In many aspects of chemistry it is of interest to find out whether a given integral $\int \psi \,\mathrm{d}\tau$ vanishes, i.e. becomes zero. Group theory provides us with a handy condition:

If $\Gamma_\psi$ does not include the totally symmetric irreducible representation (TSIR) of the point group, then $\int \psi \, \mathrm{d}\tau$ must be equal to zero by symmetry.

The utility of this is that it can be easily expanded to different types of integrals, such as matrix elements $\langle i | A | j \rangle$, where the integrand transforms as $\Gamma_i \otimes \Gamma_A \otimes \Gamma_j$. If the direct product does not contain the TSIR, then the matrix element must be zero.

What is the theoretical basis of this statement, i.e. is there a proof of it?

In particular I'd like to see a proof that:

  1. doesn't make assumptions about which symmetry operator(s) the integrand is not symmetric under; and

  2. is able to deal with irreps with dimensionality larger than one.

But if such a proof doesn't exist, I'd be happy to compromise on one or both of these criteria.

I checked FA Cotton's book, but he skirts the issue by saying that "it is a generalisation of the case where $\int y(x)\,\mathrm{d}x = 0$ if $y$ is an odd function of $x$" (p 107, 3rd ed.). That does makes sense, and I have an idea of how it could be mathematically formalised, but it doesn't satisfy either of the two criteria above.


4 Answers 4


By Maschke's Theorem, every direct product of representations is decomposable into a direct sum of representations, that is, the function you are integrating can be rewritten as a sum of functions with symmetry properties equal to specific irreducible representations.

As you note, integrating an odd function (which has a certain symmetry, namely antisymmetry with respect to $\sigma_y$, the mirror "plane" of the y-axis) over all space is zero, while integrating an even function (which is symmetric with respect to $\sigma_y$) is generally not. Since integration is linear (that is, the integral of a sum is the sum of an integral), we can consider the integral of each irrep separately.

If the function being integrated is antisymmetric with respect to some symmetry, it will be zero when integrated over all space. The only way for the integral not to be zero is if it is symmetric with respect to all possible symmetries, i.e. it is the totally symmetric irrep.

Expressed in equation form: $$\int{\prod_{i}\Gamma_i\,\mathrm d\tau}=\int{\sum_j\Gamma_j\,\mathrm d\tau}=\sum_j\int{\Gamma_j\,\mathrm d\tau}$$ and $$\int{\Gamma}\,\mathrm d\tau=\left\{ \begin{array}{@{}ll@{}} 0, & \text{if}\ \Gamma \text{ contains any odd (antisymmetry)} \\ \text{not necessarily } 0, & \text{otherwise} \end{array}\right.$$

Then, it is clear that the only time the integral is nonzero is when, amongst the irreducible representations of the direct product (regardless of its dimensionality), the totally symmetric one is present, all other ones integrate out to zero.

I should clarify that when I write the integral of $\Gamma$, I really mean the integral of some awful, complicated looking function that has the symmetry properties of $\Gamma$, which is why I used $\prod$ and $\sum$ instead of $\otimes$ and $\oplus$.

If your question was instead how do you know that the functions with a given antisymmetry really do integrate to zero, let's consider what it means to have a certain symmetry. It means that there is a particular line or plane or point about which the value of the function is redundant. That is, if you specify it on one side, then the value is exactly determined on the other. We are integrating over all space, which means we can choose bounds of integration that utilize this symmetry. This then separates your integral into a non-symmetric piece and a piece using the symmetry. You first integrate over the non-symmetric piece and get something, and then you integrate that over the symmetry-including part. If that symmetry is antisymmetry, as will be the case with at least one symmetry of any irrep other than the totally symmetric one, this will cancel out the non-symmetric part with an equal and opposite non-symmetric part and the whole integral will be zero. Regardless of the dimensionality of the irrep being integrated it has at least one symmetry element over which it can be integrated that is antisymmetric. Hence, only integrals of functions that are totally symmetric are non-zero.

As a basic example, consider your $\int{x\,\mathrm dx}$ integral that you started with. Since this function has a symmetry (namely antisymmetry with respect to $\sigma_y$), I can make the following choice of bounds: $$\int_{-\infty}^\infty{x \,\mathrm dx} = \int_{-\infty}^0{x \,\mathrm dx}+\int_{0}^\infty{x \,\mathrm dx} =-\int_{0}^\infty{x \,\mathrm dx}+\int_{0}^\infty{x \,\mathrm dx}=0$$

I can choose to integrate only over the nonsymmetric part and then use the symmetry for the rest of space. The presence of symmetry guarantees that I can do that. Remember, the irrep label is just telling you the symmetry properties of some complicated function. If $x$ in the above equation were some function $f$ that I knew nothing about except that is has this symmetry, I could still make this choice of bounds and use this property to get a 0 integral.

EDIT: To address orthocresol's question below:

How does this work for the $E$ set in $C_{3v}$? Let us first think of a function with $E$ symmetry. The $p_x+p_y$ orbitals are such a candidate. Let's consider a top-down view of a $C_{3v}$ molecule. Now, I will consider one of the mirror planes to lie in the $xz$-plane (see Figure 1). I can do this because I can rotate the $p_x+p_y$ ensemble anywhere in the $xy$-plane. That is, there is some linear combination of $p_x+p_y$ such that there is a new set $p_{x'}+p_{y'}$ where the mirror plane lies on the $x'y'$-plane.

Figure 1, top-down view of an $E$-symmetry function

Now we set about integrating the upper half plane. Clearly, the half of the negative lobe of the $p_x$ orbital will cancel the half of the positive lobe, leaving only the integral of the positive lobe of $p_y$. Now we consider the lower half plane, the two halves of the $p_x$ lobes cancel again and we have only the integral of the negative lobe of $p_y$. Since the integral of the upper-half plane is the negative of the lower-half plane, the integral over all space is 0.

Now, why was it important to consider $p_x+p_y$ together? This is what allowed us to orient the orbitals with respect to the mirror plane. Consider a $p_x$ orbital alone rotated away from the axis given. It will have no defined symmetries with respect to the plane (reflecting across it will not map the orbital to itself or its negative) and there is no way to rotate it without the $p_y$. This is why they need to transform together in $C_{3v}$. Indeed, the reason why $E$ functions have 0 character under this symmetry is because the $p_x$ got sent to $p_x$ and the $p_y$ to $-p_y$ and so the whole ensemble was sent to $1-1=0$.

Now, hold on you say, can't I use similar arguments to imply that if I integrated a $p_z$ orbital (which is $A_1$ in $C_{3v}$) in all space, I will also get zero. The answer is yes. The $p_z$ orbital is not zero by symmetry in $C_{3v}$. I alluded to this above, the integral of a totally symmetric function in a given point group can still be zero. However, by describing the point group as $C_{3v}$, we are indicating that there is something special about the $z$-axis that breaks the symmetry and therefore precludes coming to this conclusion based on symmetry alone. Consider the deformation of planar $\ce{NH_3}$ to pyramidal $\ce{NH_3}$. In the former ($D_{3h}$), nothing distinguishes the positive $z$ from negative $z$ axis and a $p_z$ orbital is not totally symmetric ($A_2''$) and so we could conclude that it is zero by symmetry. As we pyramidalize to $C_{3v}$ though, the negative $z$ axis becomes different (it's got hydrogens there while the postive $z$ does not) and we can no longer conclude by symmetry that the integral of the $p_z$ orbital will be 0 (indeed, if it mixes even slightly with those hydrogens, it won't be). If it didn't mix though, it's integral would still be 0 (it's still just a $p_z$ orbital), we just couldn't tell that assuming $C_{3v}$ symmetry.

  • 2
    $\begingroup$ I like your answer. Referring to 2nd eqn, if $\Gamma$ is 'odd' then the integral is always zero, but if not then is it true that it is generally not zero but could be? For example, $\cos(x)$ is even about y axis but the integral is zero over $\pm 2\pi$ but $x^2$ is also even but integral not zero. $\endgroup$
    – porphyrin
    Jul 14, 2017 at 8:48
  • 1
    $\begingroup$ Yes, absolutely. An even function need not integrate to a non-zero number. I sort of wrote that in the text, but I will revise the equation. $\endgroup$
    – levineds
    Jul 14, 2017 at 18:53
  • 1
    $\begingroup$ Thanks! I understand what you're saying; the only part is your statement that "regardless of the dimensionality of the irrep being integrated it has at least one symmetry element over which it can be integrated that is antisymmetric." - how does that apply to, for example, the E irrep in $C_{3v}$? $\endgroup$ Jul 17, 2017 at 16:30
  • $\begingroup$ @orthocresol As an example, the triple product is needed when trying to determine selection rules, transition i to f e.g. if $\Gamma_i \times \Gamma_\mu \times\Gamma_f $ contains the totally symmetric representation the transition integral is not zero. So if the product is E only then transition not allowed, integral is zero. $\endgroup$
    – porphyrin
    Jul 17, 2017 at 19:21
  • $\begingroup$ @porphyrin yes, I understand that bit. Sorry, I guess should have been clearer - I meant, if the integrand transforms as E, then why would the integral evaluate to zero? It's not obvious (to me, at least) how the integral can be split up into components that are equal in magnitude but opposite in sign (which was the strategy adopted for $\int x \,\mathrm dx$ above). $\endgroup$ Jul 18, 2017 at 6:08

EDIT: Made this more rigorous. (I think. Feel free to critique).

First, let's decompose $\psi$ into components that correspond to how its representation reduces. That is, if

$$\Gamma_{\psi} = \Gamma_{1}\oplus\Gamma_{2}+\ldots$$

we write

$$\psi = \phi_{1} + \phi_{2} + \ldots $$

where the symmetry of $\phi_{i}$ is captured by the $\Gamma_{i}$.

Suppose $\Gamma_{i}$ is not fully symmetric. There is some symmetry element $x$ whereby $x$ is antisymmetric in this representation. In other words, $$x\phi_{i} = -\phi_{i}$$

Now, consider this operation. You could perform the change of variable $x\tau \rightarrow \tau'$ for the integral.

$$\int \phi_{i}\,\mathrm d\tau = \int x \phi_{i} x^{-1}\,\mathrm d\tau'$$

This step is straightforward because we're basically arguing that if we convert our basis to something else (spcifically, transformed under $x$), the integral will be the same. This must be true since rotating the coordinate system, for example, does not change the value of the integral.

Now, we can substitute our definitions above:

$$\int \phi_{i}\,\mathrm d\tau = \int x \phi_{i} x^{-1}\,\mathrm d\tau' = \int (-1)\phi_{i} \,\mathrm d\tau = -\int \phi_{i}\,\mathrm d\tau$$

where I have used $x^{-1}\tau' = \tau$.

The equality means that the integral must be zero.

Now, if the irreducible representation is the totally symmetric representation, then antisymmetric element $x$ does not exist, which invalidates the rationale above, though the other components of $\psi$ will integrate to zero.

This argument works quite well when the irreducible representation $\Gamma_{i}$ is 1-dimensional. What happens when $\Gamma_{i}$ is higher dimensional?

I'm not sure I can rigorously prove this part, but I think I can sketch it out for a simple example.

Consider point group $C_{3}$.

\begin{array}{|c|c|c|} \hline & E & 2C_{3} \\ \hline A & 1 & 1 \\ E & 2 & -1 \\ \hline \end{array}

However, you could also write this character table as:

\begin{array}{|c|c|c|} \hline & E & C_{3} & C_{3}^{-1} \\ \hline A & 1 & 1 & 1 \\ E_{1} & 1 & \varepsilon & \varepsilon^{*} \\ E_{2} & 1 & \varepsilon^{*} & \varepsilon \\ \hline \end{array}

where $\varepsilon = -\frac{1}{2} + \frac{\sqrt{3}}{2}i$. Note that a linear combination of $C_{3}$ and $C_{3}^{-1}$, namely $\frac{\sqrt{2}}{2}C_{3} + \frac{\sqrt{2}}{2}C_{3}^{-1}$, provides a linear combination that is equal to operation $-E$. Wait, you ask, but that is not an element of the $C_{3}$ point group. That's right, but a $p_{x}$ orbital does not have 3-fold symmetry either. But if you rotate it both $120^{\circ}$ and $-120^{\circ}$, you get two orbitals that when mixed, gives the $p_{x}$ orbital rotated $180^{\circ}$.

I think this generalizes out to other 2-dimensional and higher dimensional irreducible representations in that you either have a matrix that inverts (for example character -2 in a 2-dimensional representation) or you are guaranteed that a linear combination of representations gives the inversion (like here). My background here is too weak at present to provide anything more concrete.

  • $\begingroup$ This isn't quite right. Let's be specific: say $\phi=x$ and the symmetry $x$ is negation (sending $x$ to $-x$, i.e $\sigma_y$). I agree that (-1)x = -x. The problem comes when you try to change variables. Make the substitution $u=-x$, then $du=-dx$ and $\int{xdx}=\int{-u(-du)}=\int{udu}$ and you've gotten nowhere. I think you conflated change of basis of matrices and change of variable. You applied $x$ and $x^{-1}$ to either side, which is conjugation by some operator and not variable change (but it does resemble basis change in vector spaces). $\endgroup$
    – levineds
    Jul 18, 2017 at 10:39
  • $\begingroup$ Consider $\phi_2=x^2$. Then $x\phi x^{-1}$ is the same for both $\phi$ and $\phi_2$ even though one is odd and one is even, so clearly this isn't doing what you think. The way to use symmetry for integrals, as I described in my answer, is to manipulate the bounds of integration and then use change of variables to show that the integral is sum of two integrals (with the same bounds) that are equal and opposite. $\endgroup$
    – levineds
    Jul 18, 2017 at 10:39
  • $\begingroup$ But $x^{2}$ is not antisymmetric. Doesn't it belong to the totally symmetric irreducible representation? $\endgroup$
    – Zhe
    Jul 18, 2017 at 13:06
  • $\begingroup$ But, a change of basis in the vector space is exactly what I want, @levineds. This means that the integral should be the same regardless of the coordinate system... But I see that I didn't take the chain rule correctly to compute $\mathrm{d}\tau'$. There needs to be an invocation of the fact the the volume element is unchanged under the transformation because $x$ needs to be in $O(3)$. $\endgroup$
    – Zhe
    Jul 18, 2017 at 13:13

I consulted my coworker who has a Ph.D. in algebraic topology. The guidance I received was pretty helpful, so here's the new way I understand the problem. There are a few steps that I might not be able to answer questions about because I'm not a mathematician, but hopefully, you can read up on these if my explanation is not clear enough.

First, let's decide to look at arbitrary point group $G\subseteq O(3)$.

For an arbitrary function $\psi:\mathbb{R}^{3}\rightarrow\mathbb{C}$, we decompose it as follows: $$\psi = \sum_{i}\phi_{i}$$ where each of the $\phi_{i}$ has the same symmetry as irreducible representation $\Gamma_{i}$ of $G$ and the symmetry of $\psi$ corresponds to representation $\Gamma_{\psi}$ where

$$\Gamma_{\psi} = \bigoplus_{i}\Gamma_{i}$$

We need to show that $\int \phi_{i}$ vanishes if $\Gamma_{i}$ is not the fully symmetric irreducible representation.

So let's consider a function $f : \mathbb{R}^{3}\rightarrow\mathbb{C}$. My coworker told me that the representation of $f$ from a math perspective is $G$ acting on the $\mathbb{C}G$-module generated by $f$: $$\mathrm{Rep}(f) = G\text{ acting on } (\mathbb{C}G)[f]$$

$\mathbb{C}G$ is the free vector space on $G$. It is composed of vectors $\sum\omega_{g}g$ for $\omega_{g}\in \mathbb{C}$ and $g\in G$. If $|G|=n$, then $\mathbb{C}G\cong\mathbb{C}^{n}$.

We want to prove the following theorem: Suppose $\mathrm{Rep}(f)$ is irreducible and not trivial, then $\int f = 0$.

If the representation is irreducible, it means that $\mathrm{Rep}(f)\neq V\oplus W$, where $GV \subseteq V, GW \subseteq W$.

Now, we know because of the irreducibility condition that $\mathbb{C}G[f]$ has no fixed points other than 0, or else, we could decompose it two subspaces that were disjoint under $G$. Therefore, if $v=\sum_{g}\omega_{g}(gf)$ and $gv=v$ for all $g\in G$, then $v = 0$.

Consider the vector $w = \sum_{g\in G} g f$. Note that

$$g'w = g'\sum_{g\in G} g f = w$$

This is because when you construct the multiplication table for the group, each row or column has distinct elements (otherwise, elements in the group would turn out not to be distinct). There are two conditions under which this statement is true. One is the conclusion above that $w = 0$, and the other is that $gf = f, \forall g$. The second condition is false by the statement of the problem: if $f$ is fixed $\forall g\in G$, it implies that $G$ is the totally symmetric irreducible representation. Therefore, the only valid conclusion is that $$w = \sum_{g\in G}gf = 0$$.

Now consider the integral of $f$ under a transformation $g$. Because $g$ is a transformation is $O(3)$, this integral must be just the integral of $f$. $$\int gf = \int f$$

We now sum the integrals over all elements of $G$:

$$\sum_{g\in G}\int g f = \int \sum_{g\in G}g f = 0 = \sum_{g\in G}\int f = |G|\int f$$

Since $|G|\neq 0$, we conclude that the integral of $f$ is indeed zero.

Circling back, $\int \psi$ reduces to $$\sum_{i\in S}\int \phi_{i}, i\in S \Rightarrow \Gamma_{i}\text{ totally symmetric}$$

This solution works for all point groups, doesn't require a definite basis, and works for arbitrary functions. Just the kind of answer you would expect from an abstract mathematician...

  • $\begingroup$ Does this work for groups of infinite order? Forming the G-invariant vector $w$ requires summing over all the elements in the group. Can I really do that for an infinite subgroup of $O_3$? $\endgroup$
    – levineds
    Jul 25, 2017 at 7:51
  • $\begingroup$ I love your answer! What does it mean of "G acting on the $\Bbb{C}$ G-module" generated by $f$? $\endgroup$ Nov 6, 2017 at 6:43
  • $\begingroup$ @Zhe and what does it mean of representation of a function in this context? $\endgroup$ Nov 6, 2017 at 7:15
  • $\begingroup$ @KangyuZhang As I hinted at, my colleague, who is an algebraic topologist and much better at mathematics than I am, presented this solution to me. Representations, as I understand it, are just ways to map objects to other kinds of objects with a similar property, the property here being symmetry. I can only provide very limited explanations about details here since I'm not a mathematician. I suggest consulting the internet for actual details. $\endgroup$
    – Zhe
    Nov 6, 2017 at 14:29
  • $\begingroup$ @KangyuZhang As for your other comment, see here. $\endgroup$
    – Zhe
    Nov 6, 2017 at 14:30

I was not completely happy with any answer so far, so here is my take on this: We consider some integral

$$ \int_{\Omega} f d\Omega $$

over some volume $\Omega$, ideally over $\mathbb{R}^d$, where $d$ is the dimension in which the function lives.

We next decompose the integrand into functions $f^{\alpha}_q$ spanning row $q$ of irrep $\alpha$ of the relevant (point) group, that is

$$ f = \sum_{\alpha}\sum_{q} f^{\alpha}_q c^{\alpha}_q,\quad(1) $$

where $c^{\alpha}_q\in\mathbb{C}$ are expansion coefficients.

The behaviour of function $f^{\alpha}_q$ under symmetry operation $\hat{G}$ is given by

$$ \hat{G}f^{\alpha}_q = \sum_{p}f_p^{\alpha} D^{\alpha}_{pq}(\hat{G}) $$

The representation matrices $D^{\alpha}(\hat{G})$ satisfy the great orthogonality theorem (GOT)

$$ \frac{1}{h}\sum_G D^{\alpha}_{pq}(\hat{G}) D^{\beta\ast}_{rs}(\hat{G})=\frac{1}{d_{\beta}}\delta_{\alpha\beta}\delta_{pr}\delta_{qs} $$

where $h$ is the order of the group and $d_{\beta}$ is the dimension of irrep $\beta$.

Consider now the special case that $\beta$ is the totally symmetric irrep; I will indicate this by $\beta=0$. This is a one-dimensional irrep with representation matrices $D^0_{11}(\hat{G})=1$. Inserted into the GOT we therefore get the relation

$$ \frac{1}{h}\sum_G D^{\alpha}_{pq}(\hat{G}) =\delta_{\alpha 0}\delta_{p1}\delta_{q1}\quad(2) $$

which is central in the proof.

An integral is a scalar and is thereby invariant under any symmetry operation of the relevant group. Hence

$$ \int_{\Omega} f d\Omega = \frac{1}{h}\sum_G\int_{\Omega} \hat{G}f d\Omega $$

Consider now a term in our expansion (1)

$$ \begin{array}{lcl} \int_{\Omega} f^{\alpha}_q d\Omega &=& \frac{1}{h}\sum_G\int_{\Omega} \hat{G}f^{\alpha}_q d\Omega\\ &=&\frac{1}{h}\sum_G\sum_p\int_{\Omega} f_p^{\alpha} D^{\alpha}_{pq}(\hat{G}) d\Omega\\ &=&\sum_p\int_{\Omega} f_p^{\alpha}d\Omega\left[\frac{1}{h}\sum_G D^{\alpha}_{pq}(\hat{G})\right] \end{array} $$

Using relation (2) for the term in square brackets, we immediately see that the integral is zero unless $\alpha=0$, that is, the integrand has a totally symmetric component.


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