The criteria for vanishing integrals
المؤلف:
Peter Atkins، Julio de Paula
المصدر:
ATKINS PHYSICAL CHEMISTRY
الجزء والصفحة:
ص419 -421
2025-12-02
66
The criteria for vanishing integrals
The key point in dealing with the integral I is that the value of any integral, and of an overlap integral in particular, is independent of the orientation of the molecule (Fig. 12.24). In group theory we express this point by saying that I is invariant under any symmetry operation of the molecule, and that each operation brings about the trivial transformation I → I. Because the volume element dτ is invariant under any symmetry operation, it follows that the integral is nonzero only if the integrand itself, the product f1f2, is unchanged by any symmetry operation of the molecular point group. If the integrand changed sign under a symmetry operation, the integral would be the sum of equal and opposite contributions, and hence would be zero. It follows that the only contribution to a nonzero integral comes from functions for which under any symmetry operation of the molecular point group f1f2 → f1f2, and hence for which the characters of the operations are all equal to +1. Therefore, for I not to be zero, the integrand f1f2 must have symmetry species A1 (or its equivalent in the specific molecular point group).
We use the following procedure to deduce the symmetry species spanned by the product f1f2 and hence to see whether it does indeed span A1.
1 Decide on the symmetry species of the individual functions f1 and f2 by reference to the character table, and write their characters in two rows in the same order as in the table.
2 Multiply the numbers in each column, writing the results in the same order.
3 Inspect the row so produced, and see if it can be expressed as a sum of characters from each column of the group. The integral must be zero if this sum does not contain A1.
For example, if f1 is the sN orbital in NH3 and f2 is the linear combination s3 = sB − sC (Fig. 12.25), then, because sN spans A1 and s3 is a member of the basis spanning E, we write
The characters 2, −1, 0 are those of E alone, so the integrand does not span A1. It follows that the integral must be zero. Inspection of the form of the functions (see Fig. 12.25) shows why this is so: s3 has a node running through sN. Had we taken f1 = sN and f2 = s1 instead, where s1 = sA + sB + sC, then because each spans A1 with characters 1,1,1:
The characters of the product are those of A1 itself. Therefore, s1 and sN may have nonzero overlap. A shortcut that works when f1 and f2 are bases for irreducible representations of a group is to note their symmetry species: if they are different, then the integral of their product must vanish; if they are the same, then the integral may be nonzero.
It is important to note that group theory is specific about when an integral must be zero, but integrals that it allows to be nonzero may be zero for reasons unrelated to symmetry. For example, the N-H distance in ammonia may be so great that the (s1, sN) overlap integral is zero simply because the orbitals are so far apart.
In many cases, the product of functions f1 and f2 spans a sum of irreducible representations. For instance, in C2v we may find the characters 2, 0, 0, −2 when we multiply the characters of f1 and f2 together. In this case, we note that these characters are the sum of the characters for A2 and B1:
To summarize this result we write the symbolic expression A2 × B1 = A2 + B1, which is called the decomposition of a direct product. This expression is symbolic. The × and +signs in this expression are not ordinary multiplication and addition signs: formally, they denote technical procedures with matrices called a ‘direct product’ and a ‘direct sum’. Because the sum on the right does not include a component that is a basis for an irreducible representation of symmetry species A1, we can conclude that the integral of f1f2 over all space is zero in a C2v molecule. Whereas the decomposition of the characters 2, 0, 0, −2 can be done by inspection in this simple case, in other cases and more complex groups the decomposition is often far from obvious. For example, if we found the characters 8, −2, −6, 4, it would not be obvious that the sum contains A1. Group theory, however, provides a systematic way of using the characters of the representation spanned by a product to find the symmetry species of the irreducible representations. The recipe is as follows:
1 Write down a table with columns headed by the symmetry operations of the group.
2 In the first row write down the characters of the symmetry species we want to analyse.
3 In the second row, write down the characters of the irreducible representation Γ we are interested in.
4 Multiply the two rows together, add the products together, and divide by the order of the group. The resulting number is the number of times Γ occurs in the decomposition.
Fig. 12.24 The value of an integral I (for example, an area) is independent of the coordinate system used to evaluate it. That is, I is a basis of a representation of symmetry species A1 (or its equivalent).
Fig. 12.25 A symmetry-adapted linear combination that belongs to the symmetry species E in a C3v molecule such as NH3. This combination can form a molecular orbital by overlapping with the px orbital on the central atom (the orbital with its axis parallel to the width of the page; see Fig. 12.28c).
Fig. 12.26 The integral of the function f = xy over the tinted region is zero. In this case, the result is obvious by inspection, but group theory can be used to establish similar results in less obvious cases. The insert shows the shape of the function in three dimensions.
Fig. 12.27 The integration of a function over a pentagonal region. The insert shows the shape of the function in three dimensions.
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