Does the weak nuclear force play a role (positive or negative) in nuclear binding?

Normally you only see discussions about weak decay and flavour changing physics, but is there a contribution to nuclear binding when a proton and neutron exchange a $W^\pm$ and thus exchange places? Or do $Z$ exchanges / neutral currents contribute?

Is the strength of this interaction so small that it's completely ignorable when compared to the nuclear binding due to residual strong interactions?

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    $\begingroup$ Well, intuitively, even repulsive EM force between protons is weak compared to the attractive nuclear force that holds nucleus together. So it should not be surprising that weak force has even much lesser say. Of course there will be some contribution but probably completely negligible one. $\endgroup$ – Marek Jan 3 '11 at 11:20
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    $\begingroup$ This is a GREAT question. Its answer is not obvious to me and it could shed light on beta decay - which is mediated by the weak force. As for the relative strengths of the various forces, keep in mind that when in a strong-field regime, as inside a nucleus, such distinctions can break down. At high-energies the strong force becomes weaker and the weak force becomes stronger :) $\endgroup$ – user346 Jan 3 '11 at 19:57

A back of the envelope calculation suffices, meaning all factors of $2,3, \pi$ etc. have been ignored.

The residual strong force is mainly due to pion exchange and can be modeled by this plus a short range repulsion due to exchange of $\omega$ mesons. The Yukawa potential from pion exchange is $e^{-m_\pi r}/r$. The weak interactions arise from $W^\pm$ and $Z$ exchange and will give rise to a similar potential with $m_\pi$ replaced by $m_Z$ (in the spirit of the calculation I take $m_W \sim m_Z$). At typical nuclear densities of $.16/(\rm fermi)^3$ nuclei are separated by distances which are within a factor of two of $1/m_\pi$. Thus the ratio of the Yukawa potential due to weak exchange to the Yukawa potential due to pion exchange at these densities is roughly $e^{-m_W/m_\pi} \sim e^{-640}$. I've ignored the fact that there are different coupling constants in front of the two potentials, but this difference is irrelevant compared to the factor of $e^{-640}$. So yeah, you can ignore the weak contribution to the binding energy.

  • $\begingroup$ Nice! You've made it look simple enough that I'm kicking myself for not having thought of that. $\endgroup$ – Simon Jan 4 '11 at 2:32
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    $\begingroup$ @Harvey: While the ratio of the weak to the strong force between two nucleons is $e^{-640}$, there's still the weak force acting upon a single nucleon as in beta decay. If the resulting nucleus is unstable due to beta decay, this would show up as a broadening of its mass spectrum with a Breit-Wigner distribution. The contribution is still small, but is no longer that exponentially suppressed. $\endgroup$ – QGR Jan 13 '11 at 23:07
  • $\begingroup$ On other hand, short-range potentials are very peculiar when we look to semiclassical solutions. For each bound state inside the potential range there is another state far outside, is there? This is because as we increase the coupling (not the range) the states appear paired in the S-matrix. Yep, of course the outside states have an almost null energy. $\endgroup$ – arivero Sep 25 '15 at 1:38
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    $\begingroup$ Is that right? e^a / e^b is equal to e^(a-b) not e^(a/b). $\endgroup$ – zooby Sep 27 '15 at 12:21

This question is experimentally accessible, despite the feebleness of the weak interaction, because the strong and electromagnetic interactions are symmetric under parity transformations and the weak interaction is not.

The contribution to the binding energy is small enough that it's not a good way to think of things. Better is to continue the process of trying to describe nuclear energy eigenstates as linear combinations of different spin-orbit states. For instance, the deuteron ground state has isospin zero and spin, parity $J^P=1^+$, and so must be a linear combination of the even-$L$ spin triplets $\left|{}^3S_1{}^{T=0}\right>$ and $\left|{}^3D_1{}^{T=0}\right>$; the d-wave component famously contributes about 4% of the wavefunction and was the first evidence for the tensor nature of the nuclear force. But because the weak interaction contributes to the nuclear interaction, the ground state isn't an exact eigenstate of the parity operator (or, for that matter, of isospin) and there's a little bit of p-wave mixed in: $$ \left|\text{deuteron}\right> = \sqrt{0.96}\left|{}^3S_1{}^{T=0}\right> + \sqrt{0.04}\left|{}^3D_1{}^{T=0}\right> + \epsilon_0\left|{}^3P_1{}^{T=0}\right> + \epsilon_1\left|{}^1P_1{}^{T=1}\right> $$

In the formation of deuterium by neutron capture on hydrogen, you get interference between parity-allowed capture to the $S$- and $D$-wave states and parity-forbidden capture to the $P$-wave states. These interferences manifest as asymmetries or spontaneous polarizations in the photons emitted during capture which are more or less linear in the amount of $P$-wave mixing; typical asymmetries are a few parts per billion.

In heavier nuclei (e.g. helium & beyond) you lose the luxury of a ground-state wavefunction which can be described in a paragraph, or even at all. However, a perturbation-theory way of describing the influence of the weak interaction is to say that a particular physical eigenstate with, say, positive parity $\left|\psi^+_\text{physical}\right>$ will be mostly given by a strong-force eigenstate with definite parity, but contain contributions from nearby opposite-parity states due to the weak interaction: $$ \left| \psi^+_\text{physical} \right> = \left| \psi^+ \right> + \sum_i \left| \psi^-_i \right> \frac{ \left< \psi^-_i \middle| H_\text{weak} \middle| \psi^+ \right> }{ E_i - E_+ } $$

In heavy nuclei with a dense forest of excited states, you sometimes find same-spin, opposite-parity states which have very different lifetimes and very similar energies; these states are prime candidates to exhibit parity mixing due to the weak interaction. There's a famous excitation in lanthanum which decays by emitting photons with a 10% parity-forbidden directional asymmetry.

Microscopically, your other answers are correct that the nucleus is too large and the overlap between nucleons too small for appreciable exchange of $W$ and $Z$ bosons. But you can of course say the same thing about nucleons and exchange of gluons. The effective theory of the weak interaction between nucleons models the nuclear force as an exchange of strong mesons (the $\pi,\rho,\omega$) where each nucleon-nucleon-meson vertex with a given set of quantum numbers has a particular parity-nonconserving amplitude. (There was some effort a few years ago to move into the twenty-first century and come up with an "effective field theory" which described the nucleon-nucleon weak interaction without mesons; a big pile of work seems to have produced a one-to-one relationship between the coupling constants in the modern effective field theory and the coupling constants in the old meson theory.)

This has been a pretty long-winded preparation for my answer to your question: the contribution of the weak interaction to the energy of any particular nuclear state is pretty small, for the same reason that the Coulomb-force contribution to the energies of light nuclei can generally be neglected. What's more interesting is to try an use the short-range nature of the weak interaction to peek at high-energy physics hiding inside of stable nuclei.

  • $\begingroup$ @arivero Re your bounty question, I wouldn't consider the charged current in nuclear beta decay terribly different from electromagnetism in the transitions of nuclei where isospin is a good symmetry: the interaction isn't very important to the steady-state nuclear dynamics (c.f. mirror nuclei) but its bosons may be (virtually) emitted in strong transitions. I suspect, but haven't checked, that a shell-model picture of beta decay should typically have a single nucleon which changes flavor and orbital, like an electronic transition can sometimes be attributed to a transition of "an" electron. $\endgroup$ – rob Sep 23 '15 at 5:54
  • $\begingroup$ but there is a fundamental difference with electromagnetism, namely the mass of W, that allows to ignore the momentum in the propagator. $\endgroup$ – arivero Sep 23 '15 at 17:34

From this blog post of mine, one should be inclined to think that for large nuclei it can at least contribute to the spin-orbit force, and then to the correction to N=50 and N=82 nuclear shells.

As QGR noted above, the exponential suppression of the potential does not need to be all the history. Note that usually the total potential in a nucleus is Woods-Saxon, where the exponential does not factor out.

Independently of particle mass, a low-momentum exchange is able to see all the nucleus; when a beta decay happens, the electron carries a momentum of less than 100 MeVs and then it is delocalized across all the nucleus. This is the same that when a electronic transition happens in an atom: the photon carries a momentum of only some electron-Volts, and then it is delocalised across all the orbital.

Consider that a typical contribution to nuclear shells must be about 2 or 3 MeVs. Most (well, a lot) of the nuclear shell correction comes from exchange of $\omega$ and other mesons, which have a mass around the 750 MeV range. Quotient by W mass, and we could expect linear corrections of the order of a 1%.

So, if some collective effect can do a linear correction to appear, it could be more or less in the range to be noticeable: (750 MeV)^2/81GeV=6.9 MeV.

To look for collective effects was the idea of the post I have referred in the first line: check the favored masses when the "liquid drop" nucleus splits in two pieces. The small fragment of fission channel S3 happens to have an average mass of 79.21 $\pm$ 1.14 GeV across a range of nuclei from 233Pa to 245Bk; the small frament of channel S2 happens to have an average mass of 92.34 $\pm$ 2.91 GeV.

  • $\begingroup$ And yes, there is an S1 channel, and yes, its average is the value of a boson of the electroweak model. $\endgroup$ – arivero Sep 18 '15 at 20:43

The answer is, there is a very small affect. If you take a look at the Wikipedia article, you will note that the weak force is about 10^-13 weaker then the strong force. Therefore it's just not going to have much of an affect.

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    $\begingroup$ I think that 10^-13 is compared to the strong force that binds quarks, not to the residual strong force that binds nuclei... $\endgroup$ – Simon Jan 3 '11 at 21:02

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