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Chapter 12: Problem 8

Show group theoretically that in a tetrahedral molecule (a) the mean hyperpolarizability is zero, (b) the only non-zero components are \(\beta_{x y z}\) and the permutations of its indices. Hint. The mean is defined as \(\frac{3}{3}\left(\beta_{x x z}+\beta_{y z z}+\beta_{z z z}\right)\) and so (b) implies (a). For (b) consider the symmetry characteristics of \(E=-(1 / 3 !) \Sigma_{a, b, c} \beta_{a b c} E_{a} E_{b} E_{c}\) the generalization of eqn 12.11.

Short Answer

Expert verified
The only non-zero components of the hyperpolarizability tensor of a molecule with Td symmetry are those that belong to \(T_1\) or \(T_2\). This implies \(\beta_{xyz}\) and its permutations (antisymmetric with respect to inversion) are the only non-zero components. The mean hyperpolarizability, calculated as the average of \(\beta_{x x z}\), \(\beta_{y z z}\), and \(\beta_{z z z}\), is zero since these components are symmetric with respect to inversion and thus zero.

Step by step solution

01

Understand the Problem

A tetrahedral molecule possesses Td symmetry, which is characterized by three 8-dimensional representations: \(A_1\), \(A_2\), \(E\), and two 3-dimensional representations: \(T_1\), \(T_2\). The given expression for energy \(E\) indicates third-rank tensor \(\beta_{abc}\), so we need to identify the transformation properties of this tensor that corresponds to the \(xyz\) components mentioned.
02

Break down \(\beta_{abc}\)

The tensor \(\beta_{abc}\) is a third-rank tensor and can be decomposed into irreducible representations of Td group. Using group theory, it can be shown that the reducible representation, \(\Gamma_{\beta_{abc}}\) of \(\beta_{abc}\) can be written as : \[ \Gamma_{A1} + 2\Gamma_{E} + 3\Gamma_{T1} + 3\Gamma_{T2} \] where \(\Gamma\) structures stand for different symmetry representations.
03

Determine Non-Zero Components Based on Symmetry Arguments

Knowing that tetrahedral molecule has inversion symmetry (\(i\)-symmetry), we can infer that only antisymmetric components of tensor with respect to inversion survive. The antisymmetric components are those which change sign upon inversion. Examining the symmetry types, we see that \(T_1\) and \(T_2\) are antisymmetric with respect to inversion. Thus, we are only interested in \(T_1\) and \(T_2\) components of third-rank tensor. Therefore, we can conclude that \(\beta_{xyz}\) and its permuted indices are the only non-zero components
04

Show the Mean Hyperpolarizability Is Zero

Since \(E\) is proportional to the sum of \(\beta_{x x z}\), \(\beta_{y z z}\), and \(\beta_{z z z}\) which are not antisymetric with respect to inversion, all these components are zero (as they are not part of \(T_1\) or \(T_2\)), thus the mean hyperpolarizability is zero

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Most popular questions from this chapter

Evaluate the rotational strength of a transition of an electron from a \(2 \mathrm{p}_{x}\) -orbital to a \(2 \mathrm{p}_{z}, 3 \mathrm{d}_{x y}\) -hybrid orbital. Assume the orbitals are on a carbon atom. Estimate the optical rotation angle for 590 nm light. Hint. Follow Example \(12.4,\) with changes of detail. For carbon, take \(\zeta_{p}=1.57 / a_{0}\) and \(\zeta_{d}=0.33 / a_{0}\) and use \(\lambda_{k 0}=200 \mathrm{nm}\).

Prove the sum rule \\[ \Sigma_{p} x_{m f} x_{f m} \omega_{f n}=\left(\hbar / 2 m_{e}\right) \delta_{m n}+\frac{1}{2} \omega_{m n}\left(x^{2}\right)_{m n^{+}} \\] Hint. Consider the matrix elements of the commutator \\[ \left[H, x^{2}\right] \\].

Evaluate the polarizability \(\alpha_{z z}\) and polarizability volume of a hydrogen atom; for simplicity, confine the perturbation sum to the 2 p-orbitals. Use the following results for matrix elements: \\[ \begin{array}{l} \left\langle n^{\prime}, l+1, m_{l}|z| n l m_{l}\right\rangle=\left\\{\frac{(l+1)^{2}-m_{l}^{2}}{(2 l+3)(2 l+1)}\right\\}^{1 / 2}\left\langle n^{\prime}, l+1|r| n l\right\rangle \\ \langle n p|r| 1 \mathrm{s}\rangle=\left\\{\frac{2^{8} n^{7}(n-1)^{2 n-5}}{(n+1)^{2 n+5}}\right\\}^{1 / 2} a_{0} \end{array} \\]
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