How does an external electric field induce a symmetry lowering perturbation to the $D_{3h}$ point group? I have the following homework style question which I will typeset 'word for word'. I am having trouble understanding the authors' solution to part $\mathrm{ii}.$ of the following questions:


*

*An electron is confined in a prism-shaped quantum dot of $D_{3h}$ symmetry (please use the character table available in the recommended literature and online). Assuming the potential energy outside the quantum dot is infinite, one can solve the Schrődinger equation exactly and obtain the energy levels of the system:
$$_{,,} = (^2 +  + ^2)_1 + _2_2,\quad _1 = \frac{8^2ħ^2}{3^2},\quad _2 = \frac{^2ħ^2}{2^2},$$
where $a$ is the side of the equilateral triangle (prism base) and $L$ is the height of the prism. The quantum numbers $p,\, q,\, s$ are restricted to the following values:

$q = 0, 1, 2, \cdots$ for the states of $A_1$ symmetry (${A_1}^\prime$ or ${A_1}^{\prime\prime}$),
$q = 1, 2, 3 \cdots $for  the states of $A_2$ symmetry (${A_2}^\prime$ or ${A_2}^{\prime\prime}$),
$q = 1/3, 2/3, 4/3, 5/3, \cdots $ for the states of $E$ symmetry (${E}^\prime$ or ${E}^{\prime\prime}$),
$p = q+1, q+2, \cdots$
$s = 1, 2, \cdots$
i. Consider the following properties of the exact solution:
a) the energy level $E = 49E_1 + s_2E_2$ is triply degenerate for any $s$,
b) for each permissible $p,\, q$, an $E$-symmetry wavefunction and its complex conjugate are linearly independent.
Which of the above properties is a consequence of the $D_{3h}$ symmetry? Explain.


ii.    Are the $D_{3h}$-induced degeneracies lifted as a result of:
a) applying a constant homogeneous electric field along the $C_3$ symmetry axis?
b) applying a constant homogeneous electric field along one of the $C_2$ symmetry axes?
c) distorting the quantum dot geometry such that the triangular base of the prism becomes isosceles rather than equilateral?
Explain your answer in each case.


For completeness, and in order for the whole question to make sense I will, unfortunately, have to typeset the solutions to part i) also:


*

*The character table of the group $D_{3h}$ is shown below.

$\begin{array}{c|c|c|c}
 D_{3h} & E
& 2C_3 & 3{C_2}^{\prime\prime} & \sigma_h & 2S_3 & 3\sigma_v & \text{Basis (Linear, Rotations)} \\\hline
 {A_1}^{\prime} & 1 & 1 & 1 & 1 & 1 & 1 \\\hline
 {A_2}^{\prime} & 1 & 1 & -1 & 1 & 1 & -1 & R_x \\\hline
 {E}^{\prime} & 2 & -1 & 0 & 2 & -1 & 0 & (x,y) \\\hline
 {A_1}^{\prime\prime} & 1 & 1 & 1 & -1 & -1 & -1  \\\hline
 {A_2}^{\prime\prime} & 1 & 1 & -1 & -1 & -1 & 1 & z  \\\hline
  {E}^{\prime\prime} & 2 & -1 & 0 & -2 & 1 & 0 & (R_x,R_y)   \\\hline
\end{array}
$
i. a) This property is not a consequence of $D_{3h}$ symmetry and this degeneracy is accidental. $D_{3h}$ point group has only one and two-dimensional IRREPs. Thus the highest degree of degeneracy that can be induced by the $D_{3h}$ symmetry of system is 2. Therefore triply degenerate levels are not a consequence of $D_{3h}$ symmetry.
i. b) For $E$-symmetry, a pair of allowed ($p$, $q$) quantum numbers gives a single possible energy level for any value of $s$. This level must be doubly degenerate and the wavefunctions $\psi_{p,q,s}$ and ${\psi}^{∗}_{p,q,s}$, which are linearly independent, form bases of the two-dimensional IRREPs ${E}^{\prime}$ or ${E}^{\prime\prime}$. This is indeed a consequence of the $D_{3h}$ symmetry.


ii. a) No, the degeneracies are not lifted. An electric ﬁeld along the $C_3$ axis changes the symmetry of the environment from $D_{3h}$ to $C_{3v}$. The IRREPs ${E}^{\prime}$ and ${E}^{\prime\prime}$of the $D_{3h}$ group decompose into the IRREP $E$ of the new point group $C_{3v}$ which is still 2-dimensional and therefore doubly degenerate. Hence the doubly degenerate states do not split. This is shown in the table below
$\begin{array}{c|c|c|c}
 C_{3v} & E
& 2C_3 & 3\sigma & {} \\\hline
 A_1 & 1 & 1 & 1 & z  \\\
 A_2 & 1 & 1 & -1 & R_z \\\
 E & 2 & -1 & 0 & (x,y),(R_x,R_y)  \\\hline
 E^{\prime} & 2 & -1 & 0 & E   \\\
 E^{\prime\prime} & 2 & -1 & 0 & E
\end{array}
$
Table 1: The character table of the group $C_{3v}$ and the splitting of $E$-symmetry states.
ii. b) Yes, the degeneracies are lifted. The horizontal electric ﬁeld changes the previous $D_{3h}$ symmetry of the electron’s environment into a $C_{2v}$ environment. The doubly degenerate levels would split in the new lower-symmetry environment and the splitting is shown in the table below.
$\begin{array}{c|c|c|c}
 C_{2v} & E
& 2C_2 & \sigma_h & \sigma_v & {} \\\hline
 A_1 & 1 & 1 & 1 & 1  \\\
 A_2 & 1 & 1 & -1 & -1 \\\
 B_1 & 1 & -1 & 1 & -1 \\\
 B_2 & 1 & -1 & -1 & 1 \\\hline
 E^{\prime} & 2 & 0 & 2 & 0 & A_1\oplus B_1  \\\
 E^{\prime\prime} & 2 & 0 & -2 & 0 & A_2\oplus B_2 
\end{array}
$
Table 2: The character table of the group $C_{2v}$ and the splitting of the $E$-symmetry states.
ii. c) This has the same eﬀect as part (b). The symmetry of the environment of the electron is changed from $D_{3h}$ to $C_{2v}$. Therefore the $E$-symmetry levels will split according to the table above too.


These questions are so hard to answer without a physical picture, so for the $D_{3h}$ prism this link will lead you to the best controllable and interactive model I could find:

So for part ii) a) when applying a constant homogeneous electric field along the $C_3$ symmetry axis I imagine the situation by looking at the molecule from above (through the highest-order rotation axis):

so I observe that:

But in the answer to ii) a), the author states that

No, the degeneracies are not lifted. An electric ﬁeld along the $C_3$ axis changes the symmetry of the environment from $D_{3h}$ to $C_{3v}$.

Does this mean that the electric field has 'destroyed' one end of the molecule (or 'cut it in half') so it now looks like this?

This has the shape of a tetrahedron, which has a $C_{3v}$ symmetry environment. So my question is, why would applying an electric field change its symmetry environment like this?

For part ii. b) for the case of 'applying a constant homogeneous electric field along one of the $C_2$ symmetry axes', the author states that

Yes, the degeneracies are lifted. The horizontal electric ﬁeld changes the previous $D_{3h}$ symmetry of the electron’s environment into a $C_{2v}$ environment.

I'm very confused by this, but using that link from earlier my interpretation is as follows: the electric field represented with the red horizontal line is aligned with one of the three $C_2$ axes, and 'somehow' this changes the symmetry environment to $C_{2v}$ - which is essentially a water molecule:

But how does this make any sense, and what happened to the 3rd bond linking the other hydrogen atom?
With regards to ii. c) 'when distorting the quantum dot geometry such that the triangular base of the prism becomes isosceles rather than equilateral'
for which the author states

This has the same eﬀect as part (b). The symmetry of the environment of the electron is changed from $D_{3h}$ to $C_{2v}$. Therefore the $E$-symmetry levels will split according to the table above too.

I really don't get this, I don't see how changing to an isosceles triangle would move the $D_{3h}$ point group to one of its subgroups, $C_{2v}$. Each half of the original $D_{3h}$ molecule had 2 vertices, each of which has 3 atoms linked, so how does $C_{2v}$ (water molecules) factor into this with only two atoms joined to the vertex?

The images used in this post are from chemtube3d.com
 A: I think you missed an important point in the very first line of the question:

An electron is confined in a prism-shaped quantum dot of $D_{3h}$ symmetry

i.e. the quantum dot they're referring to is a triangular prism, or a quantum Toblerone. A point group is not a molecule, it is a description of a finite set of operations (rotations, reflections and inversions) that can be performed on space, which happen to fix a certain geometric structure. In chemistry, this structure is usually an arrangement of atoms, but it may also be a cavity, a crystal field etc.
If we choose the triangle of the Toblerone lying in the xy plane, the point group $D_{3h}$ has the representation $\langle R_z, P_{xz}, P_{xy} \rangle$, where $R_z$ is a $2\pi/3$ rotation about the $\hat{z}$ axis, and $P_{xz}, P_{xy}$ are mirror symmetries about the subscripted planes. The notation $\langle... \rangle$ indicates that the group is generated by successive application of these elements.
When the electric field is applied along the $C_3$ axis (which, for my choice of prism orientation, is the $\hat{z}$ axis), we break the $P_{xy}$ symmetry of the system reducing the point group to $C_{3v}$, which is isomorphic to $D_3$, the group of symmetries of an equilateral triangle in 2D. Think of this as "the prism can still be rotated in the electric field, but it can no longer be flipped on its face". 
Though this breaks a symmetry, it does not lift the energy degeneracy for the representation-theoretic reasons outlined in the textbook answer.
When the electric field is applied along one of the triangle's axes (i.,e.e in the x-y plane, we instead have this situation:

The situation is no longer symmetric under rotations about the $C_3$ axis, but it preserves the symmetries of mirroring in the xy plane and mirroring in the xz plane, which is precisely the $C_{2v}$ point group (isomorphic to $D_2$, the symmetry group of a 2D rectangle).
