15
$\begingroup$

My understanding is that the exchange of Z bosons could yield an attractive or repulsive "force" between two fermions. For most combinations the electromagnetic or strong interactions will take place.

However, if we have two neutrinos, or a neutrino and a charged lepton, only the weak interaction comes into play. Are these possible bound states?

$\endgroup$
  • 1
    $\begingroup$ In one dimension potential well, however shallow, will always have a bound state. But in 3D this is not so: if the well is too shallow, there will be no bound states. So weak force, which falls exponentially with distance, would be too weak for bound states between neutrinos. However, there could be yet unknown WIMP's, for them the situation could be different. $\endgroup$ – user23660 Dec 6 '13 at 9:09
19
$\begingroup$

The interaction mediated by a massive boson will produce interaction potential of Yukawa type: $$ V(r) = - \frac{\alpha_\chi}{r} e^{-mr}, $$ As is well known, a scalar exchange leads to a universally attractive potential, whereas a vector exchange will attract particles to antiparticles, but repel pairs of particles or pairs of antiparticles. For Z-boson eхchange $\alpha_\chi=g^2/4\pi$, $g$ is gauge coupling constant.

The existence of the bound states of two particles of mass $M$ in this potential (assuming $\alpha_\chi>0$) could be established in term of 'Bohr radius' $a$:

$$ a^{-1} = \frac {\alpha_\chi M}{2} $$

Comparing this radius with the range of interaction $m$ yields dimensionless parameter:

$$ D =\frac{1}{am} = \frac {\alpha_\chi M}{2 m}. $$

In order to have a bound state at all (the 1$s$ state), the potential must satisfy $D \gtrsim 0.8 $. So for a bound state through a weak force mass of the particle should be comparable (or greater) than mass of weak boson.

So for neutrino (or neutino + lepton) the mass $M$ is too small to allow any bound states.

However, various extensions to Standard model consider the possibility of additional weakly interacting massive particles: WIMP's, with masses much greater than that of neutrino, and which could constitute (part of) dark matter. So this particles potentially could form bound states based on weak force: WIMP-onium.

This states has been considered in literature. Take for instance paper:

Shepherd, William, Tim MP Tait, and Gabrijela Zaharijas. "Bound states of weakly interacting dark matter." Physical Review D 79.5 (2009): 055022. arXiv:0901.2125.

From the abstract:

We explore the possibility that weakly interacting dark matter can form bound states—WIMPonium. Such states are expected in a wide class of models of particle dark matter, including some limits of the minimal supersymmetric standard model. We examine the conditions under which we expect bound states to occur and use analogues of nonrelativistic QCD applied to heavy quarkonia to provide estimates for their properties, including couplings to the standard model. We further find that it may be possible to produce WIMPonium at the LHC and explore the properties of the WIMP that can be inferred from measurements of the WIMPonium states.

$\endgroup$
  • $\begingroup$ Yes, the D ratio is unforgiving, cf. Yongyao et al , and e-nu bound states won't help, as the reduced mass will again be the neutrino mass. What makes nuclear physics work is the nucleons being larger in mass , m than the pions, M. $\endgroup$ – Cosmas Zachos Feb 26 '16 at 13:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.