# What is the total cross section for $H \rightarrow W^{+} W^{-}$?

I need to know the total cross section for the process $$H \rightarrow W^{+}W^{-}$$, but cannot find it anywhere. Decay widths for Higgs decays are mentioned in Peskin and Schroeder (page 776), but I gather that one cannot compute cross sections directly from decay widths.

Edit: looks as if the answer is on page 6 ofthis presentation.

• A decay doesn’t have a cross-section. Cross sections measure how likely two particles are to scatter. – G. Smith Jul 1 at 23:57
• That's right, but someone asked me for the total cross section for the process above, so I'm not sure what they mean. Is it some thing on page 6 of this presentation? uvic.ca/science/physics/vispa/research/projects/atlas/… Are they perhaps referring to the cross section for fusion of two W bosons to a Higgs boson? – Tom Jul 2 at 0:01
• The real Higgs is below threshold to decay to two Ws, so one off-shell W is indicated by W*, and P&S distinctly speculate on above threshold! I guess you could cook up a real Higgs production cross section and multiply it with the BR for your process which includes all decay channels of the two Ws. The p5 -mentioned preprint ATLAS-CONF-2018-004 doesn't have enough background info? Some reference yellow book? – Cosmas Zachos Jul 2 at 0:57
• @Tom if you have troubles calculating the decay widths, i can give you some hints. – ApolloRa Jul 2 at 8:58
• @ApolloRa Some hints or starting point would be useful for the decay widths, as they're not something I compute very often. – Tom Jul 3 at 22:21

A decay doesn't have a cross section associated with it, only a decay width. Most likely they are referring to the production cross section for the process $$\rm X+X\to H\to W^+W^-$$ at some specific collider. Note that this depends on the identity of the initial particles and their momenta. It can also depend on other properties such as the polarization of the beams and the geometric acceptance of the detector.

For instance, the fiducial cross section $$\rm pp\to H\to W^+W^-\to e\nu\mu\nu$$ has been measured at the CMS experiment at a center of mass energy $$\sqrt{s}=13~\rm TeV$$ to be $$85.0^{+9.9}_{-9.3}~\rm fb$$, as detailed in this paper.

The amplitude for the decay is:

$$M = \cfrac{g^{αβ}g^2 v}{2}ε^{*}_{α}(p)ε^{*}_{β}(q)$$

You can obtain the vertex factor from the assosiated Lagrangian. $$ε$$ here are the polarisations of the outgoing spin $$1$$ bosons and $$p$$ and $$q$$ denote the momenta. You want to calculate:

$$Γ = \frac{p_{f}}{32π^2 m_{H}} \int |M^2|dΩ$$

For the $$M^2$$ you use the fact that $$W$$'s are spin one massive bosons, so when you sum over polarisations:

$$Σ_{pol}(ε^{*}_{α}(p)ε_{μ}(p)) = -g_{αμ} + \cfrac{p_αp_μ}{Μ_w}$$

Assuming that we are in Higgs center of mass frame:

$$p^{α} = (Ε,0,0,p)$$, $$q^{α} = (E,0,0,-p)$$,

$$Μ_{H} = E +E \Rightarrow E = \cfrac{M_{H}}{2}$$,

$$p^{α}p_{α} = Ε*Ε - p*p = E^2 - p^2 = M_W^2$$,

$$q^{α}q_{α} = Ε*Ε - p*p = Ε^2 - p^2 = M_{W}^2$$, $$p^{α}q_{α} = Ε*Ε - (-p)*(p) = E^2 + p^2 = E^2 + E^2 -M_{W}^2 = 2*\cfrac{M_{H}^2}{4} - M_{W}^2 = \cfrac{M_{H}^2}{2} - M_{W}^2$$

$$p_{f}^2 = E^2 - M_{W}^2 = \cfrac{M_{H}^2}{4} - M_{W}^2$$

and of course $$\int dΩ = 4π$$. With all the above it is eazy to calculate the decay rate for $$H\rightarrow M^{+}M^{-}$$. Note that if you encounter the $$H\rightarrow ZZ$$ decay the final result should be devided by $$2$$ because $$Z$$ are identical particles and there is no way to distinguish which is which.