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Given i.e., a fourth order tensor $T_{ijkl}$ with spatial indices $i,j,k,l\in\{x,y,z\}$, and rotational invariance around the $z$ axis, how do I show that $T_{ijkl}=0$ because of said rotational invariance?

Let $R_{mn}$ be a rotation around the $z$ axis, then we know that the $T_{ijkl}$ transforms according to, $$T^\prime_{ijkl}=R_i^aR_j^bR^c_kR^d_lT_{abcd}.$$ If $T_{ijkl}$ respects the rotational invariance, we have $T^\prime_{ijkl}=T_{abcd}$. From this I would conclude that for every $(ijkl)$ where $R_i^aR_j^bR^c_kR^d_l\neq\delta^a_i\delta^b_j\delta^c_k\delta^d_l$ we need to have $T_{ijkl}=0$.

*what is the general form of a fourth order tensor with rotational invariance along one axis? How does one prove that it is the most general form?**

I tried to use the matrix entries of the rotation matrix in tensor notation, $$R_{mn}=(\delta_m^x\delta_n^x+\delta^y_m\delta^y_n)\cos\theta+(\delta^y_m\delta^x_n-\delta^x_m\delta^y_n)\sin\theta+\delta^z_m\delta^z_n,$$ but was not able to simplify $R_i^aR_j^bR^c_kR^d_l$ through this.

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If i’m not mistaken, the most general 4-rank tensor, invariant under rotations about z-axis should be of the form

$$ I_{ijkl} =A_1 \delta^{ij}\delta^{kl}+ A_2 \delta^{ik}\delta^{jl}+ A_3 \delta^{il}\delta^{jl}+$$

$$ +Bz^iz^jz^kz^l+$$

$$C_1z^iz^j\delta^{kl}+ C_2z^iz^k\delta^{jl}+ C_3z^iz^l\delta^{jk}+ C_4z^jz^k\delta^{il} +C_5z^jz^l\delta^{ik} + C_6z^kz^l\delta^{ij} +$$

$$+D_1z^i\epsilon^{jkl}+ D_2z^j\epsilon^{ikl}+ D_3z^k\epsilon^{jil}+ D_4z^l\epsilon^{jki}$$

Where the $A_i$, $B$, $C_i$, $D_i$ are constants;

(For the last row, we must limit ourselves to proper rotations of $SO(3)$)

So, in principle there’s no reason to assume that such an invariant tensor should have all components equal to zero;

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  • $\begingroup$ How did you find this general form? $\endgroup$ – bodokaiser Dec 15 '18 at 19:52
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    $\begingroup$ I built it out of everything I know to be invariant under rotations: the Kronecker delta and the Levi-Civita tensor which are isotropic tensors under orthogonal proper transformations, and given the specific problem, all possible combinations of z^i which could be built in combination with delta and Levi-Civita, i.e. linear, quadratic, and quartic; cubic is to be discarded because I could not find a rank 1 tensor which is invariant under rotations; $\endgroup$ – Francesco Bernardini Dec 15 '18 at 19:58
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I'm not sure why you suggest $T_{ijkl}$ should be zero just because it's symmetric under $z$ rotations. Here is how you how to construct a non zero 4-th rank tensor that is invariant under z-rotations:

Step 1: Start with Cartesians version of the spherical tensors:

$$ e^+ = -\frac 1 {\sqrt 2}[\hat x + i \hat y]$$ $$ e^0 = \hat z $$ $$ e^- = +\frac 1 {\sqrt 2}[\hat x - i \hat y]$$

Step 2: Use Clebsch-Gordan coefficients to find which combination of:

$$ |1,+1\rangle $$ $$ |1,0\rangle $$ $$ |1,-1\rangle $$

give you:

$$|J=4, M=0\rangle$$

Take those coefficients and use them to weight outer products of spherical vectors, and you will get a pure rank-4 cartesian tensor that is non-zero, and is invariant under z-rotations.

In fact, it will rotate just like a spherical harmonic:

$$ Y_4^0 \propto 35z^4 - 30z^2r^2 + 3r^4 $$

The ohter option is to start from the $Y_4^0$ and wing it using:

$$ z^2 \rightarrow {\bf \vec z \vec z} $$ $$ z^4 \rightarrow {\bf \vec z \vec z \vec z \vec z }$$

$$ r^2 \rightarrow \frac 1 3 ({\bf \vec x \vec x} + {\bf \vec y \vec y }+ {\bf \vec z \vec z})$$

Just make sure you symmetrize all the indices.

One the other hand, if you don't care about a pure trace-free rank-4 tensors, just rotate:

$$ T_{ijkl} = z_iz_jz_kz_l $$

It is invariant.

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Consider $$0\neq T = e_z\otimes e_z \otimes e_z \otimes e_z\:,$$ in components $$T_{abcd}= \delta_{az}\delta_{bz}\delta_{cz}\delta_{dz}\:.$$ $T\neq 0$ but it is invariant under any rotation $R$ leaving $e_z = (0,0,1)^t$ invariant: $$(R\otimes R\otimes R \otimes R) T = R\otimes R\otimes R \otimes R (e_z\otimes e_z \otimes e_z \otimes e_z)$$ $$ = Re_z\otimes Re_z \otimes Re_z \otimes Re_z = e_z\otimes e_z \otimes e_z \otimes e_z=T\:.$$ So your claim is untenable.

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  • $\begingroup$ How do I find a general form of a rotational invariant fourth order tensor? $\endgroup$ – bodokaiser Dec 15 '18 at 19:55

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