Why is the Higgs boson spin 0? 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.
 A: 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.
$$\delta_G S = 0$$
however, the vacuum state of the quantum field theory isn't symmetric under the generators of this symmetry,
$$ G_i|0\rangle \neq 0$$
If we want to satisfy these conditions at the level of classical field theory, there must exist a field $\phi$ such that the vacuum expectation value 
$$\langle \phi\rangle \equiv \langle 0 | \phi (x)| 0 \rangle$$ isn't symmetric under $G$,
$$\delta_G \langle \phi\rangle \neq 0 $$
 However, if the field $\phi$ 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 $g_{\mu\nu}$ and $\epsilon_{\lambda\mu\nu\kappa}$, 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. $SU(2)\times U(1)$. 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).
A: 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 $\psi$ 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 $A^\mu$ is a spin-1 field which therefor transforms like a vector going from one reference frame to another.
Hans.   
A: Because Hogs Boson are only related with any particle mass. Not with angular moment or dipole moment. So it is needless for it to have any spin. and particles mass is never changed whether it rotate (spin) or not. So Hgs Boston need not any spin value.
