I Why can't atoms accept energy from static electric fields?

1. Jan 6, 2017

Electric to be

Though I am definitely not an expert in any way in QM, I understand that on a basic level quantum systems can be solved using schrodinger's equation.

For a hydrogen atom, the wave function of the electron is found by using the hydrogen proton as an external potential and proceeding from there to find discrete energy levels eigenvalues.

Then, this electron can be excited by accepting photons of discrete energies. Sure, makes sense. Now although I may be speaking nonsense, I know that all electric and magnetic fields have an energy density that is proportional to the field strength squared. I also know that a photon is roughly modeled as the particle manifestation of an electromagnetic wave, which is basically a changing electric and magnetic field, propagated according to Maxwell'a equation.

Why is it that the energy from this electric and magnetic field, which is propagating can be accepted by the electron, but the energy stored in the static electric field of the potential field which is used to solve the schrodinger equation, cannot be?

Thank you.

2. Jan 6, 2017

hilbert2

If you put an electron in a constant electric field $\mathbf{E}(x,y,z)=E \mathbf{u}_x$ (where $\mathbf{u}_x$ is the unit vector to x direction and $E$ is a scalar constant), the electron will accelerate indefinitely by accepting energy from the field. The reason for this is that the electron in that kind of field isn't "confined", so it doesn't have a ground state.

EDIT: Just don't confuse this with the "confinement" of quarks in a nucleon (proton or neutron), which means that they can't be found in free form. It might be better to say that in an atom, electron is in a "binding" potential and therefore has some lowest possible energy.

Last edited: Jan 6, 2017
3. Jan 6, 2017

Staff: Mentor

Because you can't use a classical model to understand a quantum phenomenon. The model of an EM wave as a propagating EM field is a classical model. It can't explain the phenomenon you are describing--in fact it can't explain the existence of discrete energy levels in atoms at all.

The proper general way to incorporate the EM field into quantum mechanics, at least in a non-relativistic context, is to use the scalar and vector potential to modify the appropriate operators (for example, we replace the momentum operator $- i \hbar \nabla$ with $- i \hbar \nabla - e A$, where $A$ is the vector potential and $e$ is the charge). The usual Schrodinger equation for the hydrogen atom, which just has a potential energy term corresponding to the static Coulomb potential of the proton, is an approximation that works for a static system.

4. Jan 6, 2017

blue_leaf77

Static E field has zero frequency and thus the associated photon has zero energy. It has nothing to transfer to electron.

5. Jan 6, 2017

Staff Emeritus
Yes, but there is still a classical problem here. If I put an atom in an electric field, the nucleus gains energy and the electrons lose it (or the other way around) and these almost exactly cancel. (Depending on the atom and field configuration, this cancelation can be exact)

6. Jan 6, 2017

Staff: Mentor

Can you be more specific about what you're referring to? I'm familiar with the Stark effect, the splitting of atomic energy levels in an external electric field, but I'm not sure that can be described as the nucleus gaining energy and the electrons losing it, or the other way around.

7. Jan 7, 2017

ZapperZ

Staff Emeritus
I think people are missing the central question of the OP. I'm going to try and condense it, and the OP can correct me if I'm wrong:

1. QFT states that EM interaction can be modeled as being carried by virtual photons (there's an Insight article about the misunderstanding of this).

2. So why can't these virtual photons be absorbed and cause a transition in an atom?

Zz.

8. Jan 7, 2017

blue_leaf77

Because static E field has no photon energy, I think.

9. Jan 7, 2017

ZapperZ

Staff Emeritus
Static field is not the electromagnetic wave radiation. That isn't the issue here based on what I think the OP wrote.

Zz.

10. Jan 7, 2017

Staff Emeritus
To first order, the external electric field produces a potential V. The nucleus gains energy ZeV and the Z electrons each gain -eV. The net is zero. That's what I was talking about. (Non-uniformities in E, and thus V, are going to be very small: of order the size of an atom divided by the size of your experiment)

At second order, you have the Stark effect, which moves some orbitals up and others down. If the orbitals are all full, this exactly cancels. If not, the atom adjusts its ground state accordingly. This is usually a reduction in energy, so the atom gives up energy to the field, not the reverse.

11. Jan 7, 2017

Ah, got it.