- #1
Antarres
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In chapter 20 of Peskin&Schroeder about spontaneous symmetry breaking, he considers and example on page 696 of spontaneous symmetry breaking of SU(3) gauge group with generators taken in adjoint representation.
Covariant derivative is defined with:
$$D_\mu\phi_a = \partial\phi_a + gf_{abc}A^b_\mu\phi_c$$
The model considered is of a multiplet of real scalar fields coupled to gauge bosons. He's looking for mass matrix of bosons, which he gets from the term in Lagrangian given in (20.33):
$$\Delta\mathcal{L} = \frac{g^2}{2}(f_{abc}A^b_\mu\phi_c)^2$$
Now he defines a variable ##\Phi = \phi_c t^c##, where ##t^c## are the generators of SU(3) which are traceless 3x3 Hermitian matrices.
He says (equation 20.35), that we can then rewrite this term in form:
$$\Delta\mathcal{L} = -g^2tr([t^a,\Phi][t^b,\Phi])A^a_\mu A^{b\mu}$$
However, checking this trace, and substituting the definition of commutators of generators, there is a residual term that is the trace of two generators which arise from those two commutators. But that trace equals to the invariant C(r) multiplied by a Kronecker delta over the gauge indices. The invariant C(r) is equal to quadratic Casimir operator in adjoint representation of SU(N), ##C(G) = C_2(G) = N##. Therefore, a factor of 3 should arise there, and so I don't see how that factor transfers into the factor of one half in the mass term given in the first equation.
Any help would be appreciated. I suspect maybe I'm missing some kind of symmetrization step, but I'm not sure, I can't see a mistake in my calculations.
Covariant derivative is defined with:
$$D_\mu\phi_a = \partial\phi_a + gf_{abc}A^b_\mu\phi_c$$
The model considered is of a multiplet of real scalar fields coupled to gauge bosons. He's looking for mass matrix of bosons, which he gets from the term in Lagrangian given in (20.33):
$$\Delta\mathcal{L} = \frac{g^2}{2}(f_{abc}A^b_\mu\phi_c)^2$$
Now he defines a variable ##\Phi = \phi_c t^c##, where ##t^c## are the generators of SU(3) which are traceless 3x3 Hermitian matrices.
He says (equation 20.35), that we can then rewrite this term in form:
$$\Delta\mathcal{L} = -g^2tr([t^a,\Phi][t^b,\Phi])A^a_\mu A^{b\mu}$$
However, checking this trace, and substituting the definition of commutators of generators, there is a residual term that is the trace of two generators which arise from those two commutators. But that trace equals to the invariant C(r) multiplied by a Kronecker delta over the gauge indices. The invariant C(r) is equal to quadratic Casimir operator in adjoint representation of SU(N), ##C(G) = C_2(G) = N##. Therefore, a factor of 3 should arise there, and so I don't see how that factor transfers into the factor of one half in the mass term given in the first equation.
Any help would be appreciated. I suspect maybe I'm missing some kind of symmetrization step, but I'm not sure, I can't see a mistake in my calculations.