Selection rules: Why is $\frac1m\sum_g\sum_{\lambda}^{\oplus} \underline{\underline{I}}_{\,\lambda}\otimes\underline{\underline{D}}^{(\lambda)}(g)=0$?

Question

Here is the derivation from some notes given to me. I uploaded these handwritten notes for 2 reasons:

1. To show you that this is the only source of information I have available to me (and it's incredibly difficult to learn from it).
2. In the hopes that you can make more sense of them than I can.

There are two expressions in these notes for which I would like to understand, which I have marked with red question marks above the relevant expressions.

For the first expression,

$$\frac1m\sum_g\underline{\underline{D}}^{(\lambda)}(g)=\begin{cases} 0, & \text{for rest of $\lambda$'s} \\ 1, & \text{$\lambda$ for fully symmetric IRREP} \end{cases}$$ The $\underline{\underline{D}}^{(\lambda)}(g)$ in this expression (I think) is supposed to represent the triple direct product at the top of the page, namely, $$\underline{\underline{D}}^{(\lambda)}(g)\equiv\Big[{\underline{\underline{D}}^{(1)}}^*\otimes {\underline{\underline{D}}}^{\prime} \otimes {\underline{\underline{D}}^{(2)}}\Big]_{n,\,\beta,\,p,\, i,\,\alpha,\,j}$$ This $\frac1m\sum_g\underline{\underline{D}}^{(\lambda)}(g)$ as I understand it is not an orthogonality expression (but it should be), else how can it possibly be equal to zero or $1$?

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While in the middle of constructing this question I was able to find a PDF version of a book$^{\large\zeta}$ that has a specific chapter that derives selection rules in a remarkably similar way to the notes I have above:

I would like to know why "We note that the sum of matrices of any irreducible representation, other than the identity representation, over the group, is equal to the zero matrix (see Exercise 3.7)". Which is written in the paragraph underneath equation $(21.7)$ above. This exercise 3.7 has been asked about and answered here which I found earlier and put a comment below the question to indicate the source. User @Gerry Myerson answered that question, but I am unable to understand his proof.

So in summary, I would like to know why $\sum_g\sum_{\lambda}^{\oplus} \underline{\underline{D}}^{(\lambda)}(g)=0?$

Or, if you prefer, could someone please prove why the sum over a group of the matrix elements of any irreducible representation other than the identity/fully symmetric/trivial representation is equal to zero?

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Update:

Although this update possibly should be asked as a separate question and if I am told to do this then I will comply, the reason I ask it here is that it is simply regarding applications of the expression given in the title to this post. I have a few small questions regarding the final page of the authors' written lecture notes (page 74), embedded as an image:

As ever, I have scribbled red question marks over the parts I don't understand.

I know that for a direct product of two matrices, say $\underline{\underline{A}}=\begin{pmatrix}a_{11}& a_{12}\\a_{21}& a_{22}\\ \end{pmatrix} \,\, \text{and} \,\,\,\underline{\underline{B}}=\begin{pmatrix}b_{11}& b_{12} & b_{13}\\b_{21}& b_{22}& b_{23}\\b_{31}& b_{32}& b_{33} \end{pmatrix}\implies\underline{\underline{A}}\otimes \underline{\underline{B}}=\begin{pmatrix}a_{11}\,\underline{\underline{B}}& a_{12}\,\underline{\underline{B}}\\a_{21}\,\underline{\underline{B}}& a_{22}\,\underline{\underline{B}}\\ \end{pmatrix}$

But that was the direct product for matrices, how does this direct product work for characters?

So looking at the character table for $C_{3v}$ the author of these notes written that for $z$- polarised light $\langle\psi_{A_1}|\hat z | \psi_E\rangle$ for which the decomposition, ${\color{red}{\chi_{A_1}\otimes\chi_{A_1}}}\otimes\chi_E=\chi_E\ne\chi_{A_1}$. Although I know that for a transition to be allowed, its matrix element must be non-zero, as shown to me in the proofs by @Gerry Myerson and @lEm. Therefore, to be non-zero the decomposition must contain the trivial representation, which in this case (from the $C_{3v}$ character table) is the IRREP $A_1$, but why is $\chi_{A_1}\otimes\chi_{A_1}\otimes\chi_E=\chi_E$? Is the red part equal to 1? I know this information is coming from the character table somehow, but I don't understand how.

The author then determines whether the matrix element, $\langle\psi_{A_1}|\hat x | \psi_E\rangle$ for the same transition ($A_1 \to E$) for $x$-polarised light is non-zero by the decomposition, $\chi_{A_1}\otimes\chi_{E}\otimes\chi_E=\chi_E=(4,1,0)$. But, where does this $(4,1,0)$ come from and how does this 'contain $A_1$'? I looked at the $C_{3v}$ character table and naively note that summing together the totals for the 3 columns gives $(4,1,0)$, but this could just be a coincidence.

I won't ask about the last 2 red question marks for now, since I may be able to answer them once I've understood the first two.

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Edit:

@lEm Okay, I think I see it now, the $(4,1,0)$ is the whole row of $E$ multiplied by itself, not sure why though. Need to think more, any hints please anyone?

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$^{\large\zeta}$The textbook page 265 embedded in this post as an image is from "Applications of group theory in quantum mechanics" by Petrashen & Trifonov.

Answer 1

This is an elaboration of Gerry Myerson's answer in the post that you have linked. In any group $G$, multiplication by any element $g\mapsto hg$ would give you a bijection from $G$ to itself, so summing over $g$ is the same as summing over $hg$, therefore $$S=\sum_{g\in G}\rho(g)=\sum_{g\in G}\rho(hg)$$ But $\rho(hg)=\rho(h)\rho(g)$ by property of representation, and so $\rho(h)S=S$ for any $h\in G$. Now if you look at the column vectors $S_i$ of $S$. The above implies that $\rho(h)S_i=S_i$ for all $h\in G$. So the span of $S_i$ gives you a subrepresentation, so by irreducibility this is either the whole vector space or the trivial zero vector space $\{0\}.$ This cannot be the whole representation because that would imply $\rho(h)=I$ for all $h\in G$ which is not irreducible. Finally this implies that span of $S_i$ is $\{0\}$. So $S=0$.

Edit: For the follow-up question

The notation $\chi=\chi_{A_1}\otimes\chi_E\otimes\chi_E=(4,1,0)$ just means that $\chi(E)=\chi_{A_1}(E)\chi_E(E)\chi_E(E)=4$, $\chi(C_3)=1$ and so on. It signifies the value of $\chi$ on the conjugacy classes. Now by property of irreducible characters, every character (in fact class function) $\chi$ can be written as a linear combination of the irreducible ones. In our case, we have $$\chi=(4,1,0)=(1,1,1)+(1,1,-1)+(2,-1,0)=\chi_{A_1}+\chi_{A_2}+\chi_E$$

In general, you can find out the coefficients by taking inner product $\langle \chi, \chi_{A_1}\rangle$ and so on.

Answer 2

This is not an answer but is too long for a comment, given @lEm's answer above, the proof I actually understand, but it is the interpretation that is confusing me here. So the original problem statement was to:

Show that the sum over a group of the matrix elements of any irreducible representation other than the identity representation is equal to zero.

which can be found on page 47, exercise 3.7, "Applications of group theory in quantum mechanics" by Petrashen & Trifonov:

I have highlighted exercise 3.7 to show where it comes from in the context of other questions on that page.

I am now looking for an example (or application) of the proof given in the other answer by @lEm. So I choose as my example the $C_{4v}$ point group which can be found in my earlier question on constructing the character table for $C_{4v}$.

But just for brevity, I will restate the block-diagonal matrices for the group elements here:

$C_4=\begin{bmatrix} 0 & -1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{bmatrix},\,C_4^3=\begin{bmatrix} 0 & 1 & 0 \\ -1 & 0 & 0 \\ 0 & 0 & 1 \end{bmatrix},\,C_2=\begin{bmatrix} -1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 1 \end{bmatrix},$

$\sigma_x=\begin{bmatrix} -1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix},\,\sigma_y=\begin{bmatrix} 1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 1 \end{bmatrix},\,\sigma_d=\begin{bmatrix} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{bmatrix},\,\sigma_{d^\prime}=\begin{bmatrix} 0 & -1 & 0 \\ -1 & 0 & 0 \\ 0 & 0 & 1 \end{bmatrix}$

and $E=\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$

So in order to "Show that the sum over a group of the matrix elements of any irreducible representation other than the identity representation is equal to zero", I note that all matrix elements $a_{11},\, a_{12},\, a_{13}, \cdots a_{32}$ of $C_{4v}$ apart from $a_{33}$ all sum to zero, $\begin{bmatrix}a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & \color{red}{a_{33}} \end{bmatrix}$

As an example, just to make clear what I am doing; I am taking a particular matrix element and then summing the same entry in each of the 8 matrices that make up the representation for $C_{4v}$. So, taking say $a_{21}$ for example, for each matrix of elements: $C_4(a_{21})+ C_4^3 (a_{21})+C_2(a_{21})+\sigma_x(a_{21})+\sigma_y(a_{21})+\sigma_d(a_{21})+\sigma_d^\prime (a_{21})+E(a_{21})=1+-1+0+0+0+1+-1+0=0$ and this works for all matrix elements except the $a_{33}$'s, which sum up to give 8.

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My question is; why does this sum give 8 (for the $a_{33}$'s) instead of zero?

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Update:

Just for reference, I want to show why the first 2 of the red question marks (shown below) are allowed/forbidden using the information taught to me in the updated answer by @lEm:

$\color{red}{1}$: For $\langle\psi_{A_1}|\hat z | \psi_E\rangle$ $$\chi(E)=\chi_{A_1}(E)\chi_{A_1}(E)\chi_E(E)=1\times 1 \times 2=2$$ $$\chi(2C_3)=\chi_{A_1}(2C_3)\chi_{A_1}(2C_3)\chi_E(2C_3)=1\times 1 \times 1=-1$$ $$\chi(3\sigma_v)=\chi_{A_1}(3\sigma_v)\chi_{A_1}(3\sigma_v)\chi_E(3\sigma_v)=1\times 1 \times 0=0$$ So for $z$ polarised light, for transistion $A_1 \to E$: $\chi_{A_1}\otimes\chi_{A_1}\otimes\chi_E=(2,-1,0)=\chi_{E}\ne \chi_{A_1}\implies$ forbidden, (zero as does not contain indentity representation, $\chi_{A_1}$).

$\color{red}{2}$: For $\langle\psi_{A_1}|\hat x | \psi_E\rangle$ $$\chi(E)=\chi_{A_1}(E)\chi_{E}(E)\chi_E(E)=1\times 2 \times 2=4$$ $$\chi(2C_3)=\chi_{A_1}(2C_3)\chi_{E}(2C_3)\chi_E(2C_3)=1\times -1 \times -1=1$$ $$\chi(3\sigma_v)=\chi_{A_1}(3\sigma_v)\chi_{E}(3\sigma_v)\chi_E(3\sigma_v)=1\times 0 \times 0=0$$ So for $x$ polarised light, for transistion $A_1\to E$: $\chi_{A_1}\otimes\chi_{E}\otimes\chi_E=(4,1,0)=\chi_{A_1}+\chi_{A_2}+\chi_{E}\implies$ allowed, as contains $\chi_{A_1}$ (non zero as contains identity representation $\chi_{A_1}$).