# Higgs Boson: The Big Picture

First, please pardon the ignorance behind this question. I know a fair amount of math but almost no physics.

I'm hoping someone can give me a brief "big picture" explanation of how physicists were able to predict the existence of the Higgs boson. Here's a (perhaps completely wrong) example of the level of explanation I'm hoping for:

"Consider the following group [insert precise specification of a particular group here]. This group is know to have precisely [insert number here] irreducible representations. All but one of these representations corresponds to a previously-observed particle via a correspondence in which the property [insert mathematical property of representation here] corresponds to the property [insert physical property of particle here]. The Higgs Boson is the particle that corresponds to the remaining representation."

If I happen to have hit on the right story, I'd like someone to fill in the blanks. If, as is more likely, I've concocted a completely wrong story, I'd like someone to give me a more accurate story at about the same level of sophistication. Thanks!

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I tried to fill out your 'Mad Lib' style paragraph. But ran into a problem. The group in question is SU(2)xU(1), but not all (infinite number of) representations are realized in nature. I tried filling it out in my answer below. –  QuantumDot Aug 25 '12 at 23:39
This video lecture by Prof. Susskind could be useful. –  user10001 Aug 26 '12 at 0:29

No, it doesn't work like that. The Higgs boson doesn't complete a set of particles that we had some theoretical reason to expect to exist. (Other particles have been predicted in roughly that way, e.g. the charm and top quarks.) So in the sense I believe you're thinking about it, physicists had no reason to predict the existence of the Higgs boson.

Where it actually comes from is the fact that particles have mass. The Lagrangian (if you're not familiar with what a Lagrangian is, let me know, I'll edit in an explanation) of the standard model is supposed to be invariant under the gauge group $SU(3)\times SU(2)\times U(1)$, but if you explicitly include terms to make the particles have mass, they break that gauge invariance. The Higgs mechanism allows the mass terms to spontaneously appear out of a change of coordinates, and it turns out that when they appear in this way (as opposed to being explicitly added in), the gauge invariance is preserved. And if the Higgs mechanism is going to happen, it necessarily calls for the existence of at least one more particle, namely the Higgs boson.

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and, more importantly, renormalizability is preserved! –  Arnold Neumaier Aug 26 '12 at 11:20
Your blog post is very helpful. Thank you. –  WillO Aug 27 '12 at 14:41

With a few corrections, I think the paragraph should read:

"Consider the following group SU(2)xU(1). This group is known to have an infinite number of irreducible representations. All but most Some of these representations corresponds to a previously-observed particle via a correspondence in which the property Quadratic Casimir eigenvalues corresponds to the property (of) Weak-isospin and hypercharge. The Higgs Boson is the particle that corresponds to the remaining representation weak isospin-1/2 and hypercharge 1/2 ."

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