Subatomic Physics

390 Introduction to Gauge Theories
We still need to generalize the gauge transformation, Eq. (12.4), to the fields
Vµ. The isospin components of the vector fields, V (a) , do not commute with each
other, thus we rewrite Eq. (8.22) as
[I (a) ,I (b) ]=iɛabcI (c) . (12.20)
The symbol ɛabc is +1 if abc are normal-ordered or a cyclic variation thereof, and
−1 otherwise.
To derive the appropriate gauge transformation, we ask that D ′ µ ψ′ = UDµψ,
since this condition assures invariance of the equations of motion under the gauge
transformation, as we discussed earlier. With Eq. (12.19) we have
D ′ µψ ′ =(∇µ + ig � I · � V ′ µ)ψ ′
ψ ′ =exp(ig � I · � ξ )ψ.
D ′ µ ψ′ = ψ[∇µ exp(ig � I · � ξ )] + exp(ig � I · � ξ )∇µψ
+ ig � I · � V ′ µ exp(ig� I · � ξ )ψ
=exp(ig � I · � ξ ){ig � I · (∇ ′ µ � ξ )+∇µ + ig � I · � V ′ µ
+[ig � I · � V ′ µ, exp(ig � I · � ξ )]}ψ.
(12.21)
In order to make the evaluation of the commutator simpler in Eq. (12.21), we assume
ξ (a) (x,t) to be an infinitesimal and keep only linear terms in ξ. The commutator
in Eq. (12.21) is then
[ig� I · � V ′ µ, 1+ig� I · � ξ ]=−ig 2 V ′(a)
µ ɛabcξ (b) I (c)
= −ig 2� V ′ µ · � ξ × � I ≈−ig 2� Vµ × � ξ · � I.
(12.22)
Since we only keep linear terms in ξ and Eq. (12.22) already is linear, we have set
V ′ µ = Vµ in the last equality of this equation. The equality D ′ µ ψ′ = UDµψ then
leads to (6)
or
�I · � Vµ = � I · � V ′ µ + � I · � ∇µξ − g � Vµ × � ξ · � I,
�V ′ µ = � Vµ −∇µ � ξ + g � Vµ × � ξ. (12.23)
This is the desired generalization of Eqs. (12.4); note the appearance of the coupling
constant g in the additional term of Eq. (12.23).
6 It is relatively straightforward to generalize this expression to massless particles with higher
degrees of freedom. As shown earlier, a mass term would break gauge invariance. It is therefore
important that the gauge field quanta represented by V (a) retain zero mass.