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I have some questions about the authors' solution to the following homework style question:

<span class=$C_{60}$ fullerene molecule question" />

I have added the number of electrons per level into the image, the green numbers indicate levels that don't have a full outer shell, the blue numbers indicate levels with full outer shells. The list of numbers for each energy level sum up to give 60.


Here I have typed out word for word the author's solution to the question (or view pdf here):

  1. The character table of the group $I_h$ is shown below where $\tau = \left(1+\sqrt 5\right)/2$. $\color{red}{\text{The highest occupied level is }}$$\color{red}{H_u}$, and $\color{red}{\text{the dipole moments transform according to the IRREP}}\,$ $\color{red}{T_{1u}}$.

$\begin{array}{c|c|c|c} I_{h} & E & 12C_5 & 12{C_5}^2 & 20C_3 & 15C_2 & i &12{S_{10}}^3 & 12{S_{10}} & 20S_6 & 15\sigma & \\\hline A_g & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & \\\hline T_{1g} & 3 & \tau & 1-\tau & 0 & -1 & 3 & \tau & 1 -\tau & 0 & -1 & (R_x,R_y,R_z) \\\hline T_{2g} & 3 & 1-\tau & \tau & 0 & -1 & 3 & 1-\tau &\tau & 0 &-1 \\\hline G_g & 4 & -1 & -1 & 1 & 0 & 4 & -1 & -1 & 1 & 0 \\\hline H_g & 5 & 0 & 0 & -1 & 1 & 5 & 0 & 0 & -1 & 1 \\\hline A_u & 1 & 1 & 1 & 1 & 1 & -1 & -1 & -1 & -1 & -1 \\\hline \color{red}{T_{1u}} & 3 & \tau & 1-\tau & 0 & -1 & -3 & -\tau & \tau-1 & 0 & 1 & (x,y,z) \\\hline T_{2u} & 3 & 1-\tau & \tau & 0 & -1 & -3 & \tau-1 & -\tau & 0 & 1 \\\hline G_u & 4 & -1 & -1 & 1 & 0 & -4 & 1 & 1 & -1 & 0 \\\hline H_u & 5 & 0 & 0 & -1 & 1 & -5 & 0 & 0 & 1 & -1 \\\hline \end{array}$

Table 3: The character table of the group $I_h$

The transition to an unoccupied level $D^{(i)}$ is forbidden if, $$\color{red}{\gamma A_g H_u T_{1u} i}=\frac1m\sum_g {\chi^{(H_u)}}^*(g){\chi^{(T_{1u})}}(g)\chi^{(i)}(g)=0\tag{1}$$ Let’s check this coefficient for each unoccupied level. For simplicity, let’s calculate $n_C {\chi^{(H_u)}}^∗(g){\chi^{(T_{1u})}}(g)$ for each class where $n_C$ is number of elements in each class. We then only need to multiply $\chi^{i}(g)$ with the corresponding class and add the terms.

$\begin{array}{c|c|c|c} I_{h} & E & 12C_5 & 12{C_5}^2 & 20C_3 & 15C_2 & i &12{S_{10}}^3 & 12{S_{10}} & 20S_6 & 15\sigma \\\hline n_C {\chi^{(H_u)}}^∗(g){\chi^{(T_{1u})}}(g) & 15 & 0 & 0 & 0 & -15 & 15 & 0 & 0 & 0 & -15 \\\hline \end{array}$

we can factor out the 15 and the summation is over 4 classes only.

$$\begin{align}\gamma A_g H_u T_{1u} i=\frac1m\sum_g {\chi^{(H_u)}}^*(g){\chi^{(T_{1u})}}(g)\chi^{(i)}(g)&=\frac{1}{120}\left[15 \chi^{(i)}(E)-15\chi^{(i)}(C_2)+15\chi^{(i)}(i)-15\chi^{(i)}(\sigma)\right]\\&=\frac18\left[\chi^{(i)}(E)-\chi^{(i)}(C_2)+\chi^{(i)}(i)-\chi^{(i)}(\sigma)\right]\end{align}\tag{2}$$

Let's check the transition to each unoccupied level:

$\underline{T_{1u}}$ $$ \gamma A_gH_uT_{1u}T_{1u} = \frac18\left[3+1-3-1\right]=0$$ $\underline{T_{1g}}$ $$ \gamma A_gH_uT_{1g}T_{1u} = \frac18\left[3+1+3+1\right]=1$$ $\underline{H_{g}}$ $$ \gamma A_gH_uT_{1u}H_{g} = \frac18\left[5-1-5-1\right]=1$$ $\underline{T_{2u}}$ $$ \gamma A_gH_uT_{1u}T_{2u} = \frac18\left[3+1-3-1\right]=0$$ $\underline{H_u}$ $$ \gamma A_gH_uT_{1u}H_{u} = \frac18\left[5-1-5+1\right]=0$$ $\underline{G_{g}}$ $$ \gamma A_gH_uT_{1u}G_{g} = \frac18\left[4+0+4+0\right]=1$$ Therefore, the dipole transitions from the highest occupied level $H_u$ is forbidden to the unoccupied levels $T_{1u}$, $T_{2u}$ and $H_u$ and is allowed to the levels $T_{1g}$, $G_g$ and $H_g$.


I have marked in red the parts of the author's solution which I don't understand. My first question is why do the dipole moments transform according to IRREP $T_{1u}$? Since the question asked which transitions from the highest occupied level, $H_u$ are allowed, then it makes sense why character, $\chi^{(H_u)}$ is needed in $(1)$, but I don't see why we need $\chi^{(T_{1u})}$ in $(1)$. Why not have say,

$$\gamma A_g H_u T_{1g} i=\frac1m\sum_g {\chi^{(H_u)}}^*(g){\chi^{(T_{1g})}}(g)\chi^{(i)}(g)$$ instead? Moreover, what is meant by this $\gamma A_g H_u T_{1u} i$ notation (marked red)?

Lastly, I'm very confused by the two diagrams given at the start of the author's question. The lower diagram seems to represent electron energy levels, and if I was correct about the number of valence electrons then no transitions should occur to levels $g_u$ or $a_g$ (as they have full outer shells). But, this means there are only four possible levels ($t_{1u}$, $h_g$, $t_{2u}$ and $g_g+h_g$) for a transition from the highest level, $h_u$. But the question asked for (and the solution presented) six transitions, not 4. How is this possible? I have no clue what the upper diagram is trying to depict; why is $t_g$ higher than $h_u$ when the solution (written in red) says the highest level is $h_u$?


Remarks

I have been taught how to work through much simpler examples of dipole transitions of point groups than $I_h$ (like $C_{3v}$ in this question for example). But, I've never seen something so complicated as this Buckminsterfullerene molecule and it is also the physical picture behind this solution that I do not understand.


Regarding the two diagrams given at the start:

Why is the topmost energy level of the upper diagram labelled by IRREP $t_g$, whereas the topmost energy level of the lower diagram is labelled by IRREP $h_u$?

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  • $\begingroup$ My interpretation of the diagram (which is a single diagram, not a higher one and a lower one) is that $h_u$ in the middle is the highest occupied level. All the lower levels, as well as this one, are fully occupied. The question doesn't ask about transitions to any lower levels, but to six lowest unoccupied ones, which are all above the occupied $h_u$ level. This just leaves the highest $g_u$ and $t_g$ unused in the question, and doesn't limit you to 4 potential target levels. $\endgroup$
    – Ruslan
    Oct 19, 2021 at 22:23
  • $\begingroup$ @Ruslan Hi, I've fixed the link, thanks for reply and explanation., how do you know which levels are occupied or not? $\endgroup$
    – BLAZE
    Oct 21, 2021 at 2:40
  • $\begingroup$ The horizontal bars represent the levels, and the vertical ones represent the electrons, two per bar means they have different spins. $\endgroup$
    – Ruslan
    Oct 21, 2021 at 7:58
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    $\begingroup$ @Ruslan Thanks, I know that the vertical lines in the lower diagram represent the electrons' spins (this is why I wrote the total no. of electrons per energy level right next to it). But the point is that all of the levels have 2 per bar (different spins); there is no unpaired electron for any energy level, so by my logic, this means that ALL levels are fully occupied. I don't think you addressed my last question so I'll say it again: How do you know which levels are occupied or not? Regards. $\endgroup$
    – BLAZE
    Oct 23, 2021 at 1:38
  • $\begingroup$ "means that ALL levels are fully occupied" — this is exactly true — for the levels with vertical bars. All levels without vertical bars are unoccupied. So $h_u$ becomes the highest occupied level, and $t_{1u}$ the lowest unoccupied. $\endgroup$
    – Ruslan
    Oct 23, 2021 at 9:13

1 Answer 1

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As was mentioned in the comments, the horizontal bars on the diagram represent energy levels and their vertical position corresponds to the value of energy they correspond to. Electrons occupy the lowest available energy levels and each electron is depicted as a vertical line. Each level has two vertical bars corresponding to two spin states. On the diagram, all levels with vertical bars are completely filled (i.e. all numbers on your diagram are supposed to be blue in your notations). The highest occupied level is $h_u$, and the six lowest levels above it are $t_{1u}$, $t_{1g}$, $h_g$, $t_{2u}$, another $h_u$, and $g_g$.

Now, let us consider, why the perturbation Hamiltonian transforms as $T_{1u}$. The perturbation Hamiltonian is proportional to the electric field $\vec{E}$, and can be expressed as $$\Delta H=E_x H_x+E_y H_y + E_z H_z=E_n H_n,$$ where I used Einstein summation convention. If you rotate (with possibility of reflection) your system in a way that the fullerene remains the same, with a transformation matrix $\mathrm{R}=R_{nm}$ , electric field will rotate accordingly $E_n\rightarrow R_{nm}E_m$, but $H_n$ will remain the same (because the system didn't change), i.e. the perturbation transforms according to $E_n H_n \rightarrow R_{nm}E_m H_n$. Notice, that the set of all matrices $R_{nm}$ that keeps the fullerene the same is a representation of the original group $I_h$. You can calculate the characters of this representation by noticing that $C_n$ is a conjugacy class of rotations by $2\pi/n$, and thus it has the character $2 \cos(2\pi/n)+1$. The characters you will obtain end up matching $T_{1u}$, and this is why we say that the perturbation transforms as $T_{1u}$ (and why you need to use it instead of $T_{1g}$). The author of the solution in the OP says that the dipole moment transforms as $T_{1u}$, which is true (dipole moment is a vector), but it is the proportionality of perturbation to electric field that is important here.

Lastly, I am not really sure about why the corresponding sum of characters has $\gamma A_gH_uT_{1g}T_{1u}$ notation. The sum checks how many times trivial representation $A_g$ is contained in ${H_u}^*\otimes T_{1g} \otimes T_{1u}$ representation. It may be the standard notation, but I am not familiar with it.

Hope this helps. Ask questions something is still unclear.

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