Things happened in the past few weeks. I was called to my parents place to deliver my desktop to them and I blew up one of my midterms around the same time. I have been planning traveling plans in the summer as well, I want to visit their parents and talk to them (partly because I want to thank them), since all happened too sudden; yet I doubt I have the courage to do so.


Lately in particle physics, we are taking a trip into the symmetry and group theory. I shall outline the components you need to make a group structure:

  1. Group needs a binary operator
  2. Closure under that binary operator
  3. Has identity element
  4. Each group element has inverse

Something to keep in mind is that members of a group are operations NOT states. For example if we're talking about rotation in 2D plane, the group contains rotation of any angle. It's easier to use an underlying object to visualize the effect of operations from the group but they are not essential.


As Jeff remarked in class, physics concerns more about group representation. This is because we want to use group to represent physical transformation (like spacetime), and by using group to describe the underlying symmetry(invariance), we can do physics better.

It happens to be the case that, local $U(1)$ symmetry imposed on Dirac equation would give the correct description of Electromagnetism physics(by introducing vector field). For readers that are less familiar with groups, $U(1)$ is a group that can be represented by unitary 1x1(basically scalar) matrix that satisfy: $U^ \dagger U  = UU^ \dagger = I$, so we usually write an element as a complex number $e^ {iq\chi}$. (q is just a constant representing charge, for physical convenience)

Before doing that, let's see a trivial demonstration with global $U(1)$, starting from Dirac equation:

\[ \begin{align} i{\partial_\mu\!\!\!\!\!\!/}~~\Psi(x) &= m\Psi(x)\\ e^{iq\chi} i{\partial_\mu\!\!\!\!\!\!/}~~ \Psi(x) &= e^{iq\chi}m\Psi(x)\\ i{\partial_\mu\!\!\!\!\!\!/}~~ (e^{iq\chi}\Psi(x)) &= m (e^{iq\chi}\Psi(x))\\ i{\partial_\mu\!\!\!\!\!\!/}~~ \Psi^\prime(x) &= m \Psi^\prime(x) \end{align} \]

We can just exchange the global complex number to the right because it's just a number. We see that obviously the form of physics stays the same, which is all we can dream of. Jeff promised us that this will lead to global charge conservation, I believe him.

But now we gonna do something crazy, and I think this is the root of all evil SUSY. We gonna demand that the Dirac equation stays its form under local $U(1)$ transformation, meaning that now we will allow the $\chi(x)$ to have dependence on spacetime coordinates. This clearly would creat issue because we cannot exchange the transformation to the right:

\[ \begin{align} i{\partial_\mu\!\!\!\!\!\!/}~~\Psi(x) &= m\Psi(x)\\ e^{iq\chi(x)} i{\partial_\mu\!\!\!\!\!\!/}~~ \Psi(x) &= m e^{iq\chi(x)}\Psi(x)\\ \end{align} \]

Now we want to know how much we missed the $\Psi^\prime(x)$ mark:

\[ \begin{align} e^{iq\chi(x)} i{\partial_\mu\!\!\!\!\!\!/}~~ \Psi(x) &= e^{iq\chi(x)} i{\partial_\mu\!\!\!\!\!\!/}~~ e^{-iq\chi(x)}e^{iq\chi(x)}\Psi(x)\\ &= e^{iq\chi(x)} i\gamma^\mu\partial_\mu e^{-iq\chi(x)}\Psi^\prime(x) \end{align} \]

Concentrate on this term:

\[ {\partial_\mu} e^{-iq\chi(x)}\Psi^\prime = [e^{-iq\chi}(-iq\partial_\mu\chi)\Psi^\prime + e^{-iq\chi}(\partial_\mu \Psi^\prime)] \]

Plug back:

\[ \begin{align} e^{iq\chi(x)} i\gamma^\mu[e^{-iq\chi}(-iq\partial_\mu\chi)\Psi^\prime + e^{-iq\chi}(\partial_\mu \Psi^\prime)] &= i\gamma^\mu[\color{red}{(-iq\partial_\mu\chi)\Psi^\prime} + \partial_\mu \Psi^\prime]\\ &=m \Psi^\prime \end{align} \]

But we really want this instead to make the form of physics remain unchanged:

\[ i\gamma^\mu\partial_\mu \Psi^\prime =m \Psi^\prime \]

So we have this extra term: $i\gamma^ \mu(-iq\partial_\mu\chi)\Psi^ \prime = \gamma^ \mu q (\partial_ \mu \chi) \Psi^ \prime$. We can re-write the equation with the extra term into:

\[ \gamma_\mu [i \partial^\mu + \color{red}{q\partial ^\mu \chi}]\Psi^\prime =m \Psi^\prime \]

Here's the move we gonna make, recall that in classical EM, if we write the scalar and vector potential into a 4-vector $A$ we would have a gauge freedom denoted by the following transformation:

\[ \begin{align} (A^{\prime0}) &= A^0 - \frac{\partial \chi}{\partial t}\\ \vec{A^\prime} &= \vec{A} + \vec{\nabla}\chi \end{align} \]

Or in compact form: $(A^ \prime)^ \mu = A^ \mu - \partial^ \mu \chi$ (note: the sign switch comes from the contra/co varient nature of partials). By making the $\chi$ in both transformation the same thing, we see that the $A$ would give off terms that cancels the extra term we did not want thus giving back us the desired formalism if we propose an additional term in Dirac equation to begin with:

\[ \gamma^\mu [i\partial_\mu -qA_\mu(x)]\Psi(x) = m\Psi(x) \]

Then under $e^ {iq\chi(x)}$ transformation:

\[ \begin{align} e^{iq\chi}\gamma^\mu [i\partial_\mu -qA_\mu(x)]e^{-iq\chi}e^{iq\chi}\Psi(x) &= me^{iq\chi}\Psi(x)\\ e^{iq\chi}\gamma^\mu [i\partial_\mu -qA_\mu(x)]e^{-iq\chi}\Psi^\prime(x) &= m\Psi^\prime(x)\\ e^{iq\chi}\gamma^\mu e^{-iq\chi}[i\partial_\mu +q(\partial_\mu\chi)-qA_\mu(x)]\Psi^\prime(x) &= m\Psi^\prime(x)\\ \gamma^\mu [i\partial_\mu -q(A_\mu(x) - \partial_\mu\chi)] \Psi^\prime(x) &= m\Psi^\prime(x)\\ \gamma^\mu [i\partial_\mu -qA^\prime(x)] \Psi^\prime(x) &= m\Psi^\prime(x)\\ \end{align} \]

Which is what we wanted!