It is well known that one reason quantum mechanics started to being developed, was because scientist wanted a model to explain electron orbits in atoms.

Borh interpreted that the for orbits to exist they would need to be quantized. Using the Schrödinger equation, this quantization arise from the bound state of the electron with the nucleus (because we want the wave function to tend to $0$, as the distance from the origin goes to infinity).

When we solve the Schrödinger equation we plug the following potential: $$ V(r) = \frac{Ze^2}{4 \pi \varepsilon_0 r} $$ Which corresponds to a Culomb Gauge solution to Maxwell's equations: $$ \nabla \cdot \mathbf A = 0 $$

If we impose a Lorentz Gauge to Maxwell's Equations: $$ \nabla \cdot \mathbf A + \frac{1}{c^2}\frac{\partial \phi}{\partial t} = 0 $$ We will get another potential which takes into account the velocity of the particle. In the framework of Quantum Mechanics, using such potential would not make sense because of the uncertainty principle. No? Can classical quantum mechanics take into account this dynamism of the electric and magnetic fields (provoked by the movement of the electron)?

I know Classical Quantum Mechanics is a very simple frame work that does not take into account the quantization of the fields. The Schrödinger equation just works for describing the wave function of the electron, but no to describe the electromagnetic fields of this electron states.

If really Classical Quantum Mechanics does not take this into account, does QED not only describe the wave-functions of this electrons, but also describes the electric and magnetic fields generated by the particles itself? As in this previous question I asked, sources for the equations feel disconnected from the laws itself, which is again happening in this situation. For example in the Dirac Equation: $$ (i \gamma^\mu (\partial_\mu - i e A_\mu) - m) \psi = 0 $$ $$ \mathbf A = \left ( \frac{\phi}{c}, \mathbf 0 \right ) $$ This equation predicts how a particle would behave in the four-potential $\mathbf A$, but won't say anything about how the vector itself interacts and changes because of the particle, no? This is what I mean with "sources for the equations feel disconnected from the laws itself". Has QED anything to say about this? If this radiation really exist, does QED predicts it?

  • $\begingroup$ Let's us go back to classical electrodynamics (leave quantum aside), would you be bothered if I write the motion of a particle independently of the field that it produces? The only case I can imagine that it is not true is Abraham-Lorentz radiation reaction. Is this about the quantum version of it? $\endgroup$
    – Mauricio
    Commented Mar 3 at 22:59

1 Answer 1


In quantum field theories the interaction between particle and field are instituted as interaction terms in the Lagrangian densities that characterize the specific theory. For the ordinary electromagnetic field, we can express it as a classical field theory with the Lagrangian density: $$\mathcal L_{Maxwell}=-{1\over 4}F_{\mu\nu}F^{\mu\nu}.$$ From the Euler-Lagrange equations one can obtain the Maxwell equations. Of course, this is strictly a classical theory. Some important examples of quantum field theories include for example the Dirac field, which is the field appropriate for describing fermions. So, in QED the particles (fermions) interact with the electromagnetic field, this is captured with the QED Lagrangian: $$\mathcal L_{QED}=\mathcal L_{Dirac}+\mathcal L_{Maxwell}+\mathcal L_{int},$$ where the last term $\mathcal L_{int}$ is the term describing the interaction of the fermion and the electromagnetic field, i.e. the Dirac field and the Maxwell field are coupled through this interaction term, an explicit form for this interaction, for an electron is given by: $$\mathcal L_{int}=-e\bar\psi\gamma^{\mu}\psi A_\mu .$$Note, however, that the Maxwell field is found in its ordinary form in the QED Lagrangian. The Maxwell field is unaltered, however, it is coupled in a local interaction to the Dirac field.

So, the QED Lagrangian is given by: $$\mathcal L_{QED}=\bar\psi(i\gamma^\mu D_\mu-m)\psi-{1\over 4}F_{\mu\nu}F^{\mu\nu}-e\bar\psi\gamma^\mu\psi A_\mu,$$ where $D_\mu$ is the "gauge covariant derivative": $$D_\mu=\partial_\mu+ieA_\mu(x).$$ Plugging this Lagrangian into the Euler-Lagrnge equations for you get the following two field equaitions: $$(i\gamma^\mu D_\mu-m)\psi(x)=0, and\quad \partial_\mu F^{\mu\nu}=e\bar\psi\gamma^\nu\psi.$$ So the second of these equations is the Maxwell equations for the local interaction and the first is the Dirac equation, with the odd derivative operator. You can understand that the operator $D$ is a sort of generalization of the ordinary momentum operator of quantum mechanics for the case when there are electromagnetic fields present, at least that is the way that Dirac motivated it in his development of the relativistic treatment of the electron. Perhaps, more precisely, it provides the QED Lagrangian and resulting field equations with gauge symmetry. When the fermionic field is coupled to the electromagnetic field in this way, i.e. with the gauge symmetry: $$\psi(x)\rightarrow e^{i\alpha(x)}\psi(x),\quad A_{\mu}\rightarrow {1\over e}\partial_\mu\alpha(x),$$ it is said to be minimally coupled.

  • $\begingroup$ Very clarifying answer. I would ask what kind of equations follow from this QED Lagrangian as Euler-Lagrange equations? $\endgroup$ Commented Mar 4 at 12:44
  • $\begingroup$ @AttilaJanosKovacs I will edit and put this as an addendum to my post when I get time later this morning. $\endgroup$ Commented Mar 4 at 13:23

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