To make the kinetic term in the Lagrangian for quantum field theories (for example qed) inveriant under local phase transformations we introduce the covariant derivative $D_{\mu} = \partial _{\mu} + iA_{\mu}$ with the gauge field $A_{\mu}$. But why is this field the electromagnetic field? Couldn't it be any field instead? I can't see in the derivation of the covariant derivative why the field $A_{\mu}$ is choosen.

  • 6
    $\begingroup$ I'm not sure what exactly your question is. If you write down a $\mathrm{U}(1)$ gauge theory, it behaves exactly like electromagnetism. So the $\mathrm{U}(1)$ gauge field is indistinguishable from the electromagnetic four-potential. Can you be more specific about what you want to know? $\endgroup$
    – ACuriousMind
    Jun 1, 2016 at 22:00
  • 3
    $\begingroup$ If it walks like a duck, quacks like a duck, then... $\endgroup$
    – Diracology
    Jun 1, 2016 at 22:17

2 Answers 2


The wave equation needs to stay invariant under local changes of phase. The gauge field $A_{\mu}$ that is introduced to enforce local gauge invariance is NOT an arbitrary function, it needs to represent something and it represents the possibility that the particle either emits or absorbs a photon, a quantum of the EM field.

The probability that it does so, at any particular spacetime point, is proportional to the coupling strength q, which is the magnitude of the electric charge of the particle.

After incorporating the function $A_{\mu}$, into the relativistic wave function, the description of the minimal interaction vertex that then occurs, is EXACTLY the same as the description given by QED.

Mininal Interaction Vertex

So A can be thought of as representing the photon. Therefore, the inclusion of QED into relativistic theory is required by the demand for invariance under local changes of phase, i.e. a U(1) transformation.

The above is based around my notes from Sean Carroll's "The Particle At The End Of The Universe." and "Deep Down Things" by B.A. Schumm


$A_\mu$ is introduced simply as a tool to assert gauge invariance of the fermion's (in this case) kinetic term. Once this field is added to our lagrangian, we recognize that we must add a kinetic term for the gauge field itself. This is the point where we will make contact with the vector potential of electromagnetism.

We introduce a field strength tensor, which can be constructed from the commutator of covariant derivatives $$[D_\mu,D_\nu]\equiv ifF_{\mu\nu}$$ where $f$ is a coupling constant (that you are missing in your definition of the covariant derivative), such that $$F_{\mu\nu}=\partial_\mu A_\nu-\partial_\nu A_\mu$$. This should look familiar. From our kinetic term $$-\frac{1}{4}F_{\mu\nu}F^{\mu\nu}$$ we get Maxwell's equations. This is no accident. We recognize gauge invariance in electromagnetism and so we assert gauge invariance in our field theory, and out pops our friendly vector potential with just a couple judicious choices of coefficients.

Sources: Donoghue, Golowich, and Holstein "Dynamics of the Standard Model" Peskin and Schroeder "An Introduction to Quantum Field Theory"


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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