Transformation matrix of a strain tensor If the stress $\sigma_{xx}$ is applied to an isotropic, three-dimensional body, the following strain tensor results:
$$\boldsymbol\epsilon=\left(\begin{matrix}\frac{1}{E}\sigma _{xx}  & 0 & 0 \\0 & -\frac{\nu}{E}\sigma _{xx} & 0 \\0 & 0 & -\frac{\nu}{E}\sigma _{xx}\end{matrix}\right)$$
Now the tensor should be rotated around the y-axis with the angle $\alpha$.

The transformation should be carried out with $ A'=QAQ^T $. How would be the transformation matrix $ Q $ in this case?
 A: I use this notation
the transformation matrix, transformed a vector components from  rotate system index B   to inertial system index I
rotation about the x-axis angle $~\alpha~$ between y and y'
$${_B^I}\,Q_x=\left[ \begin {array}{ccc} 1&0&0\\ 0&\cos \left( 
\alpha \right) &-\sin \left( \alpha \right) \\ 0&
\sin \left( \alpha \right) &\cos \left( \alpha \right) \end {array}
 \right]
$$
rotation about the y-axis angle $~\alpha~$ between x and x'
$${_B^I}\,Q_y= \left[ \begin {array}{ccc} \cos \left( \alpha \right) &0&\sin \left( 
\alpha \right) \\ 0&1&0\\ -\sin
 \left( \alpha \right) &0&\cos \left( \alpha \right) \end {array}
 \right] 
$$
rotation about the z-axis angle $~\alpha~$ between x and x'
$${_B^I}\,Q_z=\left[ \begin {array}{ccc} \cos \left( \alpha \right) &-\sin \left( 
\alpha \right) &0\\ \sin \left( \alpha \right) &\cos
 \left( \alpha \right) &0\\ 0&0&1\end {array}
 \right]
$$
vector transformation from B to I system
$$\mathbf v_I={_B^I}\mathbf Q\,\mathbf v_B$$
matrix transformation
$$\mathbf M_I=\mathbf ={_B^I}\mathbf Q \,\mathbf M_B\,{_I^B}\mathbf Q =\mathbf Q\,\mathbf M_B\,\mathbf Q^T\\
\mathbf M_B=\mathbf ={_I^B}\mathbf Q \,\mathbf M_I\,{_B^I}\mathbf Q =\mathbf Q^T\,\mathbf M_I\,\mathbf Q
$$
your matrix is
$$\mathbf \epsilon_I= \left[ \begin {array}{ccc} \epsilon_{{11}}&0&0\\ 0&
\epsilon_{{22}}&0\\ 0&0&\epsilon_{{22}}\end {array}
 \right] \\
\mathbf \epsilon_B=Q^T\,\mathbf \epsilon_I\,\mathbf Q
$$
for $~\mathbf Q=\mathbf Q_x~$ you obtain
$$\mathbf \epsilon_B=\mathbf \epsilon_I$$
for $~\mathbf Q=\mathbf Q_y~$
$$\mathbf \epsilon_B=\left[ \begin {array}{ccc}  \left( \cos \left( \alpha \right) 
 \right) ^{2}\epsilon_{{11}}+\epsilon_{{22}}- \left( \cos \left( 
\alpha \right)  \right) ^{2}\epsilon_{{22}}&0&\cos \left( \alpha
 \right) \sin \left( \alpha \right)  \left( -\epsilon_{{22}}+\epsilon_
{{11}} \right) \\  0&\epsilon_{{22}}&0
\\  \cos \left( \alpha \right) \sin \left( \alpha
 \right)  \left( -\epsilon_{{22}}+\epsilon_{{11}} \right) &0& \left( 
\cos \left( \alpha \right)  \right) ^{2}\epsilon_{{22}}+\epsilon_{{11}
}- \left( \cos \left( \alpha \right)  \right) ^{2}\epsilon_{{11}}
\end {array} \right] 
$$
for $~\mathbf Q=\mathbf Q_z~$
$$\mathbf \epsilon_B= \left[ \begin {array}{ccc}  \left( \cos \left( \alpha \right) 
 \right) ^{2}\epsilon_{{11}}+\epsilon_{{22}}- \left( \cos \left( 
\alpha \right)  \right) ^{2}\epsilon_{{22}}&-\cos \left( \alpha
 \right) \sin \left( \alpha \right)  \left( -\epsilon_{{22}}+\epsilon_
{{11}} \right) &0\\ -\cos \left( \alpha \right) \sin
 \left( \alpha \right)  \left( -\epsilon_{{22}}+\epsilon_{{11}}
 \right) & \left( \cos \left( \alpha \right)  \right) ^{2}\epsilon_{{
22}}+\epsilon_{{11}}- \left( \cos \left( \alpha \right)  \right) ^{2}
\epsilon_{{11}}&0\\ 0&0&\epsilon_{{22}}\end {array}
 \right] 
$$
A: Tensors are rotated by the same rotation matrices that rotate vectors. So if the basis vectors are rotated by $Q_{ij}$:
$$e'^{(k)}_i = Q_{ij}e^{(k)}_j$$
Then a rank-2 tensor is rotated with:
$$ A'_{il}=Q_{ij}A_{jk}Q^T_{kl}$$
A symmetric tensor has 6 components. There is an invariant trace:
$$ A'_{ii}=Q_{ij}A_{jk}Q^T_{ki}$$
$$ A'_{ii}=Q^T_{ki}Q_{ij}A_{jk}$$
$$ A'_{ii}=\delta_{kj}A_{jk}=A_{kk}$$
and 5 quadrupole components.
With your initial tensor, you can subtract off the trace to get the pure rank-2 part:
$$ N_{ij} = \epsilon_{ij} - \frac 1 3 {\rm Tr}(\epsilon) $$
With
$$ {\rm Tr}(\epsilon) = \frac{\sigma_{xx}}E\big(\frac 1 3 -\frac 2 3\nu\big)$$
$$ N_{xx} = \frac{\sigma_{xx}}E\big(\frac 2 3 + \frac 2 3\nu\big)=\frac 2 3\frac{\sigma_{xx}}E\big(1+\nu\big)$$
$$ N_{yy} = N_{zz}=\frac{\sigma_{xx}}E\big(-\frac 1 3 - \frac 1 3\nu\big)
=-\frac 1 3\frac{\sigma_{xx}}E\big(1+\nu\big)=-\frac 1 2 N_{xx}$$
If you transform that into spherical tensors, you get a two non-zero parts:
$$N_{zz} \propto Y_2^0(\theta, \phi)$$
which describes how prolate or oblate the quadrupole moment is (along the $z$-axis), and
$$N_{xx}-N_{yy} \propto (Y_2^2(\theta, \phi) + Y_2^{-2}(\theta, \phi))$$
which describes the lowest degree of azimuthal asymmetry (a bulge like the Earth's tidal bulge).
The
$$N_{xz} \propto (Y_2^1(\theta, \phi) + Y_2^{-1}(\theta, \phi))$$
term is zero. This term is caused by choosing a coordinate system that is not aligned with the principle axes of the quadrupole. It can be diagonalized away.
So: when you transform to $\epsilon'$ you get non-zero $l=2$, $m=\pm 1$ moments, meaning you have not chosen the best coordinates. There is no physical meaning.
