# Why is boson spin number related to attraction and repulsion?

The accepted answer to this question says

Since the electroweak interaction is mediated by spin 1 bosons, it is the case that "like (charge) repels like and opposites attract".

The Higgs is spin - 0 (scalar field) and the graviton is spin - 2; attractive.

a force mediated by a spin-0 scalar is always attractive.

So what's going on here? Why does the spin of the boson determine whether the force is universally attractive or follows the opposites-attract-and-like-repels-like pattern?

What would it look like for a spin-3 or spin-4 boson, if such a particle existed?

• what about the gluon? spin one always attractive – anna v Oct 10 '18 at 14:06
• "What would it look like for a spin-3 or spin-4 boson, if such a particle existed?" - See Q & A here: Why do we not have spin greater than 2? – Hal Hollis Oct 10 '18 at 16:51

For bi-linear interactions it is simply a fact that the algebra of Dirac $$\gamma$$ matrices leads to uniform attraction for spin 0 boson exchange and attraction/repulsion for unlike/like charges for spin 1 boson exchanges. The nonlinear structure of QCD may alter this relationship for gluon exchanges as pointed out in a comment by @annav. I don't remember the details of the derivations from my graduate school days 50 years ago, but I do remember that the final result was clear and convincing.
Consider two models. One involves matter particles interacting with a spin 0 (scalar) field $$\phi$$. The other involves matter particles interacting with a spin 1 abelian gauge field $$A_\mu$$. The Lagrangians are $$L = \frac{\big(\dot\phi(x)\big)^2 -\big(\nabla\phi(x)\big)^2}{2} - \phi(x) J(x) + \cdots$$ and $$L = -\frac{1}{4}F_{\mu\nu}F^{\mu\nu}-A_\mu J^\mu+\cdots$$ with $$F_{\mu\nu}=\partial_\mu A_\nu-\partial_\nu A_\mu$$. The operators $$J$$ and $$J^\mu$$ are constructed from the matter field, which I'll be more explicit about later. The dots denote the kinetic terms for the matter field. The signs of the coupling terms $$\phi J$$ and $$A_\mu J^\mu$$ are not important: they can be chosen as desired by choosing the sign-convention for the boson fields. However, the signs of the kinetic terms for $$\phi$$ and $$A_\mu$$ are important: they must be chosen so that the energy (the Hamiltonian) has a finite lower bound. This positive-energy requirement is the key to explaining the connection between the spin of the boson and the sign of the force that it mediates.
Now consider situations in which the motion of the matter and the time-dependence of the boson fields are both negligible. This simplifies things in a few ways: (1) we can neglect all terms involving time-derivatives, (2) the temporal component $$J^0$$ is the only significant component of the current, and (3) with the help of a gauge-fixing choice, the temporal component $$A_0$$ can be regarded as the only significant component of the gauge field (as in electrostatics). After these simplifications, the Hamiltonians reduce to $$H = \int d^3x\ \left(\frac{ \big(\nabla\phi(x)\big)^2}{2} + \phi(x) J(x) + \cdots\right)$$ and $$H = \int d^3x\ \left(\frac{ \big(\nabla A_0(x)\big)^2}{2} + A_0(x) J^0(x) + \cdots\right),$$ and the equations of motion for the boson fields reduce to $$\nabla^2\phi(x)=J(x) \hskip2cm -\nabla^2 A_0(x) = J^0(x).$$ Notice the opposite signs here. Using the same Green function $$G(x-y)$$ in both cases, we can express the boson fields in terms of $$J$$ and $$J^0$$ like this: $$\phi(x) = \int dy\ G(x-y)J(y) \hskip2cm A_0(x) = -\int dy\ G(x-y)J^0(y).$$ Substitute these back into the Hamiltonian to see that the energy in the spin 0 case is $$H = \frac{1}{2}\int d^3x\ J(x)G(x-y)J(y) + \cdots$$ and that the energy in the spin 1 case is $$H = \frac{-1}{2}\int d^3x\ J^0(x)G(x-y)J^0(y) + \cdots$$ Again, notice the opposite signs. Given an explicit expression for the Green function, we can determine the sign of the force between particles of matter by looking at how the energy varies with the distance $$|x-y|$$. Or, since we already know that like charges repel each other in electrodynamics, we can just compare the spin-0 and spin-1 results to conclude that like charges must attract each other in the spin-0 case.
What about opposite charges? If the matter field is a spin-1/2 fermion field, then we have $$J\propto \overline{\psi}\psi = \psi^\dagger\gamma^0\psi \hskip2cm J^0\propto \overline{\psi}\gamma^0\psi = \psi^\dagger\psi.$$ If we define $$\psi_\pm = \frac{1\pm\gamma^0}{2}\psi,$$ then $$J\propto\psi^\dagger_+\psi_+ - \psi^\dagger_-\psi_- \hskip2cm J^0\propto\psi^\dagger_+\psi_+ + \psi^\dagger_-\psi_-.$$ Use the fact that fermion fields are anticommutative to get $$J\propto\psi^\dagger_+\psi_+ + \psi_-\psi^\dagger_- \hskip2cm J^0\propto\psi^\dagger_+\psi_+ -\psi_- \psi^\dagger_-.$$ I didn't write the kinetic terms for the matter field, but thanks to the positive-energy requirement, we can (schematically!) identify the first term in each of these expressions as the charge density of the particle, and the second term as the charge density of the corresponding antiparticle. Now, substitute these expressions for $$J$$ and $$J^0$$ back into the previous expressions for the energy. The fact that both terms in $$J$$ have the same sign means that in the spin 0 case, the sign of the force is the same for particle-particle, particle-antiparticle, and antiparticle-antiparticle. The fact that the two terms in $$J^0$$ have the opposite signs means that in the spin 1 case, the sign of the force for particle-antiparticle is opposite the sign of the force for particle-particle or antiparticle-antiparticle.