= Solution
For the collinear centrosymmetric arrangement, the longitudinal coordinates transform in the <D infinity h molecular point group>. Inversion sends $q_i\mapsto-q_{5-i}$, because displacement is a polar vector. Therefore
$$
\boxed{\Gamma_z=2\Sigma_g^++2\Sigma_u^+.}
$$
An orthonormal set of <symmetry coordinates> is:
|| Coordinate
|| Displacement vector
|| Symmetry
|| $\omega^2$ at $\gamma=0$
| $t$
| $(1,1,1,1)/2$
| $\Sigma_u^+$
| $0$
| $r$
| $(-1,-1,1,1)/2$
| $\Sigma_g^+$
| $0$
| $s$
| $(-1,1,-1,1)/2$
| $\Sigma_g^+$
| $2k/m$
| $a$
| $(-1,1,1,-1)/2$
| $\Sigma_u^+$
| $2k/m$
Here $t$ translates the complete system, $r$ separates the molecular centres, $s$ stretches both molecules together, and $a$ lengthens one while shortening the other. Reversing every arrow changes the coordinate sign, not its physical mode.
\Image[/past-exam-of-the-chemistry-course-of-the-year-3-of-the-natural-sciences-course-of-the-university-of-cambridge/2021/paper-2a-chain-modes.png]
{title=Normalized longitudinal symmetry coordinates for two collinear hydrogen molecules}
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