) irreps. $\Omega$ labels the elements in SO(3). $\endgroup$ – ZeroTheHero Mar 12 '20 at 14:44

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