7.4 Physical interpretation

At this stage we examine the energy gradients derived in section 7.2. At the minimum of the total functional $Q[\rho;\alpha]$,

$$\frac{\partial Q[\rho;\alpha]}{\partial K^{\alpha \beta}} = ... ...a} + 2 \alpha \left[ SKS(1-KS)(1-2KS) \right]_{\beta \alpha} .$$ (7.44)

Making the Löwdin transformation of this gradient into the representation of a set of orthonormal orbitals (using the results of section 4.6) yields

$$2 S^{1 \over 2} \left[ {\tilde H} + \alpha {\tilde K} (1 - {\tilde K}) (1 - 2 {\tilde K}) \right] S^{1 \over 2} = 0$$ (7.45)

which (pre- and post-multiplying by $S^{-{1 \over 2}}$) simplifies to

$${\tilde H} + \alpha {\tilde K} (1 - {\tilde K}) (1 - 2 {\tilde K}) = 0 .$$ (7.46)

This result shows that at the minimum, ${\tilde K}$ and ${\tilde H}$ can be diagonalised simultaneously, and will therefore commute. The result of the variation of the density-kernel is to make the density-matrix commute with the Hamiltonian in the representation of the current support functions. Transforming to the diagonal frame by making a unitary transformation (the eigenvalues of ${\tilde K}$ being $f_i$ and those of ${\tilde H}$ being $\varepsilon_i$) we obtain the following relationship:

$$\varepsilon_i + \alpha f_i (1 - f_i) (1 - 2 f_i) = 0.$$ (7.47)

For the derivative with respect to the support functions we have

$$\frac{\delta Q[\rho;\alpha]}{\partial \phi_{\alpha}({\bf r})... ...K(1-SK)(1-2SK)]^{\alpha \beta} \right\} \phi_{\beta}({\bf r})$$ (7.48)

which can again be transformed first into an orthonormal representation defined by the Löwdin transformation:

$$\varphi_{\alpha}({\bf r}) = \phi_{\beta}({\bf r}) S_{\beta \alpha}^{-{1 \over 2}}$$ (7.49)

to obtain

$$\frac{\delta Q[\rho;\alpha]}{\delta \varphi_{\alpha}({\bf r})} = \int {\mathrm d}{\bf r'}~ \frac{\delta Q[\rho;\alpha]}{\delta \ph... ... \beta}^{1 \over 2} \frac{\delta Q[\rho;\alpha]} {\delta \phi_{\beta}({\bf r})}$$

$$= 4 S_{\alpha i}^{1 \over 2} K^{ij} \left[ {\hat H} + \alpha SK(1-SK)(1-2SK) \right]_j^k S_{kl}^{1 \over 2} \varphi_l({\bf r})$$

$$= 4 \left[ {\tilde K} \left\{ {\hat H} + \alpha {\tilde K} (1 - {\t... ...) (1 - 2 {\tilde K}) \right\} \right]_{\alpha \beta} \varphi_{\beta}({\bf r}) .$$ (7.50)

Assuming that we have performed the minimisation with respect to the density-kernel for the current support functions, transforming to the representation which simultaneously diagonalises the Hamiltonian and density-matrix, by the unitary transformation $\psi_i({\bf r}) = \varphi_{\alpha}({\bf r}) U_{\alpha i}$, yields

$$\frac{\delta Q[\rho;\alpha]}{\delta \psi_i({\bf r})} = U_{i \alpha}^{\dag } \frac{\delta Q[\rho;\alpha]}{\delta \varphi_{\alpha}({\bf r})}$$

$$= 4 U_{ij}^{\dag } \left\{ {\hat H} \delta_{jk} + \alpha \left[ SK(1-SK)(1-2SK) \right]_{jk} \right\} U_{kl} \psi_l({\bf r})$$

$$= 4 f_i \left[ {\hat H} + \alpha f_i (1 - f_i)(1 - 2 f_i) \right] \psi_i({\bf r})$$ (7.51)

which is a Kohn-Sham-like equation, but where the energy eigenvalue $\varepsilon_i$ does not explicitly appear since no orthonormalisation constraint is explicitly applied. Using equation 7.47, however, yields

$$\frac{\delta Q[\rho;\alpha]}{\delta \psi_i({\bf r})} = 0 = 4 f_i \left[ {\hat H} - \varepsilon_i \right]\psi_i({\bf r})$$ (7.52)

and so, at least for $f_i \not= 0$, we see that the support function variations are equivalent to making the related wave-functions obey the Kohn-Sham equations. The factor of $f_i$ will slow this convergence for unoccupied bands, since for $f_i \approx 0$ the gradient is small. In the next section (7.5) we therefore turn our attention to a potential method for eliminating this problem.