# Peskin Schroeder Higgs mechanism for an $SU(3)$ gauge theory with a scalar field $\varphi$ in the adjoint representation

In Peskin Schroeder pag.696 a Higgs mechanism for an $$SU(3)$$ gauge theory with a scalar field $$\phi$$ in the adjoint representation is presented. The covariant derivative of $$\phi$$: $$D_{\mu}\phi_{a} = \partial_{\mu}\phi_{a}+ g f_{abc}A^{b}_{\mu}\phi_{c}\tag{20.32}$$ defining the quantity $$\Phi=\phi_c t^c\tag{20.34}$$ We consider the expansion of $$\Phi$$ around two vacuum choice: $$1) \,\,\,\Phi_0 = v \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & -2 \end{bmatrix}$$ $$2) \,\,\,\Phi_0 = v \begin{bmatrix} 1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 0 \end{bmatrix}$$ In the first case the unbroken generators are: $$T_1,T_2,T_3,T_8$$

In the second case the unbroken generators are: $$T_3,T_8$$

Where $$T_i$$ are Gell-Mann matrices according to the normalization which here is $$T_i = \lambda_i/2$$.

In the book is said that $$SU(3)$$ breaks spontaneously to $$S U (2) \times U(1)$$ in the first case and $$U(1) \times U(1)$$ in the second case.

I really can't understand how to prove that $$T_1,T_2,T_3,T_8$$ generate $$S U (2) \times U(1)$$ and $$T_3,T_8$$ generate $$U(1) \times U(1)$$

Starting from the simplest how a 3 x 3 matrix $$T_3 = \begin{bmatrix} 1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 0 \end{bmatrix}$$ can generate $$U(1)$$, (ot the same for $$T_8$$)?

A basis for the Lie Algebra of $$SU(2)$$ (aka its infinitesimal generators):

$$\left(\ \frac{1}{2} \sigma_{1}, \ \frac{1}{2} \sigma_{2}, \ \frac{1}{2} \sigma_{3} \right)$$
I can recognize that inside the matrices that remain I can found $$\sigma_{1}, \sigma_{2}, \sigma_{3}$$ $$T_1 = \begin{bmatrix} (\sigma_1)_{00} & (\sigma_1)_{01} & 0 \\ (\sigma_1)_{10} & (\sigma_1)_{11} & 0 \\ 0 & 0 & 0 \end{bmatrix} T_2 = \begin{bmatrix} (\sigma_2)_{00} & (\sigma_2)_{01} & 0 \\ (\sigma_2)_{10} & (\sigma_2)_{11} & 0 \\ 0 & 0 & 0 \end{bmatrix} T_3 = \begin{bmatrix} (\sigma_3)_{00} & (\sigma_3)_{01} & 0 \\ (\sigma_3)_{10} & (\sigma_3)_{11} & 0 \\ 0 & 0 & 0 \end{bmatrix}$$ However the matrices are not $$2 \times 2$$ but $$3 \times 3$$. Recalling the definition of $$SU(2)$$: $$SU(2) \equiv \{ M \in GL(n, \mathbb{C}) | M^{\dagger} \mathbb{1}_2 M = \mathbb{1}_2, \,\, det(M)=1 \}$$ We notice that: $$det(T_1)=det(T_2)=det(T_3)=0$$ Moreover thare is a dimensional problem in performing $$T_1^{\dagger} \mathbb{1}_2 T_1,$$ $$\,\,$$ $$T_2^{\dagger} \mathbb{1}_2 T_2,$$ $$\,\,$$ $$T_3^{\dagger} \mathbb{1}_2 T_3$$.

I'm quite lacking in group theory so I would need a step-by-step answer.

• You need consider determinant of group element, not generators!! Group element is exponent of generators with arbitrary coefficients! – Nikita Jan 14 at 10:44
• @Nikita thank you very much, my final doubt is that to be SU(2) $e^{- i \alpha_a T^a}$ has to be in $GL(2, \mathbb{C})$, however $e^{- \alpha_a T^a}$ is a $3 \times 3$ matrix. I understood that you can represent the SU (2) group with 3 x 3 matrices but in that case the generators are no longer Pauli's matrices – Stefano Barone Jan 14 at 23:18

Using concrete realization of this generators, you can easily check that $$T_1, T_2, T_3$$ are exactly $$SU(2)$$ and $$T_8$$ commute with them and so gives you $$U(1)$$.