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Fernando Garcia

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QED from gauge invariance: An introduction to gauge theory

December 26th, 2023.

Introduction

This little entry isn't about motivating (local) gauge invariance, but rather about its implications. Starting from the free spin 1/2 Lagrangian, we will go all the way to QED by making the simple assumption that our Lagrangian (density) has to be $U(1)$ invariant.

In addition to demonstrating the impact of demanding gauge invariance on a theory, this entry also aims to fill in the gaps in standard derivations by presenting all the required manipulations.

From a free theory to QED

Let's start by recalling the free spin 1/2 Lagrangian.

$${\mathcal{L}}_{1/2}=\bar{\psi }(i{\gamma }^{\mu }{\partial }_{\mu }-m)\psi$$

Where

$$\bar{\psi }\equiv {\psi }^{†}{\gamma }^0$$

Is known as the Dirac adjoint of the field $\psi$.

We know that the spin 1/2 Lagrangian has Lorentz invariance (as it is constructed to be). Furthermore, it is easy to check that it has a global $U(1)$ invariance:

$$\begin{array}{rl}\psi (x)& \to e^{i\alpha }\psi (x)\\ \bar{\psi }(x)& \to \bar{\psi }(x)e^{-i\alpha }\end{array}$$

Where $\alpha$ is any constant:

$$\begin{array}{rl}{\mathcal{L}}_{1/2}^{\prime }& ={\bar{\psi }}^{\prime }(i{\gamma }^{\mu }{\partial }_{\mu }-m){\psi }^{\prime }\\ & =\bar{\psi }{e}^{-i\alpha }(i{\gamma }^{\mu }{\partial }_{\mu }-m)e^{i\alpha }\psi \\ & =\bar{\psi }{e}^{-i\alpha }\bar{\psi }{e}^{i\alpha }\bar{\psi }(i{\gamma }^{\mu }{\partial }_{\mu }-m)\psi \\ & =\bar{\psi }{e}^{-i\alpha }\\ & ={\mathcal{L}}_{1/2}\end{array}$$

Let's now invoke the Gauge Principle (which, once again, I won't motivate in this entry). Through this principle, we promote the global $U(1)$ symmetry to a local $U(1)$ symmetry. That is:

$$\begin{array}{rl}\psi (x)& \to e^{i\alpha (x)}\psi (x)\\ \bar{\psi }(x)& \to \bar{\psi }(x)e^{-i\alpha (x)}\end{array}$$

Where $\alpha (x)$ is now a function of spacetime. We now check if the spin 1/2 is locally $U(1)$ invariant:

$$\begin{array}{rl}{\mathcal{L}}_{1/2}^{\prime }& ={\bar{\psi }}^{\prime }(i{\gamma }^{\mu }{\partial }_{\mu }-m){\psi }^{\prime }\\ & =\bar{\psi }{e}^{i\alpha (x)}(i{\gamma }^{\mu }{\partial }_{\mu }-m)e^{i\alpha (x)}\psi \\ & =\bar{\psi }{e}^{i\alpha (x)}(i{\gamma }^{\mu }{e}^{i\alpha (x)}{\partial }_{\mu }\psi +i{\gamma }^{\mu }i[{\partial }_{\mu }\alpha (x)]e^{i\alpha (x)}\psi -e^{i\alpha (x)}m\psi )\\ & =\bar{\psi }{e}^{-i\alpha (x)}(e^{i\alpha (x)}[i{\gamma }^{\mu }{\partial }_{\mu }-m]\psi -{\gamma }^{\mu }[{\partial }_{\mu }\alpha (x)]e^{i\alpha (x)}\psi )\\ & ={\mathcal{L}}_{1/2}-[{\partial }_{\mu }\alpha (x)]\bar{\psi }{\gamma }^{\mu }\psi \end{array}$$

That is, the spacetime dependence of $\alpha (x)$ won't allow us to commute the exponential with the derivative operator, and by applying the operator on the transformed field, we obtain an extra term $-[{\partial }_{\mu }\alpha (x)]\bar{\psi }{\gamma }^{\mu }\psi$.

We have good reasons to demand local gauge invariance (a topic for a future entry!), so we need a way to get rid of that term. This task turns out to be quite simple, but it might seem artificial at a first glance. What we are about to do can be justified, and this shall be done in a future entry.

To get rid of the extra term, we will include a new field in our Lagrangian. This field will be a spin 1 field, and it will be introduce through a new derivative:

$${\partial }_{\mu }\to D_{\mu }$$

Where

$$D_{\mu }\equiv {\partial }_{\mu }+ie{A}_{\mu }$$

Where $e$ is a constant. Notice that we won't explicitly add new terms (yet), but rather change our definition of the derivative to include this field. With this new derivative definition, known as a covariant derivative, we find the following result when acting on a field:

$$D_{\mu }\psi (x)={\partial }_{\mu }\psi (x)+ie{A}_{\mu }\psi (x)$$

As the name implies, the covariant derivative has a defining property: It transforms like the field $\psi (x)$. That is:

$${(D_{\mu }\psi (x))}^{\prime }=D_{\mu }^{\prime }{\psi }^{\prime }(x)=e^{i\alpha (x)}{D}_{\mu }\psi (x)$$

It is this requirement that allows us to find how the gauge field $A_{\mu }$ should transform. Explicitly, the requirement implies:

$$({\partial }_{\mu }+ie{A}_{\mu }^{\prime }(x))e^{i\alpha (x)}\psi (x)=e^{i\alpha (x)}({\partial }_{\mu }+ie{A}_{\mu }(x))\psi (x)$$

So

$$\cancel{e^{i\alpha (x)}{\partial }_{\mu }\psi (x)}+\cancel{\psi (x)}\cancel{i}\cancel{e^{i\alpha (x)}}{\partial }_{\mu }\alpha (x)+\cancel{i}e{A}_{\mu }^{\prime }\cancel{e^{i\alpha (x)}}\cancel{\psi (x)}=\cancel{e^{i\alpha (x)}{\partial }_{\mu }\psi (x)}+\cancel{i}e{A}_{\mu }\cancel{\psi (x)}\cancel{e^{i\alpha (x)}}$$

And thus

$$\begin{array}{rl}e{A}_{\mu }^{\prime }& =e{A}_{\mu }-{\partial }_{\mu }\alpha (x)\\ A_{\mu }^{\prime }& =A_{\mu }-\frac{1}{e}{\partial }_{\mu }\alpha (x)\end{array}$$

It is then straightforward to see that the (modified) spin 1/2 Lagrangian is invariant under local $U(1)$ transformations.

$$\begin{array}{rl}\bar{\psi }(i{\gamma }^{\mu }{D}_{\mu }-m)\psi & \to \overline{{\psi }^{\prime }}(i{\gamma }^{\mu }{D}_{\mu }^{\prime }-m){\psi }^{\prime }\\ & =\bar{\psi }{e}^{-i\alpha (x)}(e^{i\alpha (x)}i{\gamma }^{\mu }{D}_{\mu }\psi -e^{i\alpha (x)}m\psi )\psi \\ & =\bar{\psi }(i{\gamma }^{\mu }{D}_{\mu }-m)\psi \end{array}$$

It is then straightforward to see that the (modified) spin 1/2 Lagrangian is invariant under local $U(1)$ transformations.

$$\begin{array}{rl}\bar{\psi }(i{\gamma }^{\mu }{D}_{\mu }-m)\psi & \to \overline{{\psi }^{\prime }}(i{\gamma }^{\mu }{D}_{\mu }^{\prime }-m){\psi }^{\prime }\\ & =\bar{\psi }{e}^{-i\alpha (x)}(e^{i\alpha (x)}i{\gamma }^{\mu }{D}_{\mu }\psi -e^{i\alpha (x)}m\psi )\psi \\ & =\bar{\psi }(i{\gamma }^{\mu }{D}_{\mu }-m)\psi \end{array}$$

But we are not done yet! We have to include a free term for the Gauge field $A_{\mu }$, as it doesn't have any (it does have interaction terms).

The simplest candidate for this free term is a mass-term:

$$A^{\nu }{A}_{\nu }$$

But it won't transform nicely. Recall that the gauge field has a transformation defined by $A_{\mu }^{\prime }=A_{\mu }-\frac{1}{e}{\partial }_{\mu }\alpha (x)$, so we will get extra terms when we transform the above contraction:

$$(A^{\nu }{A}_{\nu }{)}^{\prime }\to A^{\nu }{A}_{\nu }+\text{Extra terms}$$

Taking inspiration from the Proca Lagrangian (Lagrangian for free spin 1 particles), we now consider a term given by the contraction of $F^{\mu \nu }$ defined by:

$$F^{\mu \nu }={\partial }^{\mu }{A}^{\nu }-{\partial }^{\nu }{A}^{\mu }$$

We can easily show that $F_{\mu \nu }$ (and thus $F^{\mu \nu }$) are gauge invariant:

$$\begin{array}{rl}{F}_{\mu \nu }^{\prime }& ={\partial }_{\mu }{A}_{\nu }^{\prime }-{\partial }_{\nu }{A}_{\mu }^{\prime }\\ & ={\partial }_{\mu }(A_{\nu }-\frac{1}{e}{\partial }_{\nu }\alpha (x))-{\partial }_{\nu }(A_{\mu }-\frac{1}{e}{\partial }_{\mu }\alpha (x))\\ & ={\partial }_{\mu }{A}_{\nu }-{\partial }_{\nu }{A}_{\mu }-\cancel{\frac{1}{e}{\partial }_{\mu \nu }\alpha (x)}+\cancel{\frac{1}{e}{\partial }_{\mu \nu }\alpha (x)}\\ & =F_{\mu \nu }\end{array}$$

The contraction $F_{\mu \nu }{F}^{\mu \nu }$ is used to 1. Get a scalar, and 2. Get a scalar that is Lorentz Invariant.

We are now ready to write:

$${\mathcal{L}}_{\text{QED}}=\bar{\psi }(i{\partial }^{\mu }{D}_{\mu }-m)\psi -\frac{1}{4}{F}^{\mu \nu }{F}_{\mu \nu }$$

And as the label implies, this Lagrangian corresponds to the theory of Quantum Electrodynamics.

Final remarks

The fact that we found the QED Lagrangian from imposing a new symmetry on the theory seems like a coincidental trick, and nothing more, but it is in fact this way that we get the Lagrangian for the Standard Model.

To do so, we extend the symmetry to include $SU(2)$ and $SU(3)$. By doing this, we get a Lagrangian that describes the electroweak and strong interactions. This will be the topic of a future entry.

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