## What is an intuitive explanation of the gauge covariant derivative?

The gauge covariant derivative is easiest to understand within electrodynamics, which is a U(1) gauge theory.

When we apply a U(1) gauge transformation to a charged field, we change its phase, by an amount proportional to \(e\theta(x^\mu)\), which may vary from point to point in space-time. At the same time, the electromagnetic four-potential changes by an amount proportional to \(\partial_\mu \theta\).

The gauge transformation can be seen to change the derivative of the field by an amount proportional to \(ie \partial_\mu \theta\). (We get this from applying the chain rule to the complex exponential, nothing tricky here.) You'll notice that this is \(ie\) times the change in the electromagnetic four-potential.

A gauge transformation is not supposed to change any observable quantities, so observable quantities shouldn't directly depend on the derivative of the field. Instead we postulate that we should replace the partial derivatives with the gauge covariant derivative, \[D_\mu := \partial_\mu - ieA_\mu\] Now whenever we perform a gauge transformation, the change in the partial derivative will be cancelled by the change in the gauge field \(A_\mu\), and the gauge covariant derivative will be unchanged.

So the intuitive explanation of the gauge covariant derivative is that it's a way to modify the derivative operator so that the result doesn't change when you perform a gauge transformation. This is essential since a gauge transformation is supposed to leave the system in the same state as before, just with a different configuration of unobservable degrees of freedom.

Here's another explanation, rather different: the Lagrangian for a field contains the field's derivative, for example, the Klein–Gordon Lagrangian for a free particle is \[\mathcal{L} = \frac{1}{2}\partial_\mu \phi^* \partial^\mu \phi - \frac{1}{2}m^2 \phi^* \phi \] The derivative term is the "kinetic" term, and it corresponds to the propagator; it describes how the particle moves in free space. The derivative operator is the generator of translation; a finite translation is built up by successively compounding infinitesimal translations. When we replace the partial derivative by the gauge covariant derivative, the effect is to introduce an infinitesimal phase change proportional to \(A_\mu\) that accompanies each infinitesimal translation. As a result, a finite translation is accompanied by a phase shift proportional to \(\int A_\mu dx^\mu\). This fact will be familiar to those who have studied, e.g., the Aharonov–Bohm effect. So the gauge covariant derivative can be understood as what we need to put into the Lagrangian in order for this to happen. (This puts the cart before the horse, I suppose, but in the end, as far as building "intuition" goes, it helps to have as many links as possible in your mind.) But—if you understand this, you probably already understand the gauge covariant derivative, so maybe this isn't so useful.

When we apply a U(1) gauge transformation to a charged field, we change its phase, by an amount proportional to \(e\theta(x^\mu)\), which may vary from point to point in space-time. At the same time, the electromagnetic four-potential changes by an amount proportional to \(\partial_\mu \theta\).

The gauge transformation can be seen to change the derivative of the field by an amount proportional to \(ie \partial_\mu \theta\). (We get this from applying the chain rule to the complex exponential, nothing tricky here.) You'll notice that this is \(ie\) times the change in the electromagnetic four-potential.

A gauge transformation is not supposed to change any observable quantities, so observable quantities shouldn't directly depend on the derivative of the field. Instead we postulate that we should replace the partial derivatives with the gauge covariant derivative, \[D_\mu := \partial_\mu - ieA_\mu\] Now whenever we perform a gauge transformation, the change in the partial derivative will be cancelled by the change in the gauge field \(A_\mu\), and the gauge covariant derivative will be unchanged.

So the intuitive explanation of the gauge covariant derivative is that it's a way to modify the derivative operator so that the result doesn't change when you perform a gauge transformation. This is essential since a gauge transformation is supposed to leave the system in the same state as before, just with a different configuration of unobservable degrees of freedom.

Here's another explanation, rather different: the Lagrangian for a field contains the field's derivative, for example, the Klein–Gordon Lagrangian for a free particle is \[\mathcal{L} = \frac{1}{2}\partial_\mu \phi^* \partial^\mu \phi - \frac{1}{2}m^2 \phi^* \phi \] The derivative term is the "kinetic" term, and it corresponds to the propagator; it describes how the particle moves in free space. The derivative operator is the generator of translation; a finite translation is built up by successively compounding infinitesimal translations. When we replace the partial derivative by the gauge covariant derivative, the effect is to introduce an infinitesimal phase change proportional to \(A_\mu\) that accompanies each infinitesimal translation. As a result, a finite translation is accompanied by a phase shift proportional to \(\int A_\mu dx^\mu\). This fact will be familiar to those who have studied, e.g., the Aharonov–Bohm effect. So the gauge covariant derivative can be understood as what we need to put into the Lagrangian in order for this to happen. (This puts the cart before the horse, I suppose, but in the end, as far as building "intuition" goes, it helps to have as many links as possible in your mind.) But—if you understand this, you probably already understand the gauge covariant derivative, so maybe this isn't so useful.