Yang Mills Theory - Part - 3

We have our Lagrangian $$L = ic\hbar\tilde\psi\gamma^\mu\partial_\mu\psi - mc^2\tilde\psi\psi$$
After the gauge transformation defined as, $$ \psi' = S\psi \\ where\,\,S = e^{-i\frac{q}{c\hbar}\vec{\sigma}\cdot\vec{\lambda(x)}}$$ $$ L' = ic\hbar\tilde\psi'\gamma^\mu\partial(S\psi) \\= ic\hbar\tilde\psi{S^\dagger}\gamma^\mu{S}\partial_\mu\psi + ic\hbar\tilde\psi{S^\dagger}\gamma^\mu(\partial_\mu{S})\psi $$gives $$ L' = \left[ic\hbar\tilde\psi\gamma^\mu\partial_\mu\psi = L \right]+ {extra\,term} $$[assumed $\gamma^\mu$ commutes with S]

To compensate this extra term we use the previous mathematical tool, i.e. we define the covariant derivative as, $$ D_\mu = \partial_\mu + \frac{iq}{c\hbar}\vec{\sigma}\cdot\vec{A}_\mu$$
where the new $ \vec{A_\mu}$ term should cancel our previous extra terms on transformation. [The new term "$\vec{A_\mu}$" now has three components as to cancel each component in $\vec{\lambda(x)}$]

The purpose of this new covariant derivative is,

If we define our Lagrangian using this derivative as $$L = ic\hbar\tilde\psi\gamma^\mu{D_\mu}\psi$$ then we should have the same Lagrangian after the transformation $$L' = ic\hbar\tilde\psi'\gamma^\mu{D'}_\mu(S\psi) = ic\hbar\tilde\psi{S^\dagger}\gamma^\mu{S}(D_\mu\psi) = ic\hbar\tilde\psi\gamma^\mu{D_\mu\psi}$$ where we implied the condition, $$D'_\mu(S\psi) = S(D_\mu\psi)$$Remember, unlike the usual derivatives, Covariant derivatives also transform under the gauge transformation.

From our condition, the transformation rule for $\vec{A_\mu}$ is derived as, $$ D_\mu' \psi'=\left(\partial_\mu+\frac{iq}{c\hbar}\vec{\sigma}\cdot\vec{A'_\mu}\right)(S\psi) = S \left[\left(\partial_\mu+\frac{iq}{c\hbar}\vec{\sigma}\cdot\vec{A_\mu}\right)\psi\right] $$ Using the previous relations, $$\left(\partial_\mu{S}\right)\psi + S(\partial_\mu\psi) + \frac{iq}{c\hbar}\vec{\sigma}\cdot\vec{A'_\mu}(S\psi) = S(\partial_\mu\psi) + \frac{iq}{c\hbar}S\left(\vec{\sigma}\cdot\vec{A_\mu}\psi\right)$$ Cancelling the terms and taking out psi $$ \left[(\partial_\mu{S}) +\frac{iq}{c\hbar}\left(\vec{\sigma}\cdot\vec{A'_\mu}\right)S\right]\psi=\left[\frac{iq}{c\hbar}S\left(\vec{\sigma}\cdot\vec{A_\mu}\right)\right]\psi$$ Multiplying with the $ S^{-1}$ operator on the right hand side we get the transformation relation, $$ \vec{\sigma}\cdot\vec{A'_\mu} = S\left(\vec{\sigma}\cdot\vec{A_\mu}\right)S^{-1}+ \frac{ic\hbar}{q}(\partial_\mu{S})S^{-1}$$
Now, we can check the invariance of the Lagrangian, $$ L_{new} = ic\hbar\tilde\psi\gamma^\mu(D_\mu\psi) = ic\hbar\tilde\psi\gamma^\mu\left(\partial_\mu + \frac{iq}{\hbar{c}}(\vec{\sigma}\cdot\vec{A_\mu})\right)\psi$$ After the transformation, $$ L'_{new} = ic\hbar\tilde\psi{S^\dagger}\gamma^\mu(D'_\mu(S\psi))\\ = ic\hbar\tilde\psi{S^\dagger}\gamma^\mu\left[\partial_\mu(S\psi)+\frac{iq}{c\hbar}(\vec{\sigma}\cdot\vec{A'_\mu})(S\psi)\right] \\ = ic\hbar\tilde\psi{S^\dagger}\gamma^\mu\left[S\partial_\mu\psi+(\partial_\mu{S})\psi + \frac{iq}{c\hbar}S(\vec{\sigma}\cdot\vec{A_\mu})S^{-1}S\psi+\frac{iq}{c\hbar}\frac{ic\hbar}{q}(\partial_\mu{S})S^{-1}(S\psi)\right]\\ = ic\hbar\tilde\psi{S^\dagger}\gamma^\mu\left[S(\partial_\mu\psi)+(\partial_\mu{S})\psi+\frac{iq}{c\hbar}S(\vec{\sigma}\cdot\vec{A_\mu})\psi - (\partial_\mu{S})\psi\right]\\ = ic\hbar\tilde\psi{S^\dagger}\gamma^\mu{S}\left[\partial_\mu\psi + \frac{iq}{c\hbar}(\vec{\sigma}\cdot\vec{A_\mu})\psi\right]$$Gives finally the same, $$ L'_{new} = ic\hbar\tilde\psi\gamma^\mu\left(\partial_\mu+\frac{iq}{c\hbar}(\vec{\sigma}\cdot\vec{A_\mu})\right)\psi = L_{new} $$ Thus, we can check the invariance of our new Lagrangian.

Now, we expand our definition of S by the assumption $\lambda$ is very small so that all higher order terms can be neglected. After this approximation, substitution of the new "S" in our transformation equation for $ A'_\mu$ yields, $$ \vec{\sigma}\cdot\vec{A'_\mu} \approx \vec{\sigma}\cdot\vec{A_\mu} + \vec{\sigma}\cdot\partial_\mu\lambda + \frac{iq}{c\hbar}\left[(\vec{\sigma}\cdot\vec{A_\mu}),(\vec{\sigma}\cdot\vec{\lambda})\right]$$ Using the rule, $$(\vec{\sigma}\cdot\vec{a})(\vec{\sigma}\cdot\vec{b}) = \vec{a}\cdot\vec{b} + i\sigma\cdot(\vec{a}\times\vec{b}) $$ we get, $$\vec{\sigma}\cdot \vec{A'_\mu} =\vec{\sigma}\cdot\vec{ A_\mu} +\vec{\sigma}\cdot\partial_\mu\vec{\lambda} + \frac{iq}{c\hbar}\left(2i\vec{\sigma}\cdot (\vec{A_\mu}\times\vec{\lambda})\right) $$gives$$ \vec{A_\mu'} = \vec{A_\mu} + \partial_\mu{\vec{\lambda}}+\frac{2q}{c\hbar}(\vec{\lambda}\times\vec{A_\mu})$$ Thus, we got our new Lagrangian and the new transformation rules for corresponding terms.
 
But, from the same argument for the vector potential in previous problem invokes its field term in our Lagrangian. Once again, we take the Proca Lagrangian term without mass to explain our vector potential term.
 
You can ask, why we can't take the Proca Lagrangian with mass and redefine our $A'_\mu$ such that the invariance holds for $$ A^\nu{A_\nu}$$ Yes. But we have already just finished defining the transformation rules for $A'_\mu$ and it doesn't have the invariance. Maybe, if you can find $ A'_\mu$ such that, it satisfies both the above transformation and the invariance of $A^\nu{A_\nu}$ it would be wonderful. To make a comment, I should try and workout completely and find the reason for the impossibility (if there is any).

Not even $ A^\nu{A_\nu}$ but also the $ F^{\mu\nu}F_{\mu\nu}$ term is not invariant when we redefine our $ F_{\mu\nu}$ for three vector potentials as, $$ L_{extra} = \frac{-1}{16\pi}F^{\mu\nu}_1F^1_{\mu\nu}-\frac{1}{16\pi}F^{\mu\nu}_2F^2_{\mu\nu}-\frac{-1}{16\pi}F^{\mu\nu}_3F^3_{\mu\nu} = \frac{-1}{16\pi}\vec{F}^{\mu\nu}\vec{F}_{\mu\nu}$$ We will separately deal with its transformation rules and how it can be restated with additional terms in the next post.

Reference: Introduction to Elementary Particle Physics - Griffiths