Different cross section for W and Z boson

At hadron colliders the cross section for W production is about ten times larger than the production cross section for Z bosons (e.g. Figure 2 in this review article). I guess the dominant contribution is from Drell-Yan production so let's only consider this.

A factor of two probably comes from the fact, that there is a positive and a negative W. But what about the remaining difference of factor 5?

• The $W$ can participate in both charged- and neutral-current channels, but the $Z$ can only participate in the neutral current.
– rob
Commented Jun 22, 2014 at 16:50
• @rob: that I don't understand... aren't neutral currents represented by the exchange of a Z and charged currents represented by the exchange of a W? Commented Jun 22, 2014 at 18:07
• Oops, you are of course right. It may be that in collisions between mostly-positive hadrons there are fewer neutral-current channels available. But would suggest you'd see a smaller difference in $p\bar p$ collisions, which is apparently not the case. Interesting question!
– rob
Commented Jun 22, 2014 at 18:27
• Have you seen this article arxiv.org/abs/ARXIV:1402.0923 and references therein? Commented Jun 23, 2014 at 9:17

The Figure 2 of this paper

http://arxiv.org/pdf/hep-ph/0611148v1.pdf

doesn't show a factor-of-ten difference at all! Extract the ratio properly on the log scale and you will see it is less than four, just slightly greater than 1/2 of the height corresponding to the decade. Note that 1/2 of the weight corresponds to the factor of $\sqrt{10}\sim 3.16$.

The ratio of the cross sections is close to four rather than two mostly because of the Weinberg angle $\theta_W$. The production of neutral $W^0 W^0$ boson pair would have cross section close to one-half of the inclusive charged $W^+W^-$ cross section. However, a $W^0$ only contains $\cos\theta_W$ times $Z^0$, and this must be used for both copies of the $W^0$. So some of the processes produce $\gamma Z$ or $\gamma\gamma$. Only the fraction $\cos^2\theta_W$ produces $Z^0 Z^0$ and $\cos^2\theta_W\sim 1-0.23\sim 0.77$ is a new factor in the amplitude.

Note that we are neglecting the production of $B^0 B^0$ – which also splits to four contributions $ZZ,Z\gamma,\gamma\gamma$ – because it's suppressed by $g_Y^2$ and it's much smaller than the coupling constant for the $SU(2)$.

So the estimated ratio of the $W/Z$ inclusive cross sections is $2/0.77^2\sim 3.37$, very close to what the graph shows. I had to square the amplitude to get the probability which is why the factor $1/0.77$ appeared twice.

Other potential asymmetries that raise $W^+W^-$ relatively to $Z^0Z^0$ probably include the fact that the different $u\bar d/\bar u d$ quarks are more likely to be found inside the hadrons than $u\bar u$ and $d\bar d$. Also, if a $Z$ appears in a propagator, the diagram has a greater (in the $s$-channel) suppression $m_Z^2$ in the denominator than $m_W^2$.

• Thanks for the clear answer! However, there are two points which I don't understand: 1) e.g. the experimental result quoted by @DarioP gives a factor of 10 (maybe due to restrictions in the phase space?), 2) why are you only considering pairs of bosons in your argumentation and not single bosons? Commented Jun 25, 2014 at 7:08
• Hi, good questions - answers, sorry, in both cases it's because of my limitation. I don't know how to run a symmetry argument that would relate the inclusive production of a single W to a single Z because a single W/Z is (approximately) a triplet but the remaining hadrons that must be included in the final state aren't necessarily electroweak triplets etc. I don't think that you or someone else has presented a valid argument that the ratio of cross sections for single W/Z bosons can't be 10, for example, and I think that those are difficult calculations. Commented Jun 25, 2014 at 7:40
• (I know, 2.5 years later.) About 1), if you factor out the BR from the CMS result, you'd get $\sigma(W)$:$\sigma(Z) \sim$ 3:1? Commented Feb 13, 2017 at 18:13
• Sorry, I don't understand the question and how it differs from all the questions discussed above. Commented Feb 14, 2017 at 9:34