Im taking my first course in QFT and has stumbled upon something that I do not understand.

Given the Yang Mills lagrangian

$$\mathcal{L} = -\frac{1}{4}F^{a}_{\mu \nu}F^{a\mu \nu}$$

with $F^{\mu \nu} =F^{a\mu \nu} \frac{\sigma^{a}}{2}$ (the pauli matrices). How can I determine the number of unphysical and physical degrees of freedom of this theory?

I know this Lagrangian describes massless spin(1) gauge bosons $A^{\mu}$, which means (I think) that the gauge boson has 2 physical degrees of freedom. However, I do not understand how to count all the remaining degrees of freedom. I suspect it is related to that there are 3 generators for SU(2), although I do not know how to make the connection.

I hope that my question is clear (=

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The field stregth tensor of a Yang-Mills theory is defined as $$F_{\mu\nu}=\partial_\mu A_\nu-\partial_\nu A_\mu+ie\left[A_\mu,A_\nu\right].$$ In general, the gauge field is in the adjoint representation of the gauge group (we normally say it takes value in the algebra) so it is written as $$A_\mu=A_\mu^aT_a,\quad a=1,2,\ldots dim(G),$$ where $dim(G)$ (dimension of $G$) is the number of generator of the gauge group. Therefore the number of degrees of freedom is $4\cdot dim(G)$. However each gauge field (correspondent to each generator) has only two physical degrees of freedom (the polarization), so the number of physical d.o.f. is $2\cdot dim(G)$.

It happens that all (simple) Lie groups have been classified and we know $dim(G)$: \begin{align*} dim(SU(n))&=n^2-1,\\ dim(SO(2n+1))&=n(2n+1),\\ dim(SO(2n))&=n(2n-1), \\ dim(Sp(2n))&=n(2n+1), \\ dim(E_6)&=78, \\ dim(E_7)&=133, \\ dim(E_8)&=248, \\ dim(F_4)&=52, \\ dim(G_2)&=14. \\ \end{align*}

In your case, $G=SU(2)$ gives $2\cdot 3=6$ physical degrees of freedom.

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