H -> W+W- -> e+e- Why are the electron/positron collimated?

[Mithra]

Our lecturer explained to us that when a stationary Higgs decays into a W+W- pair, which then each decay into an electron/neutrino pair that the electrons will be collimated, however I'm struggling to understand why. He seemed to explain that the leptons would tend to have spin in one direction whilst the neutrinos would have it in the other, and this then causes the leptons to go in the same direction? (Assuming all particles here are massless) However I can't see what would cause this preference in spin direction? If anyone could help explain/point me on the right track I'd be very appreciative thanks!

 

[e.chaniotakis]

Excuse me, if the mass of the Higgs is 125 GeV it cannot decay in two W's

 

[Mithra]

Well I guess in the model we're looking at here it is heavier than that.

 

[fzero]

A light Higgs can still decay to two vector bosons if one of them is a virtual particle. Sometimes you'll see this labeled as WW^*, where the W^* indicates a virtual particle.

That said, the argument proposed in the OP seems suspicious. If the Higgs were at rest in the lab frame, then there are no preferred directions for the final state. Such decays are sometimes called "spherical," for obvious reasons. The spin argument does not help. While it is true that, in the SM, only the left-handed neutrinos participate in the weak interaction, this only translates into some correlations between the electron spins and the helicity of the neutrinos, not the direction of momentum of any particles. The leading order kinematics can be determined by ignoring the spin, or if you want, using the spin as a separate conservation law.

Collimation in decay products is mainly seen when an intermediate particle has a large momentum in the lab frame. Then it is kinematically favorable for the decay products to have similar amounts of momentum, with directions set by the direction of the parent. This collimation is the reason behind jet events. For example, when a top quark decays into a bottom quark, quite often the bottom quark has a large momentum (because it is very light compared to the top). The subsequent decays of the bottom will be collimated resulting in a hadron jet being found in the detector.

In principle, a light Higgs could be produced at the LHC with a significant momentum. We have 7 TeV in the CoM, the Higgs mass is only 125 GeV, so if 10% of the CoM energy goes into production, the Higgs will be produced with a significant momentum in the lab frame. The decay products will then favor a similar direction. However, this does not seem to be the situation that was suggested.

 

[Vanadium 50]

e.chaniotakis comment is not helpful. As fzero says, one of the W's is "off shell", and indeed this decay mode has a fairly high branching fraction: about 15% of the Higgses decay this way.

Since the Higgs is spin-0 and both linear and angular momentum are conserved, the two W's have opposite momenta and opposite spins, and therefore the same helicity. A left-handed W has its lepton emitted preferentially in the direction of the spin, so if they are both left-handed, the W- sends its electron forward and the W+ (moving in the other direction) sends its positron backwards - i.e. in the same direction of the electron.

A right-handed W has its antilepton emitted preferentially in the direction of the spin, so if they are both right-handed, the W- sends its electron backwards and the W+ (moving in the other direction) sends its positron forwards - i.e. again, in the same direction of the electron.

 

[mfb]

That said, the argument proposed in the OP seems suspicious. If the Higgs were at rest in the lab frame, then there are no preferred directions for the final state. Such decays are sometimes called "spherical," for obvious reasons.

There is no preferred direction for any particle on its own, but there is a correlation between the decay products. Those are two completely different things.

H->WW*->lvlv is one of the most sensitive decay channels at the LHC, similar to H->ZZ*->4 leptons.

 

[Mithra]

e.chaniotakis comment is not helpful. As fzero says, one of the W's is "off shell", and indeed this decay mode has a fairly high branching fraction: about 15% of the Higgses decay this way.

Since the Higgs is spin-0 and both linear and angular momentum are conserved, the two W's have opposite momenta and opposite spins, and therefore the same helicity. A left-handed W has its lepton emitted preferentially in the direction of the spin, so if they are both left-handed, the W- sends its electron forward and the W+ (moving in the other direction) sends its positron backwards - i.e. in the same direction of the electron.

A right-handed W has its antilepton emitted preferentially in the direction of the spin, so if they are both right-handed, the W- sends its electron backwards and the W+ (moving in the other direction) sends its positron forwards - i.e. again, in the same direction of the electron.

Aha brilliant, that makes a lot more sense now. Thanks very much for such a clear explanation! :)

 

[snorkack]

Note that the conserved parametres of Higgs boson - no charge, no spin - match the parametres of neutral pion.

Are the leptons formed on neutral pion decays to other than two photons collimated in the same manner and for the same reason?

 

[mfb]

Neutral pions do not decay to two leptons plus two neutrinos with any reasonable branching ratio, I think. The decay to two electrons and two positrons was observed, but it does not happen via W bosons.