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For example a simple complex scalar field theory has a global $ U(1) $ symmetry where the field $ \psi $ can be replaced by $ e^{ i \alpha } \psi $, where $ \alpha $ is just some real constant, without changing the value of the Lagrangian.

Turning this global symmetry into a local one, where $ \alpha $ depends on the location, requires the introduction of a new field. I understand that this modification somehow forces the creation of another field and provides a "mathematical reason" for the existence of, say, the electromagnetic interaction and coupling of different fields.

But besides the fact that this produces the law of physics that we expect, is there another justification for doing this? Is the consideration of local symmetries instead of global ones a consequence of some relativistic principle requiring laws of physics to be local?

Qmechanic
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    FWIW, from the Yang, Mills paper Conservation of Isotopic Spin and Isotopic Gauge Invariance: "As usually conceived, however, this arbitrariness is subject to the following limitation: once one chooses what to call a proton, what a neutron, at one space-time point, one is then not free to make any choices at other space-time points. It seems that this is not consistent with the localized field* concept that underlies the usual physical theories."* (emphasis mine) – Alfred Centauri Jun 09 '20 at 16:52
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    I recommend https://physics.stackexchange.com/questions/76477/why-gauge-theories-have-such-a-success. A contemporary view of gauge symmetry is that it is merely a technique to handle mathematical redundancy. A Lorentz invariant theory of massless spin-1 particles can only have two physical degrees of freedom (e.g. the two photon helicities). Gauge invariance allows description of this in a manifestly Lorentz invariant way using 4-vectors. – isometry Jun 09 '20 at 20:17
  • These are interesting references. I don't understand what is meant by calling a something a proton at some point in space-time while still having the freedom to "make choice" at the other points in space-time. Isn't it clear what a proton and what a neutron are with respect to each other? –  Jun 11 '20 at 13:43

1 Answers1

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You are right that gauging a global symmetry requires the addition of another field and gives rise to electromagnetic interactions (for instance) in your theory. But you can also look at it from the other direction: if you start with some basic theory and then attempt to include electromagnetic interactions, you will see that you must demand that the complex field is invariant under local $U(1)$ transformations, or else you will lose certain desirable properties of the theory (for example, gauge invariance).

To see this, consider the most elementary non-relativistic quantum theory, where the dynamics are governed by the Schrodinger equation $H\Psi=i\hbar\partial_t\Psi$ (which is invariant under global phase transformations). If you want to introduce EM interactions, you can define the Hamiltonian of a charge particle in an EM field by

$$H=\frac{(\vec{p}-\frac{e}{c}\vec{A})^2}{2m}+eV.$$

Then the Schrodinger equation becomes

$$\left(\frac{(\vec{p}-\frac{e}{c}\vec{A})^2}{2m}+eV\right)\Psi=i\hbar\partial_t\Psi$$

which can be rewritten

\begin{equation} -\frac{\hbar^2}{2m}\left({\nabla}-\frac{ie}{\hbar c}\vec{A} \right)^2\Psi=i\hbar\left(\partial_t+\frac{ie}{\hbar}V \right)\Psi. \tag{1} \end{equation}

However, we know that Maxwell's equations should be invariant under gauge transformations:

\begin{align} V\rightarrow V'&=V-\frac{1}{c}\partial_t \chi\\ \vec{A}\rightarrow \vec{A}'&=\vec{A}+{\nabla}\chi \end{align} where $\chi=\chi(t,\vec{r})$. But you can show that $(1)$ is not invariant under these transformations; extra terms will appear on the LHS and RHS that do not cancel. Gauge invariance is lost! However, if you supplement the gauge transformations by a spacetime-dependent phase change

$$\Psi(\vec{r},t)\rightarrow \Psi'(\vec{r},t)=e^{\frac{ie}{\hbar c}\chi(\vec{r},t)}\Psi(\vec{r},t)$$

then you will see that these extra terms will cancel, and gauge invariance is restored. The main point is that the demand for local $U(1)$ symmetry can arise in a natural way when you are trying to include EM interactions in your theory. Gauging the global symmetry of the theory is not just a mathematical trick; it reflects the fact that electromagnetism is fundamentally a gauge interaction, and is required to maintain gauge invariance.

Although this argument was made for a non-relativistic theory, it can be extended for the Dirac or Klein-Gordon equations in a straightforward way (although a Lagrangian approach is probably simplest). The local $U(1)$ symmetry for electromagnetism can also be generalized for non-Abelian gauge theories.

Superbee
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