On the Flavor Problem in Strongly Coupled Theories - THEP Mainz
On the Flavor Problem in Strongly Coupled Theories - THEP Mainz
On the Flavor Problem in Strongly Coupled Theories - THEP Mainz
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F<br />
gaug<strong>in</strong>g<br />
I<br />
G<br />
SSB<br />
H<br />
1.1. Solutions to <strong>the</strong> Gauge Hierarchy <strong>Problem</strong> 19<br />
Simplest Little Higgs:<br />
G = SU(3) × SU(3) × U(1)<br />
H = SU(2) × SU(2) × U(1)<br />
F = SU(3) × U(1)<br />
I = SU(2) × U(1)<br />
Figure 1.6: Diagram illustrat<strong>in</strong>g <strong>the</strong> relations between <strong>the</strong> different symmetry groups<br />
<strong>in</strong> a Little Higgs <strong>the</strong>ory. Ignor<strong>in</strong>g gauge <strong>in</strong>teractions and spontaneous symmetry<br />
break<strong>in</strong>g (SSB), G is <strong>the</strong> global symmetry of <strong>the</strong> Lagrangian. A subgroup F – larger<br />
than <strong>the</strong> SM gauge group I– is gauged, which breaks <strong>the</strong> global symmetry. Ano<strong>the</strong>r<br />
global subgroup H is left after SSB. A number of GBs make some of <strong>the</strong> F gauge<br />
bosons heavy, while <strong>the</strong> ones from I rema<strong>in</strong> massless. The rema<strong>in</strong><strong>in</strong>g GBs form <strong>the</strong><br />
Higgs.<br />
If ei<strong>the</strong>r <strong>the</strong> SM electroweak gauge bosons or <strong>the</strong> new gauge bosons couple to <strong>the</strong><br />
Higgs, <strong>the</strong>re will still be a leftover global symmetry, which makes <strong>the</strong> Higgs a Goldstone<br />
boson, i.e. massless, but if both gauge degrees of freedom couple to <strong>the</strong> Higgs a mass<br />
term is generated, which will <strong>the</strong>n only be logarithmically divergent.<br />
This is illustrated best on <strong>the</strong> basis of a model <strong>in</strong>troduced by Schmaltz, which he<br />
called <strong>the</strong> Simplest Little Higgs [39]. It will be even more simplified here, because<br />
we will ignore <strong>the</strong> U(1) gauge groups. Consider a Lagrangian with two scalar fields,<br />
<strong>in</strong>variant under <strong>the</strong> global symmetry G = SU(3)1 × SU(3)2 (ignor<strong>in</strong>g additional U(1)<br />
factors),<br />
L = (∂µφ1) † (∂ µ φ1) + (∂µφ2) † (∂ µ φ2) + V (|φ1| 2 , |φ2| 2 ) . (1.35)<br />
Both scalar fields will take on a vev f, so that <strong>the</strong> symmetry breaks down to H =<br />
SU(2)1 × SU(2)2 at that scale. A scheme of <strong>the</strong> symmetry break<strong>in</strong>g is provided <strong>in</strong><br />
Figure 1.6. We adopt <strong>the</strong> parametrization<br />
�<br />
i<br />
φ1 = exp<br />
f<br />
�<br />
i<br />
φ2 = exp<br />
f<br />
�<br />
02×2 κ<br />
κ † �<br />
+<br />
0<br />
�κ<br />
� �<br />
i<br />
√ �3×3 exp<br />
2f f<br />
�<br />
02×2 κ<br />
κ † �<br />
+<br />
0<br />
�κ<br />
� �<br />
√ �3×3 exp −<br />
2f i<br />
f<br />
�<br />
02×2<br />
h<br />
h<br />
† �<br />
+<br />
0<br />
η<br />
�<br />
√ �3×3<br />
2f ⎛ ⎞<br />
0<br />
⎝0⎠<br />
,<br />
f<br />
�<br />
02×2<br />
h<br />
(1.36)<br />
h<br />
† �<br />
−<br />
0<br />
η<br />
�<br />
√ �3×3<br />
2f ⎛ ⎞<br />
0<br />
⎝0⎠<br />
.<br />
f