# [Physics] How the electron delocalization can lower its kinetic energy

condensed-matterheisenberg-uncertainty-principlequantum mechanics

In covalent bonding, as in metallic bonding, the lowering in energy relative to the free-atom state is achieved by the reduction in the kinetic energy of valence electrons due to their delocalization.

I understand that according to Heisenberg's uncertainty principle the delocalization of valence electrons lowers their momentum uncertainty (standard deviation), but I don't see why it should lower the momentum itself and so the kinetic energy..!

The (non-relativistic) kinetic energy expectation value of a particle moving in $$\mathbb{R}^d$$ is proportional to $$\lvert \nabla \psi \rvert^2$$, so if you delocalize it, you make the gradients — and hence, the kinetic energy — smaller. So if you rescale a wave function $$\psi_{\lambda}(x) = \lambda^{d/2} \, \psi(\lambda x)$$ by $$\lambda$$, then you see that the kinetic energy scales with $$\lambda^2$$, i. e. if $$\lambda$$ is small, then the kinetic energy expectation value with respect to $$\psi_{\lambda}$$ is $$\lambda^2$$ times the expectation value with respect to $$\psi$$.
Of course, usually there is a price you pay by delocalising because decreasing the kinetic energy means you eventually increase the potential energy expectation value. Try minimizing the total energy expectation value for $$H = \frac{1}{2m} (- \mathrm{i} \partial_r)^2 - \frac{e}{r}$$ by scaling $$\psi(r)$$. You will see that there is an optimal point between $$\lambda = 0$$ (completely delocalized) and $$\lambda = \infty$$ (localized in a single point).