How is group theory used to deduce which of these integrals are equal to 0?

Question

The number of all two-electron integrals: $$ \tag{1} \langle \phi_1 \phi_2|\phi_3\phi_4 \rangle = \int d^3\mathbf r' \int d^3\mathbf r'' \, \phi_1(\mathbf r'') \, \phi_2(\mathbf r') \frac{1}{|\mathbf r' - \mathbf r''|} \, \phi_3(\mathbf r') \, \phi_4(\mathbf r''), $$

for $N$ number of basis functions (I am using real-valued ones) is $N^4$.

Not all of them are unique, but the number of unique integrals is $N(N+1)(N^2 + N + 2)/8$.

Yet this number is still huge for a reasonably accurate basis set. I have a feeling that this number can be significantly reduced if one makes use of the point-group symmetry of the molecule to determine the vanishing integrals. If yes, how is this done given the knowledge of the point group?

Answer 1

The question is too broad to be answered directly so I will provide a somewhat general scheme.

Basically in an integral like $$ \int d\mu A B C $$ one would seek to expand each part in irreducible representations of a given group, say for instance \begin{align} B=\frac{1}{\vert \mathbf{r}-\mathbf{r}^\prime\vert} =\frac{1}{r} \sum_{\ell} \left(\frac{r'}{r}\right)^\ell \sqrt{\frac{4\pi}{2\ell+1}}Y^{\ell}_0(\theta,\varphi) \end{align} where here the group would be $SO(3)$ and the irreducible representations are labelled by $\ell$. Doing the same for $C$ and $A$, v.g. \begin{align} C&=\sum_{\ell m} c_{\ell m}Y^\ell_{m}(\theta,\varphi)\, ,\\ A&=\sum_{\ell m} a_{\ell m}Y^\ell_{m}(\theta,\varphi)\, . \end{align} The integral then becomes \begin{align} \frac{1}{r}\sum_{\ell_1m_1;\ell_2m_2;\ell} a_{\ell_1m_1}c_{\ell_2m_2}\left(\frac{r'}{r}\right)^\ell \sqrt{\frac{4\pi}{2\ell+1}} \int Y^{\ell_1}_{m_1}(\theta,\varphi)Y^\ell_0(\theta,\varphi) Y^{\ell_2}_{m_2}(\theta,\varphi) \tag{1} \end{align} and the last term is automatically $0$ unless we have \begin{align} \ell_1\otimes\ell\otimes\ell_2&=\mathbf{0}+\ldots...\, , \tag{2a}\\ m_1+m_2&=0 \tag{2b} \end{align} where (2a) comes from angular momentum coupling to the representation $\mathbf{0}$ (i.e. total $L=0$) and (2b) is the condition on $SO(2)\sim U(1)$ that the resulting magnetic quantum number is $0$.

There is nothing a priori to restrict the sum over $\ell_1,\ell_2,\ell$ in (1) unless you have some prior knowledge of $A$, $B$ and $C$.

The same general principle holds for point groups. In the case of point groups you would expand each $A$, $B$, $C$ in terms of representation of the specific point group, and use the great orthogonality theorem of representations (also called Schur orthogonality relations). Probably the integral would be broken in group elements multiplied by cosets, i.e .the integration would be written as $g\cdot h$ where $g$ is in the group, and some sums over $g$ would be $0$ if the combination of representations contained in the decomposition of $A$, $B$ and $C$ can be combined to the identity (or trivial) representation. There would then remain integration over cosets. This is a bit of what happens in the example above: writing a rotation as $R_z(\varphi) R_y(\theta)$ (there is no third angle here) the condition $m_1+m_2=0$ gets rid of the $R_z(\varphi)$ integration and the result is an integration over $R_y(\theta)$ only.

Prof. Mildred Dresselhaus of MIT still has coursenotes available, and co-wrote an excellent textbook on the general topic.

Edit:

So it seems your “real solid harmonics“ are basically the same as my spherical harmonics, up to some linear combinations.

0. I do not understand your comment re: Hilbert space. The Hilbert space here is the space of all 2-particle states (as you have written your states as products of two states).

So a more or less general procedure would be as follows.

1. Find the linear combinations of your basis sets that transform by irreducible representations of your points group. For instance, if you “only” need axial symmetry, then combinations of the type $Y^\ell_m\pm Y^\ell_{-m}$ will produce cosine and sine pieces that are symmetric or antisymmetric w/r to reversal of the $\hat z$ axis. There are systematic ways of finding these, using projection operator techniques (someone already pointed that out).

2. This decomposition is usually not that bad if the group has few representations but then some irreps may occur more than once and it can be a computational headache unless one is careful. In other words, the projection technique might provide you with multiple solutions which you have to specialize and properly normalize. The projection gives you (usually) one state in the irrep and you may have to work a little more to construct the remaining states, although with point groups the matrix representations are well known so it’s not that bad.

3. Basically the step above means you are no longer working with functions $\boldsymbol{\phi_3}\boldsymbol{\phi_4}$ in your original basis set but some combinations of states. You also need to expand the Coulomb term in this way.

4. The last step is to use orthogonality of group functions to eliminate some terms. The non-zero terms that survive are those for which the tensor product $\Gamma^*_k\otimes \Gamma_r\otimes \Gamma_i$ contains the identity representation. Here, $\Gamma^*_k$ is one piece in the sum of the expansion of $\phi_1\phi_2$, $\Gamma_r$ is one piece in the sum of the expansion of $1/\vert\mathbf{r}-\mathbf{r'}\vert$, and $\Gamma_i$ is one piece in the expansion of $\phi_3\phi_4$. This type of triple product may occur more than once for $(k,r,i)$ if the irreps $\Gamma_k$ etc occur more than once in the decomposition of the old basis set in the basis set.

You get to decide if finding these combinations ends up saving time over simply evaluating the original integrals.

Answer 2

First, you have to transform all the basis functions into the irreducible representations (irreps) of the point group of the molecule. You can do this with standard projection formulas.

Once, you know the irreps of the basis functions, you have to look at the product table of the point group to find out whether the product of those four basis functions contains the totally symmetric irrep. If it does, then the integral has to be computed. Otherwise you know it is vanishing.

I think a good reference is Albert Cotton's "Chemical Applications of Group Theory"

Answer 3

To start with, while it's not as necessary, you can apply symmetry arguments to one-electron integrals. Consider $\langle\mu|O_1|\nu\rangle$, where $O_1$ is some one-electron operator. If the molecule has some point group symmetry, we can form basis functions/operators that are irreducible representations of the group. Once we have the functions expressed in terms of irreps (as described in a question here), we can easily determine which integrals must vanish just from the group multiplication table: $$\Gamma_{\text{TSR}}\notin\Gamma_\mu \otimes\Gamma_{O_1} \otimes\Gamma_\nu\to\langle\mu|O_1|\nu\rangle=0$$ Here, $\Gamma_x$ is the irrep of $x$ and TSR is the totally symmetric representation. So if the symmetric representation is not in the product of the irreps, the integral must vanish.

To give a concrete example, consider a water molecule, which has $C_{2v}$ symmetry (character table) We can look at the very simple example of overlap integrals ($O_1=1$). In this case, the integrals will only be nonzero when the irrep of $\mu$ and $\nu$ are the same, as this ensures the TSR is in their product. In the best case in terms of reducing the cost, you will have an equal number of basis functions of each irrep. For water, this would reduce the number of overlap integrals needed from $(N^2+N)/2$ to $(\frac{N^2}{4}+N)/2$ where the $4$ comes from the number of irreps.

Two electron integrals are in principle the same, but involve 4 functions rather than 2. Also, we are almost always interested in the Coulomb operator, which is totally symmetric, so we only need to consider the irreps of the 4 functions. We can reduce this to looking at the symmetry of pairs of functions/charge distributions to make the evaluation basically the same as the case of the overlap integrals, though obviously with many more integrals. Combined with integral screening (e.g Cauchy-Schwarz) you can see a fairly substantial reduction in size. While Hartree-Fock is formally $O(N^4)$ due to forming the 2e integrals, in practice with codes that's use integral screening and symmetry this can be reduced to less than $O(N^3)$.

Others have mentioned the limits of applying symmetry and that many molecules in for example biology have no symmetry. However, there are areas like inorganic chemistry, where many molecules of interest are symmetric or nearly so. This can used to accelerate, as an example, geometry optimizations by using the idealized, symmetric geometry, which can be obtained at reduced cost, as a guess to obtain the true, nearly symmetric structure.

Answer 4

Boy, you're not starting off easy. Proper implementation of symmetry is quite a job, especially since most systems of interest nowadays have no symmetry.

For a reference, you can look at e.g. Dovesi's work on the use of symmetry in CRYSTAL, which is a periodic Hartree-Fock code using Gaussian orbitals. Symmetry is much more important in the periodic case, since the periodic packing introduces many more symmetries than in molecules. However, at the $\Gamma$ point (${\bf k}={\bf 0}$) you're essentially reduced to molecular symmetry. (CRYSTAL can use symmetry with periodic boundaries in 0, 1, 2, or 3 dimensions for molecules, rods, planes, and crystalline systems.)

Int. J. Quantum Chem 29, 1755 (1986)