Chemistry textbooks on atomic orbitals typically start off with the concept of electrons (viewed as negatively charged point-particles) moving around the nucleus, attracted and bound by the Coulomb force. They then explain that in quantum mechanics one has to solve the time-independent Schroedinger equation. This leads to a discussion of orbitals. Finally they point out that the probability density for the electron can be regarded as a charge density, a stationary cloud of negative charge.

I assume that the interpretation of an orbital as a charge density might well be useful in certain applications. However, there is one aspect that puzzles me. If an electron in an orbital truly behaves like a stationary charge density, shouldn't the Hamiltonian describing the system take into account the electric field energy resulting from this charge density? In other words, the energy required to build this charge density against its own electrostatic repulsion. This would yield a positive term going with 1/r, and its effect in the Schroedinger equation would be to increase the orbital radius.

Since the Schroedinger equation works fine without this extra term, may I conclude that there is no electric field energy associated with the above described charge density?


1 Answer 1


Finally they point out that the probability density for the electron can be regarded as a charge density, a stationary cloud of negative charge.

But, this is wrong, it only a "probability density" in the sense that this is the probability to find a particle at a given location for a given associated (hypothetical) large number of identical systems prepared in the same quantum state. What I just wrote is basic quantum mechanics... only we often stop using the precise language because it takes a long time to say it and is easier to say "probability to find the electron at x".

An electron is a point particle with no physical extent and thus no associated electrostatic energy required to "build it up" out of smaller pieces. The same is not true, for example, of a proton (but the proton is somewhat complicated for other reasons).

Furthermore, if you move away from single-electron quantum mechanics you do have to take into account the interaction energy between electrons. The hamiltonian looks like this (choosing units such that $e=\hbar=m=1$): $$ \sum_i p_i^2/2 + v_{ext}(r_i)+\frac{1}{2}\sum_{i\neq j}\frac{1}{|\vec r_i-\vec r_j|}\;, $$ where the last term is the interaction term, which is famously difficult to deal with...

  • $\begingroup$ Thank you. I fully agree, of course. Unfortunately nowadays there is trend (even among theoretical physicists) to say that the particle-wave duality questions dating back to DeBroglie and Schroedinger are obsolete. Why? Because the electron is fully described by the wave function, therefore the electron is the wave function! In this view, if the wave equation and the probability density are stationary, then the electron is just a motionless negatively charged cloud. $\endgroup$
    – M. Wind
    Apr 2, 2015 at 4:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.