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I am studying the piezoelectricity in Gallium arsenide (GaAs), $[110]$ and $[\bar{1}10]$ oriented. Piezoelectricity is usually described microscopically by a 3 index tensor, $d_{ijk}$ where i,j and k equal to x,y and z.

$[110]$ and $[\bar{1}10]$ have different piezoelectric behaviour but they are linked by a parity operation on the x-axis: $$x\rightarrow -x$$ Can I assume the following identity:$$d^{[110]}_{ijk}=(-)^{\delta_{xi}+\delta_{xj}+\delta_{xk}}d^{[\bar110]}_{ijk}$$ Derived by applying the same transformation.

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I think the problem is a general tensor analysis one rather semiconductor physics (GaAs), since no particular values of any d-symbol is of any importance? Anyway, here is my effort to produce an answer.

The piezoelectric coefficient $d_{ijk}=d^{[110]}_{ijk}$ is a third rank Cartesian tensor which transforms from the reference frame $[x_1, x_2, x_3]$ to $[\bar {x}_1,\bar {x}_2,\bar {x}_3]$ giving $\bar {d}_{pqr}= d^{[\bar110]}_{pqr}$ according to the general rule:

${d}^{110}_{ijk} =\Sigma \bar {d}^{\bar {1}10}_{pqr}a_{pi}a_{qj}a_{prk} $

where $a_{\lambda\mu}$ are the cosines between coordinates in the two frames, and the summation is over repeated indices.

In your particular coordinate transformation $[- x,y,z]$ to $[x,y,z]$ which transforms $\bar {d}_{pqr}$ to $d_{ijk}$ you need to bear in mind that $a_{pi}=\delta_{pi}$ etc so you get the general relationship with the $\delta$ symbols multiplying each other in each term of the summation . They are not exponents (powers) to base (-1).

The expression you have written doesn't seem to give the correct relationship between $d^{[110]}_{ijk}$ and $d_{pqr}=d^{[\bar {1}10]}_{pqr}$.

I hope this helps.

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I think that you are right when saying that I am confused, or, at least, was confused. The question came from an article where it was stated that GaAs [110] and [\bar{1}10] have opposite component $d_{zxx}$ and when looking at the litterature I could only find the following value for this tensor: $d = \begin{pmatrix}0&0&0&a&0&0\\0&0&0&0&a&0\\0&0&0&0&0&a\end{pmatrix}$ without specified orientation (I thought that orientation was intrinsic (linked to the orientation of the two fccs matrix for example) but no, of course, you need to look at how you put the GaAs on your substrate. –  Learning is a mess Feb 15 '13 at 16:57
And by looking at this, performing the rotations according to how they set up their experiment, I do end up finding the opposite sign between the two orientation in the new component $d_{zxx}$. Thanks for clearing this out for me, I'm still frustrated and ashamed of not understanding it before, and puzzled that this subject is not so documented in the literature (first time I had to read on GaAs, piezoelectricity and orientation). –  Learning is a mess Feb 15 '13 at 17:00
Just adding that the previous matrix is given in the Voigt notation $d = d_{i\alpha}$ with $i=x,y,z$ and $\alpha=xx,yy,zz,yz,xz,xy$. –  Learning is a mess Feb 15 '13 at 17:03
@Learning is a mess Many thanks for the extra information, appreciated. I am pleased I have made some contribution to your understanding of the problem. Tensor analysis is very important in the study of crystaline matterials in Condensed matter. Thanks again. –  JKL Feb 15 '13 at 17:12

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