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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?

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The $W$ can participate in both charged- and neutral-current channels, but the $Z$ can only participate in the neutral current. – rob Jun 22 '14 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? – sinned Jun 22 '14 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 Jun 22 '14 at 18:27
Have you seen this article and references therein? – DarioP Jun 23 '14 at 9:17

1 Answer 1

up vote 6 down vote accepted

The Figure 2 of this paper

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 $m_W^2$.

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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? – sinned Jun 25 '14 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. – Luboš Motl Jun 25 '14 at 7:40

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