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..!

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    $\begingroup$ You must read this part from Feynman's Lecture where he actually points this out quite clearly: The size of an atom; also you can also read this: The hydrogen molecule. $\endgroup$
    – user36790
    Commented Sep 1, 2016 at 15:30
  • $\begingroup$ It seems that neither of the two links is talking about the delocalization of electrons. The second one is interesting, although it was not my point, and is simply saying that the potential interaction between the two protons and the two electrons gives rise to an equilibrium point at 0,74A for the interproton distance. $\endgroup$
    – Toool
    Commented Sep 3, 2016 at 14:50
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    $\begingroup$ Think of an electron in an infinite square well: all their energy is kinetic, and a bigger well has lower ground state energy. $\endgroup$
    – Roger V.
    Commented Jun 27, 2022 at 13:56

1 Answer 1


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).

  • $\begingroup$ Thank you for this answer. It does make more sense to me now ! I guess that the optimal λ corresponds to the equilibrium point in the potential energy ? $\endgroup$
    – Toool
    Commented Sep 5, 2016 at 15:03
  • $\begingroup$ It corresponds to the minimum energy state for this type of trial wave function. The above example is of course inspired by the hydrogen atom where we know that the eigenfunctions are of product form, and that the lowest energy states are purely radial. So for this particular case you can show that you indeed obtain the minimum energy state. Another example you can try is the harmonic oscillator where the potential scales as $\lambda^2$. Also there you see that to minimize the energy, you have to find the optimal point in the tug of war between kinetic and potential energy. $\endgroup$
    – Max Lein
    Commented Sep 6, 2016 at 1:05

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