In the spirit of a related inquiry, I would like to know if there's a basis for understanding why there aren't any elementary particles that have non-zero electric charge but zero spin?

Can such a quantum theory be written down and self-consistent? Do the current symmetries of our present QFTs not allow such a particle?

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    $\begingroup$ For what it's worth, there are composite particles that with these properties, such as the charged pion. $\endgroup$
    – BMS
    Sep 11 '14 at 18:45

The standard model is very successful in its group structure in ordering all observed particles. To introduce a particle with charge and zero spin, you will need a different model that would also accommodate the symmetries observed experimentally and fitted by the standard model. So the answer to "why" is "because" we have not seen any and can model well what we have seen.

That said, when one goes to string theories and the necessary supersymmetric structures where the known from experiments elementary particles are doubled in number we have the squarks which are zero spin and charged. There are a number of sfermions with the same signature, selectrons, smuons etc.

In particle physics, a sfermion is the spin-0 superpartner particle (or sparticle) of its associated fermion. In supersymmetric extensions to the Standard Model (SM) each particle has a superpartner with spin that differs by 1⁄2. Fermions in the SM have spin-1⁄2 and therefore sfermions have spin 0.

As we have not seen them, as I explained above, the super symmetry is hypothesized as a broken symmetry, which means we will see signatures of these elementary particles with very high masses. The LHC has put limits of order of TeV for the masses.

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    $\begingroup$ +1 It should be noted there seems to be no no-go on charged spin-zero particles as this corresponds to the complex scalar field. It is actually surprising that the Standard model does not include such a particle. I think the most "directly" searched for charged scalars would be the charged Higgs bosons. $\endgroup$
    – Void
    Sep 11 '14 at 19:22
  • $\begingroup$ anna v: any chance you could briefly describe the groups to which you refer, or at least name them? $\endgroup$
    – BMS
    Sep 11 '14 at 19:41
  • $\begingroup$ @BMS The standar model is built around the symmetries of SU(3)xSU(2)xU(1), SU Special Unitary Group of order (3 strong interactions,2 electroweak ), U(1) Unitary group (1 electromagnetic), en.wikipedia.org/wiki/… . These symmetries were arrived at by studying the data . $\endgroup$
    – anna v
    Sep 12 '14 at 3:35

The Higgs is part of a complex scalar doublet in the standard model. It carries both hypercharge and weak charge. So we have discovered charged scalars.

Now perhaps you are only interested in ELECTRIC charge. So does the Higgs doublet carry this? Well, once the Higgs picks up a vev, then some parts do and some parts don't. The parts that do carry electric charge are said to be "eaten" by the W bosons, and indeed they carry electric charge. While there is an electrically neutral part which is related to the Higgs boson.

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    $\begingroup$ That is true. But It is important to reinforce that if we had two scalar doublets, then a charged Higgs would also exist. $\endgroup$ Sep 12 '14 at 20:33

The Lagrangian $$\mathcal L = -\frac{1}{4}F_{\mu\nu}F^{\mu\nu} + \mathcal L_\textrm{free} + eA_\mu J^\mu \tag{1}$$ where $A_\mu$ is the 4-potential, $F_{\mu\nu} = \partial_{[\nu}A_{\mu]}$ is the field tensor, $\mathcal L_\textrm{free}$ describes fields other than $A_\mu$, and $J^\mu$ is the 4-current density expressed in these other fields, describes a QED-like theory. When $\mathcal L_\textrm{free}$ describes a free Dirac field $\psi$ and $J^\mu = \overline\psi\gamma^\mu \psi$, it is precisely QED. The Dirac field has spin $\frac 1 2$

We can instead take $$\mathcal L_\textrm{free} = \frac{1}{2}\big( (\partial_\mu a)(\partial^\mu a^\dagger) + m^2 aa^\dagger)$$ with $$J^\mu = i(a\partial^\mu a^\dagger - (\partial^\mu a)a^\dagger). \tag{2}$$ The field $a$ describes spin $0$ particles.

The theory described by (1) is as self-consistent as QED is, that is, it is renormalizable. This is because the necessary and sufficient ingredient in the renormalizability of QED is that the constant $e$ is dimens.ionless (in natural units). With $J^\mu$ according to (2), this is the case.


The Heisenberg Uncertainty principle forbids it.

Just as all Quantum Particles cannot have less than zero-point energy, for spin nothing can have Angular Momentum of less than 1/2 in units of h-bar.

The commutation relation for Angular Momentum is [L,Lz] >= h/2π

( apologies for the poor maths notation )

  • $\begingroup$ This does not seem right. There do exist particles which have zero spin: the Higgs boson, pions. How would you explain their existence? $\endgroup$
    – mpv
    Mar 7 '16 at 16:06
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    $\begingroup$ While $[\vec S, S_z] = i\hbar \vec e_i \epsilon_{izj} S_j$, this does not imply a uncertainty relation which forbids spin 0 particles (this implies an inequality $\sigma_{S_x}\sigma_{S_z} \ge \frac \hbar 2 \left|\left<S_y\right>\right|$, for $S_z$ eigenstates the rhs is zero. Spin 0 means $S^2 \left|\psi\right> = 0$. The relation $[S^2, S_z] = 0$ holds in general, so there are states, that have definite spin qunatum-number $s$ ($s$, such that $S^2\left|\psi\right> = \hbar^2 s(s+1) \left|\psi\right>$ and $S_z\left| \psi \right> = \hbar m \left|\psi\right>$, $-s \le m \le s$) also for $s = 0$. $\endgroup$ Mar 7 '16 at 16:08
  • $\begingroup$ @mpv firstly, the Higgs Boson has a tentative spin of 0, it has not yet been measured , and due to extremely short lifetime unlikely to happen for a long time. Secondly the Higgs Boson could be a composite particle like para-positronium, ( a 'binary star' of positron and electron, with anti-parallel spins ). Neither Quantum Theory nor the Uncertainty Principle forbid a composite particle having zero overall spin. The same is true for π° , composed of quark-antiquark. $\endgroup$ Mar 9 '16 at 14:18
  • $\begingroup$ Also the question is on charged particles. Now speculating on a link between charge and angular momentum myself. $\endgroup$ Mar 9 '16 at 14:31

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