# Determining overtones bands of degenerate modes

I have a question of a more mathematical nature on the mathSE (Symmetric Direct Product Distributive?) that received a good answer, but I think an answer more oriented to chemists would be a useful resource here.

I'm trying to determine the symmetry of the second overtone band of the degenerate $\Pi_u$ bend of $\ce{CO2}$. I'm told I need to use the formula$$\chi_v(\hat{R})=\frac{1}{2}[\chi(\hat{R})\chi_{v-1}(\hat{R})+\chi(\hat{R}^v)]$$ where $v$ is the number of quanta in the mode, $\hat{R}$ is an operation and $\chi_{v}$ is the character of that operation for the $(v-1)^{\text{th}}$ overtone. I arrive at the correct answer, which is that the second overtone has symmetry $\Pi_u\oplus\Phi_u$.

But this formula seemed complicated so I wanted to see if I could get the result just from taking the symmetric direct product of the state three times like so: $$\Pi_u\otimes\Pi_u\otimes\Pi_u=(\Sigma^+_g\oplus\Delta_g)\otimes\Pi_u=2\Pi_u\oplus\Phi_u$$

Following the direct product tables I seem to get this, which has an extra $\Pi_u$.

I know this is wrong because the representation should only be quadruply degenerate, corresponding to the 4 different ways of distributing the quanta between the original degenerate modes: $(3,0) (2,1) (1,2) (0,3)$.

Why does this approach fail? Does the symmetric direct product not distribute over the direct sums?

• A freebie for anyone looking to write an answer: books.google.com/… (Wilson, Decius, and Cross Molecular Vibrations pages 151-155) – Tyberius Mar 13 '18 at 19:16

The issue with this approach is that the symmetric direct product doesn't commute, but it is a bit more subtle than that. Summarizing in somewhat less mathematical terms the answer given to my question on MathSE, the issue stems from the fact that once you move past doing a single direct product, there is no longer just a symmetric and an antisymmetric part, but rather $n+1$ parts, where $n$ is the number of direct products being taken. So, trying to drop any antisymmetric terms that arise at each step will still leave terms that aren't in the symmetric part of the representation. (To say this in mathematical language for those interested, the representation for $n$ direct products of a particular irreducible representation of a group can be decomposed into $n+1$ parts by projecting onto the symmetric group $S_{n+1}$).
Physically, the reason for needing only the symmetric part rather than the whole direct product is that we are filling quanta into one particular mode. For example, if we wanted to find a combination band composed of two different $\Pi_u$ modes, we could just take the direct product to find the representation. But, filling two into a single $\Pi_u$ mode limits the number of possible distributions of the quanta to only those given by the symmetric direct product.
Let $Q_a$ and $Q_b$ be the normal coordinates of a doubly degenerate mode. We choose these coordinates such that for an operation $R$ $$Q_a \xrightarrow{R}R_aQ_a$$ $$Q_b \xrightarrow{R}R_bQ_b$$ While the same coordinates will not diagonalize each $R$ in the group, we can find a set of coordinates for each $R$ that will. The character of $R$ for the $(v-1)^{\mathrm{th}}$ overtone, $\chi_v(R)$ can then be determined by summing the contribution of $R$ acting on each possible distribution of quanta between the modes $$\chi_v(R)=R_a^v+R_a^{v-1}R_b+...+R_aR_b^{v-1}+R_b^v$$ We can see that recursion formula holds by looking at the form of the terms on the right hand side $$\chi(R)\chi_{v-1}(R)=(R_a+R_b)(R_a^{v-1}+R_a^{v-2}R_b+...+R_aR_b^{v-2}+R_b^{v-1})=R_a^v+2R_a^{v-1}R_b+...+2R_aR_b^{v-1}+R_b^v$$ and $$\chi(R^v)=R_a^v+R_b^v$$ Averaging these two terms, we obtain the recursion formula from the question. $$\chi_v(\hat{R})=\frac{1}{2}[\chi({R})\chi_{v-1}({R})+\chi({R}^v)]$$
By a similar argument, we can obtain a recursive expression for triply degenerate states $$\chi_v({R})=\frac{1}{3}\bigg[2\chi({R})\chi_{v-1}({R})+\chi({R}^v)+\frac{1}{2}\big(\chi({R^2})+[\chi({R}]^2\big)\chi_{v-2}({R})+\chi({R}^v)\bigg]$$