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Understanding $rm SU(3)$ Gauge Invariance Through a Perturbation Theory

Physics Asked on June 9, 2021

Here I am modifying the field theory approach, since I never taken a course on Quantum Field Theory. I am exploring Gauge Invariance in $rm SU(3)$ by the following approach (which technically is incorrect but nevertheless I think it is useful):

Let $(1,0,0)^T = text{Red}$, $(0,1,0)^T = text{Blue}$, and $(0,0,1)^T = text{Green}$. Then a wavefunction can be represented as $(Psi_1, Psi_2, Psi_3)^T$. Let $alpha$ be a vector that represents a small change. Consider the following transformation:

$(Psi_1, Psi_2, Psi_3)^T to(1 – ialphacdotfrac{lambda}{2})(Psi_1, Psi_2, Psi_3)^T$

Here $lambda$ is a 8 component vector; each component $i$ is the $i^{th}$ Gelmann matrix (i.e. a generator of SU(3)). Hence we see that this transformation is perturbative.

Now the Gauge freedom is given by:

$partial_{mu} to partial_{mu} + ifrac{lambda}{2}cdot S_{mu}$

Here $S_{mu}$ is the strong potential; it has 8 components and each component has 4 "slots" (one for time and 3 for space)

Consider the transformation: $S_{mu} to S_{mu} + partial_{mu}alpha + alpha times S_{mu}$ . Here the cross product is defined by the $rm SU(3)$ structure constants $f_{ijk}$ by the following: $(A times B)_i = sum_{i,j}^{8}f_{i,j,k}A_j B_k$

My goal is to show that following:

$big[partial_{mu} + ifrac{lambda}{2}cdot {S_{mu} + partial_{mu}alpha + alpha times S_{mu}}big](1 – ialphacdotfrac{lambda}{2})(Psi_1, Psi_2, Psi_3)^T approx (1 – ialphacdotfrac{lambda}{2})[partial_{mu} + ifrac{lambda}{2}cdot S_{mu}] (Psi_1, Psi_2, Psi_3)^T$

I tried this approach in $rm SU(2)$ and it worked via the usual vector identities (because the cross product in $rm SU(2)$ had the same form as the cross product learned in school) and by ignoring $alpha^2$ terms. However, I am not sure how to show this here. (I think it should work for small perturbations)

One Answer

What you have can be rewritten as begin{eqnarray} && left( partial_mu +i A_mu + i partial _mu alpha + [alpha,A_mu] right) left(1-i alpha right) psi approx left(1-i alpha right) partial _mupsi-ipartial _mu alpha psi+iA_mupsi+A_mualphapsi+ipartial_mualpha psi+[alpha,A_mu]psi && left(1-i alpha right) partial _mupsi+iA_mupsi+A_mualphapsi+[alpha,A_mu]psi=left(1-i alpha right) partial _mupsi+iA_mupsi+alpha A_mupsi=left(1-i alpha right) partial _mupsi+ileft(1-i alpha right)A_mupsi &&=left(1-i alpha right)left(partial _mu+i A_mu right)psi end{eqnarray}

Correct answer by nwolijin on June 9, 2021

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