# Question about S(3) spontaneous symmetry breaking in Peskin & Schroeder

• A
• Antarres
Antarres
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.

Homework Helper
If I remember correctly, the proportionality constant in ##\Tr{t^at^b}\propto \delta^{ab}## is arbitrary, so one must specify it when defining the conventions adopted. A very used case (at least I think in particle physics, maybe not so commit in other areas) is to fix ##\Tr{t^at^b}=\frac{1}{2} \delta^{ab}##. P&S uses such convention.
With this convention, the equality follows immediately.

Antarres
Okay well, it is true that rescaling the generators changes this number, since it is a normalization convention. The equation is:
$$\text{tr}(t^at^b) = C(r)\delta^{ab}$$
where C(r) is a constant that is dependent on representation, and t's are generators. The representation is taken to be irreducible, and it is said that if we fix this number in one representation, then we fix it in all other representations(although I haven't tried to prove that). That means, from what I get, that this is indeed a convention, however that convention is not giving the same constant for every representation. It is conventional for fundamental representation of SU(N) to choose:
$$\text{tr}(t^at^b) = \frac{1}{2}\delta^{ab}$$
However since he mentioned that in this model we take the scalar field to transform according to the adjoint representation, I figured these generators in (20.35) would also be in adjoint representation. In adjoint representation, for the convention above, we have that:
$$\text{tr}(t^at^b) = 3\delta^{ab}$$
That is calculated in Peskin chapter 15.4. It is possible though, that when he rewrote the equation like this, he chose the generators to be in the fundamental representation, it didn't cross my mind he would do that, but I guess it's possible, since this substitution doesn't seem to have anything to do with the transformation of the fields. So maybe you're correct, I just wasn't sure that is the case.

Homework Helper
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.
Mmm... I think this sounds like the fundamental representation.

• vanhees71
Antarres
Yeah, you're right. Adjoint representation of SU(3) isn't three dimensional. I guess my mind slipped on that sentence. Thank you, anyways, now it all makes sense.

• vanhees71