I am pretty familiar with the Lagrangian formulation of quantum electrodynamics and perturbation theory techniques; however, I am hoping to move into QCD and other Yang-Mills Theories. As I do, I am struggling to understand the abstract algebra behind gauge theories. Wikipedia states:

"The standard model is furthermore a gauge theory, which means there are degrees of freedom in the mathematical formalism which do not correspond to changes in the physical state."

What are the physical meanings of the degrees of freedom we have within $U(1)$, $SU(2)$, and $SU(3)$ theories? I am believe that correct in saying that the symmetry of the wave function with respect to local phase shifts can be see as the gauge transformation that provides the single degree of freedom in $U(1)$, but does such a simple answer exist for $SU(2)$ and $SU(3)$?

Source: https://en.wikipedia.org/wiki/Mathematical_formulation_of_the_Standard_Model


1 Answer 1


Gauge symmetries are internal symmetries of the field that we are considering. So, instead of "symmetry of the wave function" we study the symmetry of the Lagrangian of the field. The Lagrangian should be invariant under Lorentz transformation and gauge transformation.

U(1) symmetry is a phase with a tweak. In QFT, it is claimed that the Lagrangian should be invariant under LOCAL gauge transformation. This word "local" is very important and it has important consequences.

Let's look at the local U(1) and $SU(2)_L \times U(1)$ gauge transformations

Local U(1):

$\mathscr{L_o} = i \bar{\psi} (x) \gamma^\mu \partial_\mu \psi (x)$

This is Dirac Lagrangian, as you can easily check it is invariant under $\psi \rightarrow e^{iQ\theta} \psi$ (global U(1) transformation). However, it is not invariant under $\psi \rightarrow e^{iQ\theta (x)} \psi$ (local U(1) transformation). Note that in the later $\theta(x)$ is a function of the $x$

This means that we need to change our Lagrangian such that it remains invariant under local U(1).

Let's define the Lagrangian as follow.

$\mathscr{L} = i \bar{\psi} (x) \gamma^\mu D_\mu \psi (x) \;\;\;\;\;\;\;\; Eq.A$

where $D_\mu = \partial_\mu + ieQA\mu(x)$ and $A_\mu(x)$ (which is a new field we are introducing into the theory) transforms as $A_\mu(x) \rightarrow A_\mu(x) - \frac{1}{e}\partial_\mu \theta(x)$ under U(1).

Eq.A is now invariant under LOCAL U(1) and if you expand it you will get,

$\mathscr{L} = \mathscr{L_o} - eQA_\mu(x) \bar{\psi} (x) \psi(x)$

because we required the Lagrangian to ne invariant under LOCAL U(1) , we found the interaction of the $\phi (x)$ field with a new field $A_\mu (x)$ which is photon.

Local $SU(2)_L \times U(1)$:


$\psi(x) = \begin{pmatrix} \nu_e \\ e\end{pmatrix}_L$

where $\nu_e$ is a electron neutrino field and $e$ is electron field and $L$ means that we are only looking at the left-handed ones.

$\psi(x)$ transforms like

$\psi(x) \rightarrow exp(iy_1\beta(x)) exp(i\frac{\sigma_i}{2}\alpha^i(x))\psi(x) $

where $exp(iy_1\beta(x))$ is the U(1) transformation $exp(i\frac{\sigma_i}{2}\alpha^i(x))$ is a $2\times 2$ matrix of SU(2) because $\sigma_i$ are Puili matrices.

A Lagrangian which is invariant under Local $SU(2)_L \times U(1)$ is.

$\mathscr{L} = i \bar{\psi} (x) \gamma^\mu D_\mu \psi (x) \;\;\;\;\;\;\;\; Eq.B$

where $D_\mu$ is,

$D_\mu = \partial_\mu + ig \frac{\sigma_i}{2}W_\mu^i(x) + ig'y_1 B_\mu (x)$

where three $W_\mu^i$'s and $B_\mu$ are four new gauge fields (boson fields) which will correspond to photon, W+, W- and Z bosons. (I will leave it to you to figure out how they tranform under gauge transformation)

Local gauge invariance gives us all the interactions between fermions and bosons in SM. The interaction between the bosons come from the fact that these new boson field do not commute with themselves (non-Abilian groups).

For more info read this: The Standard Model of Electroweak Interaction by A.Pich


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.