Why is the Higgs boson spin 0?

Question

Why is the Higgs boson spin 0? Detailed equation-form answers would be great, but if possible, some explanation of the original logic behind this feature of the Higgs mechanism (e.g., "to provide particles with mass without altering their spin states") would also be appreciated.

It doesn't have to be spin-0 AFAIK, but the simplest prediction, the "standard model Higgs" just happens to be. Are you asking for an explanation of why the SM Higgs turns out to have zero spin, or were you thinking about more general requirements that would mean any Higgs would have had to be spinless? — David Z♦, Mar 31 '12 at 16:44
My intent was the latter: Were there drivers other than simplicity and Occam's razor at work back when Higgs was first postulated to be spin 0? Since no known examples of point-like spin 0 bosons exist, that seemingly trivial assumption actually steps a wee bit outside the boundaries of known physics. I wondered if that point might have driven any early exploration of more complicated Higgs options. For example, pion-like compositions of two spin 1 bosons comes to mind as the next simplest option if spin 0 is avoided. Hmm. Maybe I just need to do a good literature search on my own question?... — Terry Bollinger, Mar 31 '12 at 17:44
I think it came out of the calculation, but I don't remember the details - I'd have to look some things up to address that. As for Higgs models beyond the SM Higgs, someone else would have to address that. — David Z♦, Mar 31 '12 at 17:57
David, thanks. The recent LHC concerns hopefully have re-activated multiple "other options?" lines of Higgs thinking, so that may make my literature scan easier. — Terry Bollinger, Mar 31 '12 at 18:45

Luboš Motl · Accepted Answer · 2012-03-31 18:56:55Z

The Higgs boson is, by definition, the excitation of the field behind the Higgs mechanism. The Higgs mechanism is a spontaneous symmetry breaking. Spontaneous symmetry breaking means that the laws of physics, or the action

S , is symmetric with respect to some symmetry

G , i.e.

δ G S = 0

however, the vacuum state of the quantum field theory isn't symmetric under the generators of this symmetry,

G i | 0 ⟩ \neq 0

If we want to satisfy these conditions at the level of classical field theory, there must exist a field

ϕ such that the vacuum expectation value

⟨ ϕ ⟩ \equiv ⟨ 0 | ϕ (x) | 0 ⟩

isn't symmetric under

G ,

δ G ⟨ ϕ ⟩ \neq 0

However, if the field

ϕ with the nonzero vev had a nonzero spin, the vacuum expectation value would also fail to be symmetric under the Lorentz symmetry because particular components of a vector or a tensor would be nonzero and every nonzero vector or tensor, except for functions of

g μν and

ϵ λμνκ , breaks the Lorentz symmetry.
Because one only wants to break the (global part of the) gauge symmetry but not the Lorentz symmetry, the field with the nonzero vev has to be Lorentz-invariant i.e. singlet i.e. spin-zero

j=0 field, but it must transform in a nontrivial representation of the group that should be broken, e.g.

SU(2)×U(1) . The Standard Model Higgs is a doublet under this

SU(2) with some charge under the

U(1) so that the vev is still invariant under a different "diagonal"

U(1) , the electromagnetic one. The Higgs component that has a vev is electrically neutral, thus keeping the electromagnetic group unbroken, photons massless, and electromagnetism being a long-range force.
Aside from the Higgs mechanism, there exist other, less well-established proposed mechanisms how to break the electroweak symmetry and make the W-bosons and Z-bosons massive. They go under the names "technicolor", "Higgs compositeness", and so on. The de facto discovery of the 125 GeV Higgs at the LHC has more or less excluded these theories for good. The Higgs boson seems to be comparably light to W and Z-bosons and weakly coupled, close to the Standard Model predictions, and the Higgs mechanism sketched above has to be the right low-energy description (up to energies well above the electroweak scale).

Hans de Vries · Answer 2 · 2012-03-31 20:47:45Z

Just to highlight one simple but important argument.
If Higgs is to be responsible for giving particles mass then it has to be a scalar (spin-0) particle because a particle's mass is reference frame independent, just like the values of the field

ψ of a spin-0 particle are reference frame independent.
Compare this for instance with charged particles at rest which gain potential energy in a static electromagnetic potential field

A o (which is the same anywhere). As long as you stay in the rest frame then it looks like the particles have gained an extra mass

qA o . However, as soon as you change to another reference frame then the illusion of extra mass breaks down. The reason it breaks down is that

A μ is a spin-1 field which therefor transforms like a vector going from one reference frame to another.
Hans.

@LubošMotl, Hans, my thanks to both of you. These are both superb explanations, very much of the type for which I was hoping. I will attempt to Accept both of you answers, Luboš first for the level of detail enabling deeper study, then Hans for his deft highlighting of a nicely critical point. If I can only do one, Hans, my intent was there even if the software was uncooperative. [And I see I really am a newbie: Stack Exchange would not let me accept either answer yet...!] — Terry Bollinger, Mar 31 '12 at 22:32

phymath999

Tuesday, December 29, 2015

hartree iteration ab initio 量子化学的概念和方法第三章計算化學理論方法簡介 Planck h (普朗克常數) 來表示相空間運動的一個量子

固体物理第四章总结_图文_百度文库

[PDF]R

[PDF]第三章計算化學理論方法簡介

ab initio计算的理论[转]_Coffee_新浪博客

[转]ab initio计算的理论- Happy的日志- 网易博客

[PDF]C

第一性原理是什么？ - 第一原理- 小木虫- 学术科研第一站

[PDF]量子化学的概念和方法 - 中国科学院上海有机化学研究所

结构化学 - 精品课程 - 牡丹江师范学院

第五章固体电子论_图文_百度文库