# Recently discovered ZZ diboson

1. Jan 17, 2009

### Dmitry67

Why 2 Z bosons interact (contrary to 2 photons)?
How do we know that we detected 1 ZZ boson, not 2 independent interactions with single Z bosons?

2. Jan 17, 2009

### bomanfishwow

I presume you are meaning something related to this measurement?

http://www.fnal.gov/pub/today/archive_2008/today08-08-21.html

Nobody has claimed that di-boson production implies a bound state. What is happening is illustrated by the feynman diagram here:

http://www.pd.infn.it/~dorigo/zzdecay.jpg

So you see, the initial fermion line (from quark anti-quark annihilation) radiates two Z bosons, which each in turn decay into whatever Zs can decay into. If each Z decays into the same thing (i.e. you get 4 electrons, 4 muons or 4 quarks), they are difficult to untangle. However, if you look for events where one Z decays to quarks, and the other decays to leptons (i.e. electron or muons, as taus look rather like quark jets), then you can find these events as quark jets and leptons are, generically speaking, detected in different subdetectors of an experiment. I have not mentioned neutrinos as you can't detect these at all (in a colliding HEP experiment at least)...

It is wrong to say that the Z bosons interact with themselves. The Z can, however, interact with the W+ and W-. As to why these gauge bosons can interact and the photon can't is a very interesting one. It boils down to the gauge structure of the underlying theory. The U(1) group of electromagnetism is an abelian group (meaning the generator - the fundamental description of the group - commute, i.e. G(1)G(2) = G(2)G(1)). However, in SU(2) x U(1), the electroweak group structure, the generators no longer commute (i.e. G(1)G(2) != G(2)G(1)). The non-commutation implies self interaction terms are allowed.

3. Jan 17, 2009

### Dmitry67

Thank you for the good explanation!

4. Jan 17, 2009

### daschaich

Actually, the photon can interact with the W+ and W- just like the Z can. The interaction is almost identical, except that the ZWW coupling has an extra factor of $$\cot\theta_w$$. Recall that in the electroweak theory, the U(1) of electromagnetism is a subgroup of the full SU(2) x U(1).

5. Jan 18, 2009

### bomanfishwow

Yes indeed, I was talking explicitly about Z/W interaction terms - I didn't want to get into the fact the U(1)s are different in each case!