What are the basic ingredients of the standard model (SM) and their applications?
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$\begingroup$ In what sense would you have the standard model, without the assumptions of the standard model? What does that mean? ... Anyway, it all follow from the choice of symmetry groups and their representations, working in the context of relativistic quantum field theory. $\endgroup$– Mitchell PorterCommented Aug 3, 2019 at 12:45
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$\begingroup$ Thank You. I want to list out all the essential requirements like gauge invariance and symmetry breaking. Anything more? $\endgroup$– TwinklestarCommented Aug 3, 2019 at 12:50
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$\begingroup$ The SM explains experimental results. That has implications, but I’m not sure what you mean by assumptions. Could you say a bit more, for example why you think symmetry breaking is an assumption? $\endgroup$– Bob JacobsenCommented Aug 3, 2019 at 16:43
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$\begingroup$ You may want to reduce your question to something a bit less broad. I mean, if you ask something like, "What assumptions is the Standard Model (SM) based on?", that'd be a pretty big question already. Probably best not to add anything else to that, e.g. also asking for applications. $\endgroup$– NatCommented Aug 5, 2019 at 13:22
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1 Answer
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The basic ingredients that make up the standard model are the following :
(i) A local (gauge) symmetry $SU(3)\times SU(2)\times U(1)$
(ii) Three fermion generations , consisting of the following representations: $$Q_{L_i}(3,2)_{\frac{1}{6}},U_{R_i}(3,1)_\frac{2}{3},D_{R_i}(3,1)_{-\frac{1}{3}},L_{L_i}(1,2)_{-\frac{1}{2}},E_{R_i}(1,1)_{-1}\ \ \ \ \ i=1,2,3$$ where the $i$ represents the three generations, the numbers in the parentheses $(X,Y)_Z$ denotes the fact that the field sits in the $X,Y,Z$ representations of the $SU(3),SU(2),U(1)$ symmetry groups respectively, $L,R$ denotes the chirality.
(iii) A single scalar multiplet $\phi(1,2)_\frac{1}{2}$
(iv) It exhibits the following SSB, $$SU(3)\times SU(2)\times U(1)\rightarrow SU(3)\times U(1)_{EM}$$
Now, once you bring in the gauge fields and write the most general renormalisable Lagrangian, you will end up with the SM Lagrangian, after SSB.
