Gauge Symmetries
================================

U(1) Global Gauge Invariance
----------------------------------

Assuming the Lagrangian is invariant under U(1) global gauge transformation

.. math::
   \psi(x) \to e^{i\alpha} \psi(x),

we obtain the Noether current

.. math::
   j^\mu = - e \bar \psi \gamma^\mu \psi,

which is conserved

.. math::
   \partial_\mu j^\mu = 0.



.. admonition:: Derivation of Noether Current
   :class: toggle

   QED



U(1) Local Gauge Invariance
----------------------------------


Introduce the U(1) local gauge transformation

.. math::
   \psi(x) \to e^{i\alpha(x) } \psi(x)

to the Lagrangian, we notice that the Lagragian is generally not invariant. However, if we require it to be invariant, a new field :math:`A_\mu` should be introduced, so that

.. math::
   A_\mu \to A_\mu + \frac{1}{2} \partial_\mu \alpha(x).

The way this new field comes into the Lagriangian is to replace all the derivatives :math:`\parital_\mu` with :math:`\mathrm D_\mu`,

.. math::
   \mathrm D_\mu = \partial_\mu - i e A_\mu.


What about the kinetic term for this new field? It is constructed from the field strength tensor

.. math::
   F_{\mu\nu} = \partial_\mu A_\nu - \partial_\nu A_\mu,

so that the kinetic term is

.. math::
   - \frac{1}{4} F_{\mu\nu}F^{\mu\nu}.


Finally we reach the new (QED) Lagrangian that is invariant under U(1) gauge transformation

.. math::
   \mathcal L = \bar \psi ( i\gamma^\nu \partial_\mu - m )\psi + e\bar \psi \gamma^\mu \partial_\mu \psi - \frac{1}{4} F_{\mu\nu} F^{\mu\nu} .



Non-Abelian
-------------------------------


Introduce the non-abelian local gauge transformation

.. math::
   q(x) \to e^{i\alpha_a (x) T_a} q(x).

The Lagrangian that is invariant under this transformation is

.. math::
   \mathcal  = \bar q (i\gamma^\mu \partial_\mu - m) q - g(\bar q \gamma^\mu T_a q) G_{\mu}^a - \frac{1}{4} G_{\mu\nu}^a G^{\mu\nu}_a.