Yang Mills Theory - Part - 2

We previously constructed our Lagrangian in a simplified form using the matrix notation for the combined spinor field.

$$L = ic\hbar\tilde\psi\gamma^\mu\partial_\mu\psi - mc^2 \tilde\psi\psi$$ But, now we cannot define transformation as previous, since the new $\psi$ here is a column matrix. The transformation should be redefined in the matrix representation. So, we introduce an unitary operator U acting on our wave function to represent gauge transformation. $$\psi' = U\psi \\ \tilde\psi' = \tilde\psi U^\dagger \\ where U^\dagger{U} =\mathbb{I}$$

For this transformation, the Lagrangian is invariant, if the $\gamma^\mu$ matrices commute with the Unitary matrix.

To apply the same case for Local gauge invariance, where we used to get an extra term, we need a new representation and Unitary matrix in general can be represented using Hermitian matrices as, $$U = e^{iH}$$ where a general Hermitian matrix is defined using four parameters using Pauli matrices, $$H = a_0\mathbb{I} + a_1\sigma_1 + a_2\sigma_2 +a_3\sigma_3 = a_0\mathbb{I}+a\cdot\sigma$$ where $ a\cdot\sigma = a_1\sigma_1+a_2\sigma_2+a_3\sigma_3$ in shorthand notation.

Thus, we get our Unitary matrix, $$U = e^{ia_0\mathbb{I}} e^{ia\cdot\sigma}$$ where $ e^{ia\cdot\sigma}$ has determinant "1" and corresponds to SU(2).

With the same strategy as used in previous, we define $$\lambda(x) = - \frac{\hbar{c}}{q} \vec{a}$$
actually it is not a vector in the usual sense. But, the notation is preferred here to differentiate it from others.

So, we have, $$\psi' = e^{-\frac{q}{c\hbar}\vec{\sigma}\cdot\vec{\lambda(x)}}\psi$$ "$e^{ia_0\mathbb{I}}$" being a simple phase factor doesn't affect the final result even though it is a function of the coordinates.

For eg: We redefine $$U = e^{ia_0\mathbb{I}}e^{ia\cdot\sigma} = e^{ia_0}S$$ where "S" is also unitary matrix and Identity is implied in the exponential of $a_0$. Under the unitary transformation.
$$L' = i\hbar{c}\tilde\psi'\gamma^\mu\partial_\mu\psi'$$
where $mc^2\tilde\psi'\psi$ term doesn't affect the invariance in both local and global gauge tranformation. So, we have, $$L' = ic\hbar\tilde\psi{U^\dagger}\gamma^\mu\partial_\mu(U\psi)$$ $$L' = ic\hbar\tilde\psi{S^\dagger}e^{-ia_0}\gamma^\mu\partial_\mu(e^{ia_0}S\psi)$$ Expanding the differential (call $ic\hbar = k)$ , $$L' =  k \tilde\psi{S^\dagger}e^{-ia_0}\gamma^\mu\left( (\partial_\mu{e}^ia_0)S\psi + e^{ia_0}\partial_\mu(S\psi)\right)$$ $$L'= k\tilde\psi{S^\dagger}e^{-ia_0}\gamma^\mu{i}e^{ia_0}\partial_\mu(a_0)(S\psi) + k\tilde\psi{S^\dagger}e^{-ia_0}\gamma^\mu{e^{ia_0}}\partial_\mu(S\psi)$$ Defining $S\psi$ as our new $\psi'$ we have $ \psi' = S\psi$ we have, $$L = ic\hbar\tilde\psi'\gamma^\mu{i}(\partial_\mu(a_0))\psi' + ic\hbar\tilde\psi'\gamma^\mu\partial_\mu(\psi')$$
with the same definition of $\lambda$, $$L' = -q\tilde\psi'\gamma^\mu\psi'\partial_\mu\lambda(x) + ic\hbar\tilde\psi'\gamma^\mu\partial_\mu\psi'$$ where the first term is exactly what we got as additional term in the previous lagrangian for which we have already defined transformation rules
$It has its own lagrangian and own vector potential - $A_0$$.
So, we just need to focus on the second term.

Thus, we have our new transformed Lagrangian as , $$L' = ic\hbar\tilde\psi'\gamma^\mu\partial_\mu\psi' - mc^2\tilde\psi\psi$$
We will focus entirely on this new Lagrangian in the next post.

Reference: Introduction to Elementary Particle Physics - Griffiths