Print publication date: 2014

Print ISBN-13: 9780199670888

Published to Oxford Scholarship Online: April 2014

DOI: 10.1093/acprof:oso/9780199670888.001.0001

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(p.385) A12 Vanishing integrals

Source:
Symmetry of Crystals and Molecules
Publisher:
Oxford University Press

A12.1 Introduction

If the sine function

(A12.1)
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is integrated, the result is zero for any limit ±p, because of the antisymmetry of the sine function across the origin. This type of function could not span the totally symmetric A1 irreducible representation. Consider next the integral

(A12.2)
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where $ψ1andψ2$ are wave functions and the integration is over the space of τ. Since the value of the integral $I.$ is independent of molecular orientation, any symmetry operation acting on $I.$ is equivalent to the transformation $I→I$, that is, a product $ψ1ψ2$ can be the basis for the A1 irreducible representation. For example, in point group C3v, suppose that $ψ1andψ2$span A1 and E respectively, then the table hereunder follows:

E

2C3

v

$ψ1(A1)$

1

1

1

$ψ2(E)$

2

−1

0

The product $ψ1ψ2$ is 2 −1 0, which has the symmetry of E: A1 is not present ($A1⊗E=E,$ Appendix A10.2.1) and the integral, Eq. (A12.2), vanishes because $ψ1andψ2$ do not transform as the same irreducible representation of the point group. Had $ψ2$ also spanned A1, then $ψ1ψ2$ would have had the symmetry of A1, and the integral would not necessarily have vanished; it could however, vanish for reasons unrelated to symmetry [1]. The arguments may be extended to products of more than two functions.

Example A12.1

Can the integrals, Eq. (A12.2), of the functions (a) $x2−y2,$ and (b) $x2−y2+z2$be non-vanishing when integrated over a square centred at the origin?

A square corresponds with point group D4h. Hence, from the character table for this point group: (a) The function $x2−y2$spans B1g and the integral vanishes. (b) The function $x2−y2+z2$spans A1g + B1g, and the integral can be now non-vanishing.

(p.386) A12.2 Spectroscopic applications

In applying quantum mechanical principles to spectroscopic transitions, integrals of the form

(A12.3)
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are encountered, where i and f are initial and final states of the system, and μ is a transitional moment operator, the dipole moment operator, for example, which may be written as

(A12.4)
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where ei is the charge of the ith particle, and xi, yi and zi are the coordinates of its position vector ri, where i, j and k are unit vectors along x, y and z, respectively. Now, Eq. (A12.3) can be resolved into three integrals, that along the x axis, for example, being

(A12.5)
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The integral, Eq. (A12.3), will be non-zero if any one of its component integrals like Eq. (A12.5) is non-zero: for this condition to exist, the direct product of $ψi,μxandψf$, or its counterpart in y or z, should contain the fully symmetric type A1 representation [2].

The ground state vibrational wave function $ψi$ of an atom has the same mathematical form as that of an s atomic orbital, so it is spherically (totally) symmetric, which means that there are no symmetry restrictions attached to it.

Consider the water molecule, point group C2v. The character table shows that x, y and z have the symmetries $B1,B2andA1$, respectively. Following Section 8.3.3.2, it can be shown that this molecule has vibrational (final) states $ψf$ corresponding to A1 and B2 symmetries. Hence, the integral Eq. (A12.3) may be investigated by the direct products

(A12.6)
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With $Γψ1=A1$ and $Γψf=A1$,

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(p.387) and with $Γψ1=A1$ and $Γψf=B2$,

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Thus, Eq. (A12.3) can be non-zero in both A1 and B2 symmetries because the direct products contain the fully symmetric A1 representation in each case.

References

[1] Bishop DM. Group theory and chemistry. Clarendon Press, 1973.

[2] McWeeny R. Symmetry. Pergamon Press, 1963.