= Solution
Fix the positive lobes of the real <atomic orbitals> along the printed positive axes, and write $\phi_1=p_x(1)$, $\phi_4=p_x(4)$, $\phi_5=p_y(5)$ and $\phi_8=p_y(8)$. A coordinate model consistent with the figure is $r_1=(0,a,-b)$, $r_4=(0,-a,-b)$, $r_5=(a,0,b)$ and $r_8=(-a,0,b)$, with $a,b>0$. For the <D2d molecular point group>, take the two transverse twofold axes along $x+y$ and $x-y$, with vertical mirrors $xz$ and $yz$. The $A_2$ <character table> row on $(E,2S_4,C_2,2C_2',2\sigma_d)$ is $(1,1,1,-1,-1)$. The <character projection operator> is
$$
P_{A_2}=\frac18\sum_{g\in D_{2d}}\chi_{A_2}(g)^*\,U(g).
$$
The eight signed images needed for this projection are
$$
\begin{array}{c|rrrrrrrr}
g&E&S_4^+&S_4^-&C_2&C_2(x+y)&C_2(x-y)&\sigma(xz)&\sigma(yz)\\\hline
U(g)\phi_1&\phi_1&\phi_8&-\phi_5&-\phi_4&\phi_5&-\phi_8&\phi_4&-\phi_1\\
\chi_{A_2}(g)&1&1&1&1&-1&-1&-1&-1
\end{array}
$$
so $P_{A_2}\phi_1=(\phi_1-\phi_4-\phi_5+\phi_8)/4$. Under the standard orthonormal <atomic orbital> approximation its squared norm is $4/16=1/4$. Therefore the normalized <symmetry-adapted linear combination> is
$$
\boxed{\Phi_{A_2}=\tfrac12(\phi_1-\phi_4-\phi_5+\phi_8).}
$$
An overall minus sign describes the same <quantum state>; changing one local <atomic orbital> phase requires changing its coefficient too. If intersite <orbital overlap> were retained, normalization would instead use the overlap <matrix>.
Back to article page
