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W boson-like particle from Higgs

by tansic
Tags: boson production, higgs
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tansic
#1
Dec4-12, 10:29 PM
P: 10
Just a random question I thought of while daydreaming. Is it possible for a virtual particle that is not a W Boson, but is very like one (same mass, spin, etc.) to be produced from a Higgs decaying into two photons?



Higgs→λλ+disturbance in W field.
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ofirg
#2
Dec5-12, 06:30 AM
P: 90
I'm not sure I understand the question.

Higgs→λλ+disturbance in W field.
This writing is misleading. There is no W boson (real or virtual) present in the final state. The final state consists of two photons and thats it.

The Virtual W (The disturbance in the W field, if thats what you mean) contributes to the decay amplitude (not the only contribution in the SM. The top quark also has a sizable contribution)

If your question is whether a particle similiar to the W boson could also contribute to the decay amplitude, The answer is yes. A massive spin 1 charged particle would contribute, However there are no such particles in the SM except the W boson.
tansic
#3
Dec5-12, 07:43 AM
P: 10
Would said W boson-like particle have to share the same charge and mass or just the spin of a W boson? And there would be an anti particle to accompany, correct?

ofirg
#4
Dec5-12, 07:48 AM
P: 90
W boson-like particle from Higgs

Any particle which couples to the higgs (and thus gets mass from the higgs mechanism) and has electric charge would contribute.
It doesn't have to have the same mass, spin or charge as the W boson.
It Would have a distinct antiparticle as any electrically charged particle.
tansic
#5
Dec5-12, 07:53 AM
P: 10
Ok, thank you for your replies.
tansic
#6
Dec5-12, 07:56 AM
P: 10
Wait, I hear you saying that any charged particle can cause a disturbance in any particles field.
ofirg
#7
Dec5-12, 08:09 AM
P: 90
I'm still not sure what you mean by the word disturbance.

The fact that the particle is charged means that it can emit and absorb photons. This is required in order for the particle to mediate the [itex]h->\gamma\gamma[/itex] decay.

That that the particle is charged doesn't mean that it couples to every particle (or field)

W bosons for example don't couple to gluons
tansic
#8
Dec5-12, 11:49 AM
P: 10
Ok, my bad I misunderstood.
tansic
#9
Dec5-12, 02:04 PM
P: 10
And by 'disturbance' I mean a ripple in the W field suggesting a virtual W boson.


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