Non-abelian Yang-Mills

This is a generalisation of Electrodynamics. Here, the gauge fields (gluons) are matrices and do not commute, i.e., they will be self-interacting, unlike photons.

$(A_{\mu }{)}_i^j=A_{\mu }^a(T^a{)}_i^j;a=1,\dots ,N^2-1$,

where $T^a$ are $N\times N$ traceless Hermitian matrices corresponding to generators of $SU(N)$.

• $N=3$ corresponds to QCD, describing strong interactions.

$[T^a,T^b]=i\sqrt{2}{f}^{abc}{T}^c$ and $\text{Tr}(T^a{T}^b)={\delta }^{ab}⟹A_{\mu }^a=\text{Tr}(T^a{A}_{\mu })$.

Now, $\mathcal{L}=-\frac{1}{4}\text{Tr}(F_{\mu \nu }{F}^{\mu \nu })$ with $F_{\mu \nu }={\partial }_{\mu }{A}_{\nu }-{\partial }_{\nu }{A}_{\mu }-\frac{ig}{\sqrt{2}}[A_{\mu },A_{\nu }]$.

$F_{\mu \nu }^a={\partial }_{\mu }{A}_{\nu }^a-{\partial }_{\nu }{A}_{\mu }^a+\sqrt{2}g{f}^{abc}{A}_{\mu }^b{A}_{\nu }^c$.

Now, the gauge symmetry can be generalised to $A_{\mu }\to U{A}_{\mu }{U}^{†}+\frac{i}{g}U{\partial }_{\mu }{U}^{†}$ where $U(x)\in SU(N)$.

Note: For $U(x)\in U(1)$, e.g., $U(x)=e^{i\alpha (x)}$, we recover the gauge symmetry from electromagnetism since $\alpha (x)$ is just a function.

The action is invariant under the above symmetry. Further,

$F_{\mu \nu }=\frac{\sqrt{2}i}{g}[D_{\mu },D_{\nu }];(D_{\mu }{)}_j^i={\delta }_j^i{\partial }_{\mu }-\frac{ig}{\sqrt{2}}(A_{\mu }{)}_j^i$.

Under a gauge transformation,

$D_{\mu }\to U{D}_{\mu }{U}^{†};F_{\mu \nu }\to U{F}_{\mu \nu }{U}^{†}$,

$\text{Tr}(F_{\mu \nu }{F}^{\mu \nu })\to \text{Tr}(U{F}_{\mu \nu }{U}^{†}U{F}^{\mu \nu }{U}^{†})=\text{Tr}(F_{\mu \nu }{F}^{\mu \nu })$, using the cyclicity of trace and the fact that $U$ is unitary.

We can fix the gauge using the Fadeev-Popov procedure:

$\mathcal{L}=-\frac{1}{4}\text{Tr}(F_{\mu \nu }{F}^{\mu \nu })-\frac{1}{2}\text{Tr}(GG)+\text{ghosts}$, where $G$ is the gauge-fixing functional and we will neglect the ghosts since we are interested only in the tree-amplitudes.

We will choose the Gervais-Neveu gauge, which is convenient for amplitudes,

$G={\partial }_{\mu }{A}^{\mu }-\frac{ig}{\sqrt{2}}{A}_{\mu }{A}^{\mu }$.

The Lagrangian then becomes $\mathcal{L}=\text{Tr}\left(\frac{1}{2}{A}_{\mu }{\partial }^2{A}^{\mu }-i\sqrt{2}g{\partial }^{\mu }{A}^{\nu }{\partial }_{\nu }{\partial }_{\mu }+\frac{1}{4}{g}^2{A}^{\mu }{A}^{\nu }{A}_{\mu }{A}_{\nu }\right)$.