What particles can the Z boson couple to in the standard model?

[jc09]

Hi just wondering if anyone could tell me why the Z boson can couple to any standard model particle except forgluons and photons?

 

[Vanadium 50]

It has no color charge and the gluon has no weak charge, so that's why it doesn't couple to the gluon. The photon is more subtle: the symmetry structure of the standard model prevents it.

 

[jc09]

ok thank you for your help

 

[kaksmet]

one could add that the Z boson and the photon are two orthogonal linear combinations of the same fields in the electroweak lagrangian, and because of their orthogonality they do not couple to each other.

A second way to see it is that the photon only couple to particles with charge, but the Z boson is neutral.

 

[Vanadium 50]

A second way to see it is that the photon only couple to particles with charge

That's often written, but it's not correct. A photon can couple magnetically to an uncharged object (like a neutron).

 

[GreyBadger]

That's often written, but it's not correct. A photon can couple magnetically to an uncharged object (like a neutron).

Hmm, in what limit? See, I would think of the neutron as made of udd (in valence terms), and therefore the photon 'coupling' to the neutron would actually be interacting with one of the constituent quarks. Of course, one could write a Fermi-like term for the interaction, but that's kind of cheating.

 

[Vanadium 50]

But before we knew about quarks, we knew the photon coupled to the neutron magnetically. There is no trouble with writing a magnetic coupling.

 

[tom.stoer]

But before we knew about quarks, we knew the photon coupled to the neutron magnetically. There is no trouble with writing a magnetic coupling.

In the context of the standard model the photon coupling to neutral particles is zero at tree level. The only way the photon couples to neutral particles (including itself and the Z boson) is via higher order corrections. The standard model does not contain a neutron as a fundamental degree of freedom so it's missleading to discuss couplings to fundamental particles and to effectice degrees of fredom on an equal footing.

 

[Vanadium 50]

This is where things get complicated - the issue of what can the Z do and what the SM says it does can easily get mixed up.

I maintain that one can write a photon's magnetic coupling to a neutral particle without any problem. It is true that the standard model has no neutral particles with magnetic moments, but that doesn't mean one cannot write down a term in the Lagrangian like this - the difference is between what can happen, and what does happen. I gave the pre-quark theory example of the neutron.

In fact one doesn't have to stop here - the magnetic coupling adds one derivative. If I add two, I get an electric quadrupole coupling. Three, magnetic octupole, and so on.

In the SM, however, these couplings are nonexistent. Indeed, as tom.storer said, they are zero at tree level, but it's stronger than that: they arise only through fermion loops. In a world with just W's, Z's and photons, the photon and Z would not couple at any order.

So the Z "can" couple to the photon in the sense that I can write down a legal Z-photon coupling. However, in the SM this piece is zero at tree level, and in fact, would be zero at all orders if it weren't for matter effects in loops.

 

[tom.stoer]

Agreed.