# N2 symmetry representation in higher symmetry groups (D3h…)

I'd like to describe the symmetry of the ground state of $$N_2$$ in the higher symmetry groups like $$D_{3h}$$ and $$D_{\infty h}$$.

In this answer the topic is covered very good, but only for $$D_{2h}$$ point group, where $$N_2$$ orbitals are described by the irreducible representations like this:

$$\begin{array}{cccc} \hline \mathrm{s} & \mathrm{p}_x & \mathrm{p}_y & \mathrm{p}_z \\ \hline \mathrm{A_{g}} & \mathrm{B_{3u}} & \mathrm{B_{2u}} & \mathrm{B_{1u}} \\ \hline \end{array}$$

But, when I have a look at character tables for $$D_{3h}$$ and $$D_{\infty h}$$, there are no irreducible representations with basis functions $$x$$ and $$y$$ as can be seen below.

So, how am I supposed to describe the symmetry of the $$p_x$$ and $$p_y$$ orbitals to be able to represent the symmetry of the whole system subsequently?

### My attempt

I've tried to describe the symmetry of $$p_x$$ in $$D_{3h}$$ like this:

$$\begin{array}{cccccc} \hline E & 2C_3(z) & 3C^{'}_{2} & \sigma_h(xy) & 2S_3 & 3\sigma_v\\ \hline 1 & 0 & -1 & 1 & 0 & 0\\ \hline \end{array}$$

Having this reducible representation, I've tried to identify needed irreducible representation using the reduction formula described here:

$$a_i = \frac{1}{h}\sum_Q N\cdot \chi(R)_Q \cdot \chi_i(R)_Q$$

For the first IR I'm getting the following result

$$a_{A^{'}_1} = \frac{1}{12}\left[ 1\cdot1\cdot1 + 0 + 3\cdot(-1)\cdot1 + 1\cdot1\cdot1 + 0 + 0 \right] = -\frac{1}{12},$$

which doesn't maky any sense, as it should be an integer, if I understand it well.

What am I doing wrong here? Is my reducible representation incorrect or does the formula not apply in this situation?

• Will $p_x$ not transform as $x$ in the point group? In $D_{3h}$ this would be $E'$. – porphyrin Sep 28 '18 at 12:34
• @porphyrin $E'$ is two dimensional, $x$ is one-dimensional only. Otherwise, $E$ operation couldn't be +2. – Eenoku Sep 28 '18 at 12:37
• Yes, that’s why you need to look at p_x and p_y together. If you sum up the characters for both orbitals under each symmetry operation, it should exactly match the E’ row. You can’t look at one part without the other, which is why you’re running into issues with the reduction formula. – orthocresol Sep 28 '18 at 12:48
• @orthocresol Thank you! I'm trying that, but $C_3(x)$ is 0 for both $p_x$ and $p_y$. Then it doesn't correspond with -1 in $E'$. – Eenoku Sep 28 '18 at 12:55
• Not at all; its the projection of the unit vector representing the p orbital on to the axis. – porphyrin Sep 28 '18 at 16:10