Electric fields created by electrons

Question

If I want to calculate the electric field inside the $1s$ cloud of an atom. Do I have to consider the protons only or the electrons too? If I take the electric field created by protons only then won't electrons create their own field and distort it?

Answer 1

The total charge density of proton ($p$) and electron ($e^-$) is given by $$ \rho(\vec{x})= \rho_{ p}(\vec{x})+\rho_{e^-}(\vec{x}),$$ with $$\begin{align}\rho_p(\vec{x})= e\, \delta^{(3)}(\vec{x}) \quad {\text{ and}} \quad \rho_{e^-}(\vec{x})= -e \,|\psi_{1s}(\vec{x})|^2 = -e \, \frac{e^{-2r/a}}{\pi a^3} \, , \end{align} $$ where the proton charge was approximated by a point charge and $a$ is the Bohr radius. The electric field generated by the proton is a pure Coulomb field given by $$ \vec{E}_p(\vec{x}) = E_p(r) \, \vec{e}_r, \qquad E_p(r) =\frac{e}{4 \pi \varepsilon_0 r^2},$$ where $r=|\vec{x}|$ and $\vec{e}_r= \vec{x}/r$. The electric field generated by the $1s$ state of the electron is obtained by using the Gauss law, $$ \begin{align}4 \pi r^2 E_{e^-}(r) &=\frac{4 \pi}{\varepsilon_0} \int\limits_0^r \! dr^\prime \, r^{\prime \,2} \, \rho_{e^-}(r^\prime) \\ &=-\frac{4 e}{\varepsilon_0 a^3} \int\limits_0^r \! dr^\prime r^{\prime\, 2}\, e^{-2r^\prime/a} \\ &= -\frac{e}{\varepsilon_0} \left[ 1- e^{-2r/a} \left( 1+2r/a+2r^2/a^2 \right)\right]. \end{align}$$ Because of the linearity of the Maxwell equation ${\rm div} \vec{E} = \rho/\varepsilon_0$ (see the comment by @Ghoster), the total electric field generated by the proton and the electron is finally given by the sum $$\vec{E}(\vec{x})= \left[ E_p(r) + E_{e^-}(r) \right] \vec{e}_r, $$ falling off exponentially for $r \to \infty$.

Summary: This problem boils down to the simple fact that Maxwell's equations (in this case, only the equations of electrostatics needed), ${\rm curl} \, \vec{E}=0$ and ${\rm div} \, \vec{E}= \rho/\varepsilon_0$, are linear, meaning that finding solutions for ${\rm curl} \, \vec{E}_a=0$, ${\rm div} \, \vec{E}_a = \rho_a$ gives you at the same time a solution for the problem ${\rm curl} \, \vec{E}=0$, ${\rm div } \, \vec{E}=\sum_a \rho_a/\varepsilon_0$ given by $\vec{E} = \sum_a \vec{E}_a$.

Answer 2

In the days when I did this sort of calculation we normally worked with the potential and if we needed the field we'd get it by differentiating the potential.

The advantage of using a potential is because the electric field is linear the potentials just add. The potential from the nucleus is approximately just a point charge potential. The potential from the electrons is complicated because, as you say, the electrons do create their own electric fields and they interact with each other as well as with the nucleus.

The way we get round this is using an approximation called the self consistent field approach. We assume some form for the atomic orbitals and calculate the average potential from the electrons in those orbitals. Then we use this average potential to recalculate the orbitals, then use the recalculated orbitals to recalculate the average potential, and so on. With luck the process converges and we end up with a good approximation to the real orbitals.

So the simple answer to your question is we do need to include the fields from both the nucleus and the electrons, but since potentials just add we can calculate the two potentials separately and just add them together.