U5 1 7 I pse2 hb fl5e3

= cos 6 — ¿758sin# . Hence, the transformation law (13.77) can be written ga<j>'„ + ig*-ys#„ = (cos 9 -f- sin9) (.gad>a + ig^tyj {cos 6 + i-y5e sin0) =g<T<f>IT [cos 26 + ?75E sin 29]

+ »Sir75 cos2 9 - £<)>„.£ sin2 9} - gn A sin 29 {£, ^ } .

Using where = (4>\, ,is the three-vector in isospin space (not to be confused with the matrix (j)), gives ga4>'„ + ignl5^ =g<T<t><7 cos 29 - g„ {4>n ■ e) sin29

+ H5 [9o<Pa £ sin 29 + g^ - 2g* {4>n ■ e) £ sin2 9} .

Substituting 2sin20 = 1 - cos 29 into the last term gives finally <j/a = <J>a cos 29 - (^j {(¡>n ■ e) sin 29

The transformation is therefore a rotation in four-dimensional space through angle 29 in the plane defined by <fia and e. The components of <f>v perpendicular to e are unchanged.

The transformation above looks very complicated but is much simpler when written out to first order in f . It becomes

9*, infinitesimal.

Note that the chiral rotation mixes the a and one of the 7r components of the four-dimensional field vector (cpa,(t>\), and hence, in the linear sigma model, both fields must be present in order to have chiral invariance.

Now, examine the kinetic energy part of the Lagrangian for the tt and a fields. Since these fields are self-conjugate, the KE term must have the form

This is clearly invariant under the gauge group SU(2). Invariance under an infinitesimal chiral transformation requires

jince st is a constant, the terms proportional to g5 have the same structure and A£ke = gs(-~ [9^1 U . (13.85)

Hence these will cancel, and £ke will be invariant if ga — gn. Furthermore, as we did in our discussion of gauge invariance, we will take g5 — gn, so that there is only one coupling,

The meson matrix can therefore be written M = gn <j), where the new, simplified meson matrix is

The trace of (Jx))* in flavor space can be used to define a chirally invariant length

Was this article helpful?

0 0

Post a comment