Kinetic and pseudopotential energies

We define the kinetic energy operator ${\hat T} = -{\textstyle{1 \over 2}} \nabla^2$, whose matrix elements are

$$T_{ij} = \int {\mathrm d}{\bf r}~ \phi_i({\bf r}) {\hat T} \phi_j({\bf r}) .$$ (7.30)

Since the operator is Hermitian,

$$\frac{\delta T_{ij}}{\delta \phi_{\alpha}({\bf r})} = \delta... ...\phi_j({\bf r}) + \delta_j^{\alpha} {\hat T} \phi_i({\bf r}) .$$ (7.31)

Therefore

$$\frac{\delta T_{\mathrm s}^{\mathrm J}}{\delta \phi_{\alpha}({\bf r})} = 2 \frac{\delta}{\delta \phi_{\alpha}({\bf r})} \left( K^{ij} T_{ji} \right) = 2 K^{ij} \frac{\delta T_{ji}}{\delta \phi_{\alpha}({\bf r})}$$

$$= 4 K^{\alpha \beta} {\hat T} \phi_{\beta}({\bf r}) .$$ (7.32)

The derivation for the pseudopotential energy is identical with the replacement of ${\hat T}$ by the pseudopotential operator, and so the result for the sum of these energies is just

$$\frac{\delta E_{\mathrm{kin,ps}}}{\delta \phi_{\alpha}({\bf ... ... + {\hat V}_{\mathrm{ps,tot}} \right) \phi_{\beta}({\bf r}) .$$ (7.33)