Projection formalism and eigenvectors of normal mode [closed]

Question

Consider a water molecule:

As one can prove, the group symmetry for this molecule is $C_{2v}$, with a character table like

I am asked to use the projector operator formalism in order to determine the eigenvector associated with the normal mode which transforms like the totally symmetric representation, $A_1$.

My attempt

As my professor thaught us, I started by writing a vector

$$| \phi \rangle = (\delta x_1,\delta y_1,\delta x_2,\delta y_2,\delta x_3,\delta y_3)$$

which represents the deviations from the equilibrium positons. Then, I wrote the matrices representing each element of the symmetry group. Then, using the projector formalism, I wrote

$$P| \phi \rangle = \frac{1}{4}\left(D(E)+D(C2)+D(\sigma_v)+D(\sigma'_v)\right) |\phi \rangle$$

which yielded

$$\frac{1}{2}(\delta x_1, 2\delta y_1, \delta x_2, \delta y_3+\delta y_2, \delta x_3, \delta y_3 + \delta y_2)$$

My problems

• I can't understand why we choose the vector $| \phi \rangle$;
• What does the eigenvector we are looking for represent?
• Is there a different/other intuitive method of solving this problem?

Edit 1: After checking the maths again and fixing the axis, I got

$$\frac{1}{2}(0, 2\delta x_2, \delta x_2-\delta x_3, \delta z_2+\delta z_3, \delta x_2-\delta x_3, \delta z_2 + \delta z_3)$$

Answer 1

After reading the comments and searching a bit more, I think I've managed to find the solution.

Considering the two displacements $r_1$ and $r_2$:

applying the projection formalism should give, for $r_1$ or $r_2$,

$$P r_1 = \frac{1}{4}\left(D(E)+D(C2)+D(\sigma_v)+D(\sigma'_v)\right) r_1$$

$$\Leftrightarrow P r_1 = \frac{1}{2}(r_1 + r_2)$$

which indicates us that both Hidrogen atoms expand and contract, at the same time!