December 26th, 2023.
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.
Let's start by recalling the free spin 1/2 Lagrangian.
\begin{equation} \mathscr{L}_{1/2}=\overline{\psi }(i\gamma ^{\mu }\partial _{\mu }-m)\psi \end{equation}Where
\begin{equation} \overline{\psi }\equiv \psi ^{\dagger}\gamma ^{0} \end{equation}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{equation} \begin{split} \psi (x) &\rightarrow e^{i\alpha }\psi (x) \\ \overline{\psi }(x) &\rightarrow \overline{\psi }(x)e^{-i\alpha } \\ \end{split} \end{equation}Where $\alpha $ is any constant:
\begin{equation} \begin{split} \mathscr{L}'_{1/2} &= \overline{\psi }'(i\gamma ^{\mu }\partial _{\mu }-m)\psi ' \\ &= \overline{\psi }e^{-i\alpha }(i\gamma ^{\mu }\partial _{\mu }-m)e^{i\alpha }\psi \\ &= \overline{\psi }e^{-i\alpha }\overline{\psi }e^{i\alpha }\overline{\psi }(i\gamma ^{\mu }\partial _{\mu }-m)\psi \\ &= \overline{\psi }e^{-i\alpha } \\ &= \mathscr{L}_{1/2} \end{split} \end{equation}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{equation} \begin{split} \psi (x) &\rightarrow e^{i\alpha(x) }\psi (x) \\ \overline{\psi }(x) &\rightarrow \overline{\psi }(x)e^{-i\alpha (x) } \\ \end{split} \end{equation}Where $\alpha (x) $ is now a function of spacetime. We now check if the spin 1/2 is locally $U(1) $ invariant:
\begin{equation} \begin{split} \mathscr{L}' _{1/2}&= \overline{\psi }'(i\gamma ^{\mu }\partial _{\mu }-m)\psi ' \\ &= \overline{\psi }e^{i\alpha (x)} \left( i\gamma ^{\mu }\partial _{\mu }-m \right) e^{i\alpha (x)}\psi \\ &=\overline{\psi }e^{i\alpha (x)} \left( i\gamma ^{\mu }e^{i\alpha (x)}\partial _{\mu }\psi +i\gamma ^{\mu }i \left[ \partial _{\mu }\alpha (x) \right] e^{i\alpha (x)}\psi -e^{i\alpha (x)}m\psi \right) \\ &= \overline{\psi }e^{-i\alpha (x)} \left( e^{i\alpha (x)} \left[ i\gamma ^{\mu }\partial _{\mu }-m \right] \psi -\gamma ^{\mu } \left[ \partial _{\mu }\alpha (x) \right] e^{i\alpha (x)}\psi \right) \\ &= \mathscr{L} _{1/2}- \left[ \partial _{\mu }\alpha (x) \right] \overline{\psi }\gamma ^{\mu }\psi \end{split} \end{equation}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 $- \left[ \partial _{\mu }\alpha (x) \right] \overline{\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:
\begin{equation} \partial _{\mu }\rightarrow D_{\mu } \end{equation}Where
\begin{equation} D_{\mu }\equiv \partial _{\mu }+ieA_{\mu } \end{equation}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:
\begin{equation} D_{\mu }\psi (x) = \partial _{\mu }\psi (x) + ieA_{\mu }\psi (x) \end{equation}As the name implies, the covariant derivative has a defining property: It transforms like the field $\psi (x) $. That is:
\begin{equation} \left( D_{\mu }\psi (x) \right) '=D'_{\mu }\psi '(x)=e^{i\alpha (x)}D_{\mu }\psi (x) \end{equation}It is this requirement that allows us to find how the gauge field $A_{\mu } $ should transform. Explicitly, the requirement implies:
\begin{equation} \left( \partial _{\mu }+ieA'_{\mu }(x) \right) e^{i\alpha (x)}\psi (x)=e^{i\alpha (x)} \left( \partial _{\mu }+ieA_{\mu }(x) \right) \psi (x) \end{equation}So
\begin{equation} \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}eA'_{\mu }\cancel{e^{i\alpha (x)}}\cancel{\psi (x)} = \cancel{e^{i\alpha (x)}\partial _{\mu } \psi (x)} + \cancel{i}eA_{\mu }\cancel{\psi (x)}\cancel{e^{i\alpha (x)}} \end{equation}And thus
\begin{equation} \begin{split} eA'_{\mu } &= eA_{\mu }-\partial _{\mu }\alpha (x) \\ A'_{\mu } &= A_{\mu }-\frac{1}{e}\partial _{\mu }\alpha (x) \end{split} \end{equation}It is then straightforward to see that the (modified) spin 1/2 Lagrangian is invariant under local $U(1) $ transformations.
\begin{equation} \begin{split} \overline{\psi } \left( i\gamma ^{\mu }D_{\mu }-m \right) \psi &\rightarrow \overline{\psi '} \left( i\gamma ^{\mu }D'_{\mu }-m \right) \psi' \\ &= \overline{\psi } e^{-i\alpha (x)}\left( e^{i\alpha (x)}i\gamma ^{\mu }D_{\mu }\psi- e^{i\alpha (x)}m \psi\right) \psi \\ &= \overline{\psi } \left( i\gamma ^{\mu }D_{\mu }-m \right) \psi \end{split} \end{equation}It is then straightforward to see that the (modified) spin 1/2 Lagrangian is invariant under local $U(1) $ transformations.
\begin{equation} \begin{split} \overline{\psi } \left( i\gamma ^{\mu }D_{\mu }-m \right) \psi &\rightarrow \overline{\psi '} \left( i\gamma ^{\mu }D'_{\mu }-m \right) \psi' \\ &= \overline{\psi } e^{-i\alpha (x)}\left( e^{i\alpha (x)}i\gamma ^{\mu }D_{\mu }\psi- e^{i\alpha (x)}m \psi\right) \psi \\ &= \overline{\psi } \left( i\gamma ^{\mu }D_{\mu }-m \right) \psi \end{split} \end{equation}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:
\begin{equation} A^{\nu }A_{\nu } \end{equation}But it won't transform nicely. Recall that the gauge field has a transformation defined by $A'_{\mu } = A_{\mu }-\frac{1}{e}\partial _{\mu }\alpha (x) $, so we will get extra terms when we transform the above contraction:
\begin{equation} (A^{\nu }A_{\nu })'\rightarrow A^{\nu }A_{\nu }+\text{Extra terms} \end{equation}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:
\begin{equation} F^{\mu \nu }=\partial ^{\mu }A^{\nu }-\partial ^{\nu }A^{\mu } \end{equation}We can easily show that $F_{\mu \nu } $ (and thus $F^{\mu \nu } $) are gauge invariant:
\begin{equation} \begin{split} F'_{\mu \nu } &= \partial _{\mu }A'_{\nu }-\partial _{\nu }A'_{\mu } \\ &= \partial _{\mu } \left( A_{\nu }-\frac{1}{e}\partial _{\nu }\alpha (x) \right) -\partial _{\nu } \left( A_{\mu }-\frac{1}{e}\partial _{\mu }\alpha (x) \right) \\ &= \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{split} \end{equation}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:
\begin{equation} \mathscr{L}_{\text{QED} }=\overline{\psi } \left( i\partial ^{\mu }D_{\mu }-m \right) \psi -\frac{1}{4}F^{\mu \nu }F_{\mu \nu } \end{equation}And as the label implies, this Lagrangian corresponds to the theory of Quantum Electrodynamics.
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.