# 1.29: Appendix A

• • Contributed by Claire Vallance
• Professor of Physical Chemistry (Department of Chemistry) at University of Oxford

## Proof that the character of a matrix representative is invariant under a similarity transform

A property of traces of matrix products is that they are invariant under cyclic permutation of the matrices.

i.e. $$tr \begin{bmatrix} ABC \end{bmatrix} = tr \begin{bmatrix} BCA \end{bmatrix} = tr \begin{bmatrix} CAB \end{bmatrix}$$. For the character of a matrix representative of a symmetry operation $$g$$, we therefore have:

$\chi(g) = tr \begin{bmatrix} \Gamma(g) \end{bmatrix} = tr \begin{bmatrix} C \Gamma'(g) C^{-1} \end{bmatrix} = tr \begin{bmatrix} \Gamma'(g) C^{-1} C \end{bmatrix} = tr \begin{bmatrix} \Gamma'(g) \end{bmatrix} = \chi'(g) \tag{29.1}$

The trace of the similarity transformed representative is therefore the same as the trace of the original representative.

## Proof that the characters of two symmetry operations in the same class are identical

The formal requirement for two symmetry operations $$g$$ and $$g'$$ to be in the same class is that there must be some symmetry operation $$f$$ of the group such that $$g' = f^{-1} gf$$ (the elements $$g$$ and $$g'$$ are then said to be conjugate). If we consider the characters of $$g$$ and $$g'$$ we find:

$\chi(g') = tr \begin{bmatrix} \Gamma(g') \end{bmatrix} = tr \begin{bmatrix} \Gamma^{-1}(f) \Gamma(g) \Gamma(f) \end{bmatrix} = tr \begin{bmatrix} \Gamma(g) \Gamma(f) \Gamma^{-1}(f) \end{bmatrix} = tr \begin{bmatrix} \Gamma(g) \end{bmatrix} = \chi(g) \tag{29.2}$

The characters of $$g$$ and $$g'$$ are identical.

## Proof of the Variation Theorem

The variation theorem states that given a system with a Hamiltonian $$H$$, then if $$\phi$$ is any normalized, well-behaved function that satisfies the boundary conditions of the Hamiltonian, then

$\langle\phi | H | \phi\rangle \geq E_0 \tag{29.3}$

where $$E_0$$ is the true value of the lowest energy eigenvalue of $$H$$. This principle allows us to calculate an upper bound for the ground state energy by finding the trial wavefunction $$\phi$$ for which the integral is minimized (hence the name; trial wavefunctions are varied until the optimum solution is found). Let us first verify that the variational principle is indeed correct.

We first define an integral

$\begin{array}{rcll} I & = & \langle\phi | -E_0 | \phi\rangle & \\ & = & \langle\phi | H | \phi\rangle - \langle\phi | E_0 | \phi\rangle & \\ & = & \langle\phi | H | \phi\rangle - E_0 \langle\phi | \phi\rangle & \\ & = & \langle\phi | H | \phi\rangle - E_0 & \text{since} \: \phi \: \text{is normalized} \end{array} \tag{29.4}$

If we can prove that $$I \geq 0$$ then we have proved the variation theorem.

Let $$\Psi_i$$ and $$E_i$$ be the true eigenfunctions and eigenvalues of $$H$$, so $$H \Psi_i = E_i \Psi_i$$. Since the eigenfunctions $$\Psi_i$$ form a complete basis set for the space spanned by $$H$$, we can expand any wavefunction $$\phi$$ in terms of the $$\Psi_i$$ (so long as $$\phi$$ satisfies the same boundary conditions as $$\Psi_i$$).

$\phi = \sum_k a_k \Psi_k \tag{29.5}$

Substituting this function into our integral $$I$$ gives

$\begin{array}{rcl} I & = & \left \langle \sum_k a_k \Psi_k | H-E_0 | \sum_j a_j \Psi_j \right \rangle \\ & = & \langle\sum_k a_k \Psi_k | \sum_j (H-E_0) a_j \Psi_j\rangle \end{array} \tag{29.6}$

If we now use $$H \Psi + E \Psi$$, we obtain

$\begin{array}{rcl} I & = & \langle\sum_k a_k \Psi_k | \sum_j a_j (E_j - E_0) \Psi_j\rangle \\ & = & \sum_k \sum_j a_k^* a_j (E_j - E_0) \langle\Psi_k | \Psi_j\rangle \\ & = & \sum_k \sum_j a_k^* a_j (E_j - E_0) \delta_{jk} \end{array} \tag{29.7}$

We now perform the sum over $$j$$, losing all terms except the $$j = k$$ term, to give

$\begin{array}{rcl} I & = & \sum_k a_k^* a_k (E_k - E_0) \\ & = & \sum_k |a_k|^2 (E_k- E_0) \end{array} \tag{29.8}$

Since $$E_0$$ is the lowest eigenvalue, $$E_k -E_0$$ must be positive, as must $$|a_k|^2$$. This means that all terms in the sum are non-negative and $$I \geq 0$$ as required.

For wavefunctions that are not normalized, the variational integral becomes:

$\frac{\langle\phi | H | \phi\rangle}{\langle\phi | \phi\rangle} \geq E_0 \tag{29.9}$

## Derivation of the secular equations – the general case of the linear variation method

In the study of molecules, the variation principle is often used to determine the coefficients in a linear variation function, a linear combination of $$n$$ linearly independent functions $$\begin{pmatrix} f_1, f_2, ..., f_n \end{pmatrix}$$ (often atomic orbitals) that satisfy the boundary conditions of the problem. i.e. $$\phi = \sum_i c_i f_i$$. The coefficients $$c_i$$ are parameters to be determined by minimizing the variational integral. In this case, we have:

$\begin{array}{rcll} \langle\phi | H | \phi\rangle & = & \langle\sum_i c_i f_i | H | \sum_j c_j f_j\rangle & \\ & = & \sum_i \sum_j c_i^* c_j \langle f_i | H | f_j\rangle & \\ & = & \sum_i \sum_j c_i^* c_j H_{ij} \end{array} \tag{29.10}$

where $$H_{ij}$$ is the Hamiltonian matrix element.

$\begin{array}{rcll} \langle\phi | \phi\rangle & = & \langle\sum_i c_i f_i | \sum_j c_j f_j\rangle & \\ & = & \sum_i \sum_j c_i^* c_j \langle f_i | f_j\rangle & \\ & = & \sum_i \sum_j c_i^* c_j S_{ij} \end{array} \tag{29.11}$

where $$S_{ij}$$ is the overlap matrix element.

The variational energy is therefore

$E = \dfrac{\sum_i \sum_jci^* c_j H_{ij}}{\sum_i \sum_J c_i^* c_j S_{ij}} \tag{29.12}$

which rearranges to give

$E \sum_i \sum_j c_i^* c_j S_{ij} = \sum_i \sum_j c_i^* c_j H_{ij} \tag{29.13}$

We want to minimize the energy with respect to the linear coefficients $$c_i$$, requiring that $$\dfrac{\partial E}{\partial c_i}$$for all $$i$$. Differentiating both sides of the above expression gives,

$\frac{\partial E}{\partial c_k}\Sigma_i \Sigma_j c_i^* c_j S_{ij} + E \Sigma_i \Sigma_j \begin{bmatrix} \frac{\partial c_i^*}{\partial c_k} c_j + \frac{\partial c_j}{\partial c_k} c_i^* \end{bmatrix} S_{ij} + \Sigma_i \Sigma_j \begin{bmatrix} \frac{\partial c_i^*}{\partial c_k}c_j + \frac{\partial c_j}{\partial c_k}c_i^* \end{bmatrix} H_{ij} \tag{29.14}$

Since $$\frac{\partial c_i^*}{\partial c_k} = \delta_{ik}$$ and $$S_{ij} = S_{ji}$$, $$H_{ij} = H_{ji}$$, we have

$\frac{\partial E}{\partial c_k} \Sigma_i \Sigma_j c_i^* c_j S_{ij} + 2E \Sigma_i S_{ik} = 2 \Sigma_i c_i H_{ik} \tag{29.15}$

When $$\frac{\partial E}{\partial c_k} = 0$$ , this gives

$\begin{array}{cll} \boxed{\Sigma_i c_i (H_{ik} - ES_{ik}) = 0} & \text{for all k} & \text{SECULAR EQUATIONS} \end{array} \tag{29.16}$