# Higgs, top production

• I
Hello

What possible processes are there for a production of Higgs from t-tbar?

mfb
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What do you mean by "from t-tbar"? There is no t-tbar collider, but direct production would be possible and by far the dominant process.

Google and arXiv search should show all the relevant production modes for every process. In general, it's not about what is possible, but what is frequent enough to be relevant.

Hi and thanks for the reply.

I will try being more clear about what I'm asking. To my knowledge and from what I learned, in the LHC t quarks are formed in pairs of t tbar. Since they're massive, they decay before they're able to hadronize and we have processes such as

or

or

Now, I want to know which possible processes are there for a production of Higgs (along with other particles) from a pair t tbar.
I hope this has made it a little more clear.

mfb
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A top can radiate a Higgs boson, but that is a very unlikely process.

Decays like top -> H + charm are forbidden (at tree level) in the standard model, experiments look for them to set exclusion limits or (ideally) find such a process.

The production of a t-tbar pair together with a Higgs (lower left process, not from top decays) is much more likely. Summary of searches

MMS
ChrisVer
Gold Member

in the LHC t quarks are formed in pairs of t tbar.

Well you can have the formation of single tops too...[formation via weak interaction]
And for pp colliders, you have ~2 times more single tops than antitops...

Also what's the difference in the ttbar decays you posted?

MMS
Also what's the difference in the ttbar decays you posted?

That I uploaded the wrong images. :P
The decays i was talking about are

or

or

While I'm at it, how would a Feynman diagram look for, say, the first ttbar decay?

mfb
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There is also ##t \bar t \to b \bar b q \bar q q \bar q## but it is a tricky channel experimentally.
While I'm at it, how would a Feynman diagram look for, say, the first ttbar decay?
Both top decay to Wb, the W then decay either to lepton+neutrino or quark+antiquark.

Clever Penguin and MMS
There is also ##t \bar t \to b \bar b q \bar q q \bar q## but it is a tricky channel experimentally.
Both top decay to Wb, the W then decay either to lepton+neutrino or quark+antiquark.
Thank you!

There is also ##t \bar t \to b \bar b q \bar q q \bar q## but it is a tricky channel experimentally.
Both top decay to Wb, the W then decay either to lepton+neutrino or quark+antiquark.
I'm trying to draw the actual diagram like you said and I'm kind of struggling to do so. Can you please show me the actual diagram with what you said?

ChrisVer
Gold Member
I'm trying to draw the actual diagram like you said and I'm kind of struggling to do so. Can you please show me the actual diagram with what you said?

The diagram is exactly the same as the one you drew with the lepton final states... Remember that the W boson can decay to leptons+neutrinos, but it can also decay to q+q' (one up-type and one down-type quarks, eg $W^- \rightarrow \bar{c}d$)

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mfb
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The internet is full of Feynman diagrams of top decays.

First off thanks for the help ChrisVer and mfb. I asked about the diagrams because I wasn't feeling pretty confident with what I'm drawing. But I tried understanding this better yesterday and I hope I got this right. Attached are 3 figures of the Feynman diagrams for the 3 decays I described above. I'd be happy if you guys could take a look at them and tell me if I did this correctly or give out some remarks. I can provide an argument for each if needed.

First decay

Second decay

Third decay

Thank you!

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mfb
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Correct.
And all the other decay modes work in the same way.

Edit: Didn't check the arrows. Orodruin is right, some of them point in the wrong direction.

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Orodruin
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Your fermion flows are not correct. You should take care in which way the arrows of your fermion lines go. In the standard model, the fermion flow never changes direction.

ChrisVer
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also if you want to be extremely scholastic, writing $q\bar{q}$ on the diagrams can be misleading... the reason they write $\bar{t}t\rightarrow b \bar{b} q\bar{q} l \nu$ is to make it readable (in fact read the quarks as jets)... In the diagram you should make sure you prime the quarks coming from the Ws, in explicitely $W\rightarrow q \bar{q}~'$, because the two quarks are not each others' antiparticle (eg you don't have $W \rightarrow u \bar{u}$).

MMS
Correct.
And all the other decay modes work in the same way.

Edit: Didn't check the arrows. Orodruin is right, some of them point in the wrong direction.
Your fermion flows are not correct. You should take care in which way the arrows of your fermion lines go. In the standard model, the fermion flow never changes direction.

So the arrows on the W+ decay on each of the 3 ttbar decays is opposite?

ChrisVer
Gold Member
So the arrows on the W+ decay on each of the 3 ttbar decays is opposite?
Ws don't need an arrow...
the arrows for antiparticles [as you plot the diagrams] should look to the left (so for example some W product(s)-find them out- and $\bar{t}\bar{b}$ are wrong).
the arrows for particles to the right.

Orodruin
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Maybe this is something basic that I am missing out on but assuming I draw the antiparticle arrows in the opposite direction, how can one tell that it's a decay of ttbar that we're talking about? I thought an arrow going in and one out means that there is an interaction between the two particles. Or in general, how can I tell the difference between a decay and simply an interaction?

ChrisVer
Gold Member
Maybe this is something basic that I am missing out on but assuming I draw the antiparticle arrows in the opposite direction, how can one tell that it's a decay of ttbar that we're talking about?

Arrows don't necessariy show the flow of the interaction. They are there corresponding to some mathematical quantity, as are the Feynman Diagrams... they roughly tell you what kind of Dirac spinors you are using. In some interpretation, the antiparticles move "backwards in time", although that's an interpretation and shouldn't be taken literally in the same way as the Feynman diagrams shouldn't be literally taken as the physical interaction, so in that view it's natural to draw them like that...[in order to have a flow]

I thought an arrow going in and one out means that there is an interaction between the two particles.
well in your plot there are 2 arrows moving out from a "vertex".... [have a look at ttbar]... of course the tops don't come from the same vertex, but still.... the idea is again the one I mentioned, particles -> antiparticles <-.

Or in general, how can I tell the difference between a decay and simply an interaction?
A decay is pretty simple an interaction that looks like this:
$1 \rightarrow 2+3+...+N$
An interaction though should have at least 2 particles in the initial state....
Although a decay is an interaction too [eg the particle 1 decays via an interaction to the particles 2,3...N) ...

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mfb
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Or in general, how can I tell the difference between a decay and simply an interaction?
If one particles come in and multiple go out, the whole diagram is a decay, otherwise the more general "interaction" is used.

Antiparticles always have their arrows go against the time direction. You can tell it's a ttbar process (including the decays of the tops) because you start with ttbar production (the production process is currently not part of your diagrams) and you have different outgoing particles.

MMS
Arrows don't necessariy show the flow of the interaction. They are there corresponding to some mathematical quantity, as are the Feynman Diagrams... they roughly tell you what kind of Dirac spinors you are using. In some interpretation, the antiparticles move "backwards in time", although that's an interpretation and shouldn't be taken literally in the same way as the Feynman diagrams shouldn't be literally taken as the physical interaction, so in that view it's natural to draw them like that...[in order to have a flow]

well in your plot there are 2 arrows moving out from a "vertex".... [have a look at ttbar]... of course the tops don't come from the same vertex, but still.... the idea is again the one I mentioned, particles -> antiparticles <-.

A decay is pretty simple an interaction that looks like this:
$1 \rightarrow 2+3+...+N$
An interaction though should have at least 2 particles in the initial state....
Although a decay is an interaction too [eg the particle 1 decays via an interaction to the particles 2,3...N) ...

If one particles come in and multiple go out, the whole diagram is a decay, otherwise the more general "interaction" is used.

Antiparticles always have their arrows go against the time direction. You can tell it's a ttbar process (including the decays of the tops) because you start with ttbar production (the production process is currently not part of your diagrams) and you have different outgoing particles.

That really really helped and made things clearer. I will sketch those decays again and update in a little.

Thank you!

Orodruin
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Antiparticles always have their arrows go against the time direction. You can tell it's a ttbar process (including the decays of the tops) because you start with ttbar production (the production process is currently not part of your diagrams) and you have different outgoing particles.
I just want to point out that it is essentially pointless to interpret the "time direction" of internal lines and assign particle/antiparticle lables to them. The Feynman diagram is a pictorial representaton of terms in a perturbative expansion of a path integral. For example, we do not draw separate diagrams to represent W- exchange in one direction and W+ in the other. They are the same diagram representing the same term in the expansion.

mfb
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For internal lines, the label "antiparticle" itself is not meaningful any more, so my statement doesn't apply to those.
Here all fermions are external lines, at least if we consider t+tbar to be the inital state.

Staff Emeritus
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I just want to point out that it is essentially pointless to interpret the "time direction" of internal lines and assign particle/antiparticle lables to them.

I agree, but it's also important to get the arrow right so you write down the right wavefunction.

Orodruin
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I agree, but it's also important to get the arrow right so you write down the right wavefunction.
Indeed, never reverse the direction of a fermion line! (In the Standard Model ...)

ChrisVer
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they look OK,

MMS