How much energy is in the electric field of a single lone hydrogen atom that has no neutrons and exists at the lowest energy state possible? Well in a universe where nothing else exists.

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    $\begingroup$ The existence of hydrogen atoms relies on quantum physics, and I'm not sure the mass-energy of a charged particle can be meaningfully separated from the energy of its electric field in the context of quantum physics. (In quantum electrodynamics, gauge invariance implies that a charged particle cannot exist without its electric field, and conversely that same electric field cannot exist without the charged particle.) $\endgroup$ Sep 12, 2021 at 21:16
  • $\begingroup$ @ChiralAnomaly, Im not following. Asking about the field not the particles themselves. I feel like Im missing something obvious here but not figuring it out. $\endgroup$
    – Jason
    Sep 14, 2021 at 23:12
  • $\begingroup$ I'll reframe my earlier comment as a request for clarification: How would you define the electric field of a hydrogen atom? $\endgroup$ Sep 15, 2021 at 0:44
  • $\begingroup$ @ChiralAnomaly, Id define it with maxwells equations, taking the proton and electron as point charges, stationary electron as if its low temp or at absolute zero, (I guess with point like spin or magnetic fields but can be ignored I think for this question), and the spacing defined by the Schrodinger equation's lowest energy level. Guess I should put that in the question? Can you reprase the comment with this in mind and thanks in advance. $\endgroup$
    – Jason
    Sep 17, 2021 at 19:35
  • $\begingroup$ What you described in the comment is a pair of classical point-charges, not a hydrogen atom. (I didn't notice the "electrostatics" tag until now.) In classical physics, the energy in the electric field of a point charge is infinite because of how the field diverges at the charge's location. Putting two oppositely-charged point particles close together doesn't make that problem go away. $\endgroup$ Sep 17, 2021 at 22:59

1 Answer 1


The answer is either a Rydberg, $R_H$, or a Hartree, $E_H$. Note that:

$$ E_H/2 = R_H \approx 13.8\,{\rm eV}$$

and it's probably negative, since hydrogen is a bound state.

The electron's total energy is the sum of kinetic ($T$) and potential energy ($V$).

$$E = T+V $$

To unbind the system into a final state with zero kinetic and potential energy requires inputing the binding energy:

$$E + R_H = T + V + R_H = 0$$


$$T+V = -R$$

The virial theorem states that, in a $r^n$ central potential, the average kinetic and potential energies are related by:

$$ 2T = -nV $$

For the Coulomb potential, $n=-1$, so that:

$$ 2T = -V $$

Substituting that in:

$$-\frac 1 2 V +V = -R$$

$$\frac 1 2 V = -R $$

$$ V = -2R = -E_H $$

So the potential energy in the field is a negative Hartree.


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