According to the book Quantum Field Theory for the Gifted Amateur, on page 128 they say

A theory which had a field $A^\mu(x)$ introduced to produce an invariance with respect to local transformations is known as a gauge theory. The field $A^\mu(x)$ is known as a gauge field.

I have tried to learn more about gauge theory, but I am struggling to understand the context behind where the idea is coming from.

My Question:

What kind of systems have these invariants? Is this something that is limited to particle physics or do we see this in macroscopic systems? If so, can you provide an example?

  • 1
    $\begingroup$ I'm not entirely sure what you're asking--if you're just looking for examples then electromagnetism fits the bill. $\endgroup$
    – user21299
    Commented Jan 29, 2015 at 22:59
  • $\begingroup$ Okay so I have heard about this thing called the $U(1)$ symmetry. Is that a gauge symmetry? What makes it "gauge"? $\endgroup$ Commented Jan 29, 2015 at 23:04
  • $\begingroup$ U(1) gauge symmetry is the gauge symmetry of electromagnetism. I suggest reading up on the formulation of electromagnetism based on the 4-potential before tackling quantum field theory $\endgroup$
    – user21299
    Commented Jan 29, 2015 at 23:08
  • $\begingroup$ Okay, I will start with that then. Does that mean classical electrodynamic with the Faraday tensor? $\endgroup$ Commented Jan 29, 2015 at 23:10
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    $\begingroup$ Related: What is the basis of gauge theory?, How do we know what type of gauge field to add to a theory?. (The cat in RodVance's answer to the first is a funny, but instructive example) $\endgroup$
    – ACuriousMind
    Commented Jan 29, 2015 at 23:21

1 Answer 1


The general idea is that you are able, starting from a Lagrangian $\mathscr L$ invariant under some global (i.e. not space-time dependent) transformation, to "derive" the interaction of the field described by that theory solely by requiring that the Lagrangian is still invariant when the transformation is allowed to be local (meaning that the parameter defining the transformation is space-time dependent).

A simple example is QED. Consider the Lagrangian density for a massive Dirac field $\psi$, which reads: $$ \tag 1 \mathscr{L} = \bar \psi (i \gamma^\mu \partial_\mu - m) \psi.$$ Note that this Lagrangian is invariant under the global transformation $$ \tag{2} \psi(x) \rightarrow e^{i\alpha} \psi(x),$$ $$ \tag{2'} \bar \psi(x) \rightarrow e^{-i\alpha} \bar \psi(x),$$ where $\alpha_0 \in \mathbb{R}$ is a number (as in not a function).

But if you now try to generalize the transformations (2) and (2'), allowing $\alpha$ to depend on the space-time point, you easily notice the Lagrangian (1) is no longer invariant.

It turns out that if you add to the Lagrangian density an additional term, writing it as $$ \tag 3 \mathscr{L} = \bar \psi (i \gamma^\mu \partial_\mu - m) \psi - ieA_\mu \bar \psi \gamma^\mu \psi,$$ where $A^\mu(x)$ is a field with certain transformation properties under the phase transformation (2), then you obtain that (3) is invariant under the more general gauge transformations $$ \tag{4} \psi(x) \rightarrow e^{i\alpha(x)}\psi(x),$$ $$ \tag{4'} \bar \psi(x) \rightarrow e^{-i\alpha(x)} \bar \psi(x),$$ $$ \tag{5} A_\mu(x) \rightarrow A_\mu(x) + \frac{1}{e} \partial_\mu \alpha(x).$$

$A^\mu$ is nothing else that the photon field, and you thus see that the requirement of gauge invariance under the local phase transformations (also called U(1) transformations) written above reproduces (as in: is equivalent to) the electromagnetic interaction between charged fermions/antifermions, like electrons and positrons.

In a similar fashion, the requirement of gauge symmetry is used to "derive" all of the fundamental forces:

  • In QCD requiring invariance under the $SU(3)$ gauge group introduces the gluons as carriers of the strong force;
  • Weak interactions are introduced requiring invariance under the $SU(2)_W$ gauge group, and this produced the coupling of left-handed leptons with the $W^\pm$ and $Z^0$ gauge fields;

And these are just a couple of examples taken from a really broad subject.

Other Phys.SE questions related to the subject are:

  • $\begingroup$ So let me see if I understand: once you add the additional term to the Lagrangian density involving $A^\mu (x)$, then this equation becomes invariant under certain transformations of $A^\mu (x)$, namely in this case $$A_\mu (x) \rightarrow A_\mu (x) + \frac{1}{e}\partial_\mu \alpha(x) $$. We call these gauge transformations. And these are valuable because if we want to make $\alpha$ a parameter that is spacetime-dependent, then we need a way to ensure the Lagrangian density remains invariant. But invariant with respect to what? The symmetry? $\endgroup$ Commented Jan 29, 2015 at 23:35
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    $\begingroup$ The gauge transformation is the local transformation, acting on all the fields of the theory (eventually in the trivial way as an identity). The U(1) gauge transformation in QED acts on the Dirac field $\psi$ as in (2), and on the gauge field $A^\mu$ as in (5). The invariance is of the Lagrangian (not of the fields), and is with respect to the gauge transformation. This means that if you plug the transformation rules for $\psi, \bar \psi$ and $A^\mu$ in the Lagrangian (3), you get that $\delta \mathscr{L} = 0$. $\endgroup$
    – glS
    Commented Jan 29, 2015 at 23:40
  • $\begingroup$ I get it. That's amazing. Thank you for clarifying. So cool! $\endgroup$ Commented Jan 29, 2015 at 23:44
  • $\begingroup$ What about gravity? Are there any known gauge theories or transformations applied to that? $\endgroup$ Commented Jan 30, 2015 at 3:43
  • $\begingroup$ @StanShunpike did you see the linked post? $\endgroup$
    – glS
    Commented Jan 30, 2015 at 10:07

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