Do SALC-AOs really belong to their symmetry species?

Question

I'm working through a molecular symmetry textbook and something keeps nagging at me. If I derive the SALC-AOs for NH₃ (using the projection operator method), I'll get

A₁: $ \frac{1}{\sqrt{3}}(\phi_1+\phi_2+\phi_3)$

Doubly-degenerate E:

$ \frac{1}{\sqrt{6}}(2\phi_1-\phi_2-\phi_3)$

$ \frac{1}{\sqrt{2}}(\phi_2-\phi_3)$

where the $\phi$s are the H 1s orbitals, ie.:

(This was the best image I could find, but in the second E orbital the black s-orbital should be enlarged)

I'm having trouble seeing how these SALC-AOs would be a member of their symmetry species. The A₁ SALC-AO is identical under each operation, so it makes sense that it would belong to A₁. But if I want to confirm that the two E SALC-AOs belong to E, how would I do that?

My intuition is that I should be able to apply each C3v operation to the SALC-AO and get back the E row of the character table (Below). But if you apply a C₃ rotation to the e2 orbital, you get the black orbital taking the place of the white orbital, the white taking the place of the node, and the node taking the place of the black orbital. This doesn't seem like it could be expressed as a number in a character table, since the SALC-AO isn't being taken to itself or -itself.

Am I thinking about this all wrong? How should I understand the SALC-AO as a whole? Any guidance is appreciated, the textbook seems evasive on this.

Answer 1

To show that your two functions do represent the $\mathrm{E}$ irreducible representation, we can approach the problem algebraically.

We start by representing the operators of the $C_{3v}$ point group in the basis of the AOs. The identity element is easy in any basis: $$E=\pmatrix{1 &0 &0\\0 &1 &0\\0 &0 &1\\}$$

We can see that applying this operation to $e_1$ and $e_2$ will just return these functions, so $\chi(E)=2$. We can also represent this operation in the basis of $a_1$, $e_1$, and $e_2$, which will be important for explaining the remaining operations.

For the $C_3$ rotation, we can notice that this just has the effect of moving each AO to the next site (i.e. $\phi_1\to\phi_2$, $\phi_2\to\phi_3$, and $\phi_3\to\phi_1$), so we can write it in matrix form as $$C_3=\pmatrix{0 &1 &0\\0 &0 &1\\1 &0 &0\\}$$

Lets see what happens when we apply this to $e_1$: $$C_3e_1=\frac{1}{\sqrt{6}}\pmatrix{0 &1 &0\\0 &0 &1\\1 &0 &0\\}\pmatrix{2\\-1\\-1}=\frac{1}{\sqrt{6}}\pmatrix{-1\\-1\\2}=e_1'$$

Writing $e_1'$ in the basis of the original functions, we obtain: $$e_1'=0a_1+\frac{-1}{2}e_1+\frac{-\sqrt{3}}{4}e_2$$

Doing the same with $e_2$: $$e_2'=0a_1+\frac{-\sqrt{3}}{4}e_1+\frac{-1}{2}e_2$$

We can then take the trace of this matrix (or, to simplify, the 2x2 submatrix spanned by $e_1$ and $e_2$) to get the character of $C_3$: $$\chi(C_3)=\mathrm{Tr}\pmatrix{\frac{-1}{2} & \frac{-\sqrt{3}}{4} \\ \frac{-\sqrt{3}}{4} & \frac{-1}{2}}=-1$$

You can go through the same sort of exercise with $\sigma_v$ (determine form in AO basis, apply to each function, write resulting functions in original function basis, take the trace to get the character).