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.
[Physics] Gauge field and covariant derivative
gauge-invariancegauge-theorylagrangian-formalism
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I am not sure of what you mean by "make a difference".
The covariant derivative is introduced in the Lagrangian density to add Local Gauge Symmetry to the Action. A given field theory is described by its Action. Hence if the Action does not have a particular symmetry, you cannot introduce the symmetry later.
By "making a difference", if you mean a difference to the physical consequences, then it will make a difference if you do not introduce the covariant derivative in the Lagrangian density itself. Taking your example of the complex scalar field theory, the complete Lagrangian density is,
$(1/2)(D_{\mu}\phi)^{*}D^{\mu}\phi-(1/4)F_{\mu\nu}F^{\mu\nu}-V(\phi,\phi^{*})$
The first term gives the kinetic term for your scalar field and an interaction term with $\phi, \phi^{*}, A^{\mu}$ coupled. The second term is the kinetic term for $A^{\mu}$, while the term is an interaction term with $\phi$ and $\phi^{*}$.
If you do not have the covariant derivative in your Lagrangian density, you do not have the interaction between the complex scalar field and the gauge field, and hence it is a lot of difference to the physics. In QED that would mean that Compton scattering and a lot that happens in nature will not happen.
A gauge transformation is defined to be the simultaneous transformation \begin{align} A_\mu & \mapsto A_\mu + e^{-1} \partial_\mu A \\ \phi & \mapsto \mathrm{e}^{\mathrm{i}\Lambda}. \end{align} It is not that the transformation of the gauge field "necessarily" leads to the transformation of the charged field, but only under this joint transformation everything is properly covariant. More abstractly, we conceive of both transformations to be different representations/realizations of the same abstract gauge algebra, consisting of functions $\Lambda : \mathbb{R}^4\to\mathbb{R}$. $\Lambda$ acts in one way on gauge fields and in another on charged fields, it's perfectly analogous to how the rotation group acts differently on scalars, vectors and tensors, yet "a rotation" is transforming all scalars, vectors and tensors and not just the vectors.
In a way, the covariance of $D_\mu\phi$ is the whole reason we invent the gauge field because $\partial_\mu \phi$ would not be covariant under the transformation of just $\phi$. It's less that the transformation of $\phi$ derives from that of $A$ and most that that of $A$ derives from that of $\phi$, at least if we take the viewpoint that the local symmetry is the fundamental object here. If we want to build a theory with such a local symmetry dependening on a spacetime function $\Lambda(x)$, we need to modify the derivative in order to build an invariant Lagrangian. So the first thing to try (perhaps motivated by the addition of the Christoffel symbols to the ordinary derivative in GR, which are effectively also only a type of gauge field) is to modify $\partial_\mu$ as $D_\mu := \partial_\mu + A_\mu$ for some $A_\mu$.
Now you examine $D_\mu (\mathrm{e}^{\mathrm{i}\Lambda}\phi)$, which we want to be equal to $\mathrm{e}^{\mathrm{i}\Lambda}D_\mu\phi$, and just look at the extra terms. After staring long enough at them, you will realize that those are exactly the terms that get cancelled if we let $A_\mu$ transform as we usually do, that is, the transformation of $A_\mu$ is engineered such that the covariant derivative is truly covariant.
Best Answer
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.
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