The SM without the Higgs is not “mathematically inconsistent”. What is a problem is tree level unitarity – the theory becomes strongly coupled but it is not ill-defined. You just don’t have a small parameter to do a perturbative expansion any more (Like chiral perturbation theory close to the cut-off).
というもの。
renormalisableさんから
I think there is still one more important point being missed here which is to do with whether the theory is renormalisable. To highlight this lets look at two separate cases: with and without a Higgs boson.
With the higgs boson, the theory is renormalisable and so UV complete. However if the Higgs mass is too large the breakdown of tree unitarity signals that the theory becomes strongly coupled and so not calculable, but not inconsistent. All loop orders become relevant, keeping the theory unitary
Without the Higgs boson, the theory is indeed like a non-linear sigma model but this is non-renormalisable, So it is not a complete theory, only an effective theory up to a cut-off. The breakdown of tree unitarity in this scenario actually tells us where this cut-off is and above the cut-off we would need new physics which ultimately must be a renormalisable unitary theory.
The only caveat to the above is if the non-linear sigma model is asymptotically safe, in which case the theory could be non-perturbatively renormalisable and hence UV complete. Here the breakdown of tree unitarity again simply informs us when perturbation theory breaks down (strong coupling). This idea has received some attention recently by Roberto Percacci.
Anonymousさんからの指摘は、
返信削除The SM without the Higgs is not “mathematically inconsistent”. What is a problem is tree level unitarity – the theory becomes strongly coupled but it is not ill-defined. You just don’t have a small parameter to do a perturbative expansion any more (Like chiral perturbation theory close to the cut-off).
というもの。
renormalisableさんから
I think there is still one more important point being missed here which is to do with whether the theory is renormalisable. To highlight this lets look at two separate cases: with and without a Higgs boson.
With the higgs boson, the theory is renormalisable and so UV complete. However if the Higgs mass is too large the breakdown of tree unitarity signals that the theory becomes strongly coupled and so not calculable, but not inconsistent. All loop orders become relevant, keeping the theory unitary
Without the Higgs boson, the theory is indeed like a non-linear sigma model but this is non-renormalisable, So it is not a complete theory, only an effective theory up to a cut-off. The breakdown of tree unitarity in this scenario actually tells us where this cut-off is and above the cut-off we would need new physics which ultimately must be a renormalisable unitary theory.
The only caveat to the above is if the non-linear sigma model is asymptotically safe, in which case the theory could be non-perturbatively renormalisable and hence UV complete. Here the breakdown of tree unitarity again simply informs us when perturbation theory breaks down (strong coupling). This idea has received some attention recently by Roberto Percacci.
とコメントが入っている。なるほど。