Higgs decay to zz

Hi,

Firstly,there is new coupling called HIW/HIG in new models,What's the mesning about that?

The meaning can depend of the model, so I can not answer such question without knowing the model.
In general, HIW is related to the higgs photon photon interactions, and the HIG is related to the Higgs gluon gluon coupling

Secondly,I'm trying to compute the decay width from a Higgs to ZZ,respectively in SM and a new
model. How can i get it?

That number is actually zero since the Higgs can not decay into two on shell Z.
So you indeed need to go to the three body final state to have such contribution.
Note that the “Auto” width will automatically include such three body computation (if relevant, see : arXiv:1402.1178<http://arXiv.org/abs/arXiv:1402.1178> for more details)
Note also that in the SM, the compute_widths options, will not include the loop-induced decay (and therefore the total width of the Higgs is off)

how can i reset the decay width of higgs and z in the sm and the new model,so that i can compare the result?

I do not understand your question here. Note that when you generate your events, we never use the information about the Branching fraction of any given particle.
(We only use the total width as input)

Cheers,

Olivier

On 26 Mar 2017, at 09:27, niujj <<email address hidden><mailto:<email address hidden>>> wrote:

New question #592278 on MadGraph5_aMC@NLO:
https://answers.launchpad.net/mg5amcnlo/+question/592278

hello,

I have some questions for your help.

Firstly,there is new coupling called HIW/HIG in new models,What's the mesning about that?

Secondly,I'm trying to compute the decay width from a Higgs to ZZ,respectively in SM and a new
model. How can i get it? I tried the compute_width command to compute the higgs, but there is not zz;
and then i tried the process by generate h > z e+ e-, how can i reset the decay width of higgs and z in the sm and the new model,so that i can compare the result? the same or Auto?

all in all ,how can i get the decay width from a higgs to two z boson.

I'm waiting for your help.

sincerely,
niujj

--
You received this question notification because you are an answer
contact for MadGraph5_aMC@NLO.

Hi Olivier，
thank you very much for your help.
To follow,I have some other questions.

Firstly ,i go to the three body final state of higgs decay. However in the different models,there are different process.(for example,in the sm ,there are h- z d d~;z uu~,z ss~,z c c~,z b b~;in the new model,there are only
z u u~ and z c c~.) why? and why not z e e~ or other? how to compare the result ?

secondly, i generate h > z e+ e-,it said the method is narrow width approximation.which means first produce h> zz and then multiply by the branch ratio of z to e+ e-. that 's right? the the first two z is on shell ?

I'm waiting for your help.

sincerely,
niujj

Hi,

> Firstly ,i go to the three body final state of higgs decay. However in the different models,there are different process.(for example,in the sm ,there are h- z d d~;z uu~,z ss~,z c c~,z b b~;in the new model,there are only
> z u u~ and z c c~.) why? and why not z e e~ or other? how to compare the result ?

Not sure to understand what you are doing. Could you be more explicit?
If your question is about the Auto width (or equivalently the compute_widths function) then the answer should be in the paper I mention in the previous email.

> secondly, i generate h > z e+ e-,it said the method is narrow width
> approximation.which means first produce h> zz and then multiply by the
> branch ratio of z to e+ e-. that 's right? the the first two z is on
> shell ?

This is not correct, you can not put your two Z on shell, so you can not use Narrow width approximation for h > z z.
And therefore you can not use the branching ratio in that context.

Some people find convenient to speak about the partial width of h > Z Z*.
Since this is not a physical quantity, they defined that quantity by computing h > z e+ e- and dividing by the Z branching ratio.

Cheers,

Olivier

> On 26 Mar 2017, at 14:28, niujj <email address hidden> wrote:
>
> Question #592278 on MadGraph5_aMC@NLO changed:
> https://answers.launchpad.net/mg5amcnlo/+question/592278
>
> Status: Answered => Open
>
> niujj is still having a problem:
> Hi Olivier，
> thank you very much for your help.
> To follow,I have some other questions.
>
> Firstly ,i go to the three body final state of higgs decay. However in the different models,there are different process.(for example,in the sm ,there are h- z d d~;z uu~,z ss~,z c c~,z b b~;in the new model,there are only
> z u u~ and z c c~.) why? and why not z e e~ or other? how to compare the result ?
>
> secondly, i generate h > z e+ e-,it said the method is narrow width
> approximation.which means first produce h> zz and then multiply by the
> branch ratio of z to e+ e-. that 's right? the the first two z is on
> shell ?
>
> I'm waiting for your help.
>
> sincerely,
> niujj
>
> --
> You received this question notification because you are an answer
> contact for MadGraph5_aMC@NLO.

Hi， thanks again for your help. For the reply of the first question,i have something to say. I just follow the introduction of the article ,which you recommend to me,and then use the comand:compute-width h to get all the tree-level higgs decay width.There are many two-body and three-body decay processes.i wanna in diffrent models why higgs decay give different processes? for example:in sm model.compute-width h, three-body decay (include a z boson ) are five,like u u~ z,c c～z,d d~ z.... However in a new model,compute-width h,three-body decay (include a z boson )only have two ,like cc~z uu~z. why that happens? How can i compare the all higgs to zz contribution decay width in the sm model and new model? sincerely, niujj------------------ 原始邮件 ------------------
发件人: "Olivier Mattelaer"<email address hidden>
发送时间: 2017年3月26日(星期天) 晚上10:37
收件人: "501852601"<email address hidden>;
主题: Re: [Question #592278]: Higgs decay to zz

Your question #592278 on MadGraph5_aMC@NLO changed:
https://answers.launchpad.net/mg5amcnlo/+question/592278

    Status: Open => Answered

Olivier Mattelaer proposed the following answer:
Hi,

> Firstly ,i go to the three body final state of higgs decay. However in the different models,there are different process.(for example,in the sm ,there are h- z d d~;z uu~,z ss~,z c c~,z b b~;in the new model,there are only
> z u u~ and z c c~.) why? and why not z e e~ or other? how to compare the result ?

Not sure to understand what you are doing. Could you be more explicit?
If your question is about the Auto width (or equivalently the compute_widths function) then the answer should be in the paper I mention in the previous email.

> secondly, i generate h > z e+ e-,it said the method is narrow width
> approximation.which means first produce h> zz and then multiply by the
> branch ratio of z to e+ e-. that 's right? the the first two z is on
> shell ?

This is not correct, you can not put your two Z on shell, so you can not use Narrow width approximation for h > z z.
And therefore you can not use the branching ratio in that context.

Some people find convenient to speak about the partial width of h > Z Z*.
Since this is not a physical quantity, they defined that quantity by computing h > z e+ e- and dividing by the Z branching ratio.

Cheers,

Olivier

> On 26 Mar 2017, at 14:28, niujj <email address hidden> wrote:
>
> Question #592278 on MadGraph5_aMC@NLO changed:
> https://answers.launchpad.net/mg5amcnlo/+question/592278
>
> Status: Answered => Open
>
> niujj is still having a problem:
> Hi Olivier，
> thank you very much for your help.
> To follow,I have some other questions.
>
> Firstly ,i go to the three body final state of higgs decay. However in the different models,there are different process.(for example,in the sm ,there are h- z d d~;z uu~,z ss~,z c c~,z b b~;in the new model,there are only
> z u u~ and z c c~.) why? and why not z e e~ or other? how to compare the result ?
>
> secondly, i generate h > z e+ e-,it said the method is narrow width
> approximation.which means first produce h> zz and then multiply by the
> branch ratio of z to e+ e-. that 's right? the the first two z is on
> shell ?
>
> I'm waiting for your help.
>
> sincerely,
> niujj
>
> --
> You received this question notification because you are an answer
> contact for MadGraph5_aMC@NLO.

--
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You received this question notification because you asked the question.

Hi,

The idea of that function is not to compute ALL three body decay but rather to provide an estimation of the total width at a given precision.
This is why some three body decay might be (or not) evaluated depending of your model.

If you want to force the code to compute all three body decay. It is possible to do it as well but you need to use some options of the command.
Those options are documented in the paper or in the code by typing
help compute_widths

More in details you need to do
compute_widths h —body_decay=3

Cheers,

Olivier

> On 26 Mar 2017, at 17:43, niujj <email address hidden> wrote:
>
> Question #592278 on MadGraph5_aMC@NLO changed:
> https://answers.launchpad.net/mg5amcnlo/+question/592278
>
> Status: Answered => Open
>
> niujj is still having a problem:
> Hi， thanks again for your help. For the reply of the first question,i have something to say. I just follow the introduction of the article ,which you recommend to me,and then use the comand:compute-width h to get all the tree-level higgs decay width.There are many two-body and three-body decay processes.i wanna in diffrent models why higgs decay give different processes? for example:in sm model.compute-width h, three-body decay (include a z boson ) are five,like u u~ z,c c～z,d d~ z.... However in a new model,compute-width h,three-body decay (include a z boson )only have two ,like cc~z uu~z. why that happens? How can i compare the all higgs to zz contribution decay width in the sm model and new model? sincerely, niujj------------------ 原始邮件 ------------------
> 发件人: "Olivier Mattelaer"<email address hidden>
> 发送时间: 2017年3月26日(星期天) 晚上10:37
> 收件人: "501852601"<email address hidden>;
> 主题: Re: [Question #592278]: Higgs decay to zz
>
>
> Your question #592278 on MadGraph5_aMC@NLO changed:
> https://answers.launchpad.net/mg5amcnlo/+question/592278
>
> Status: Open => Answered
>
> Olivier Mattelaer proposed the following answer:
> Hi,
>
>> Firstly ,i go to the three body final state of higgs decay. However in the different models,there are different process.(for example,in the sm ,there are h- z d d~;z uu~,z ss~,z c c~,z b b~;in the new model,there are only
>> z u u~ and z c c~.) why? and why not z e e~ or other? how to compare the result ?
>
> Not sure to understand what you are doing. Could you be more explicit?
> If your question is about the Auto width (or equivalently the compute_widths function) then the answer should be in the paper I mention in the previous email.
>
>> secondly, i generate h > z e+ e-,it said the method is narrow width
>> approximation.which means first produce h> zz and then multiply by the
>> branch ratio of z to e+ e-. that 's right? the the first two z is on
>> shell ?
>
> This is not correct, you can not put your two Z on shell, so you can not use Narrow width approximation for h > z z.
> And therefore you can not use the branching ratio in that context.
>
> Some people find convenient to speak about the partial width of h > Z Z*.
> Since this is not a physical quantity, they defined that quantity by computing h > z e+ e- and dividing by the Z branching ratio.
>
> Cheers,
>
> Olivier
>
>> On 26 Mar 2017, at 14:28, niujj <email address hidden> wrote:
>>
>> Question #592278 on MadGraph5_aMC@NLO changed:
>> https://answers.launchpad.net/mg5amcnlo/+question/592278
>>
>> Status: Answered => Open
>>
>> niujj is still having a problem:
>> Hi Olivier，
>> thank you very much for your help.
>> To follow,I have some other questions.
>>
>> Firstly ,i go to the three body final state of higgs decay. However in the different models,there are different process.(for example,in the sm ,there are h- z d d~;z uu~,z ss~,z c c~,z b b~;in the new model,there are only
>> z u u~ and z c c~.) why? and why not z e e~ or other? how to compare the result ?
>>
>> secondly, i generate h > z e+ e-,it said the method is narrow width
>> approximation.which means first produce h> zz and then multiply by the
>> branch ratio of z to e+ e-. that 's right? the the first two z is on
>> shell ?
>>
>> I'm waiting for your help.
>>
>> sincerely,
>> niujj
>>
>> --
>> You received this question notification because you are an answer
>> contact for MadGraph5_aMC@NLO.
>
> --
> If this answers your question, please go to the following page to let us
> know that it is solved:
> https://answers.launchpad.net/mg5amcnlo/+question/592278/+confirm?answer_id=2
>
> If you still need help, you can reply to this email or go to the
> following page to enter your feedback:
> https://answers.launchpad.net/mg5amcnlo/+question/592278
>
> You received this question notification because you asked the question.
>
> You received this question notification because you are an answer
> contact for MadGraph5_aMC@NLO.

Thanks Olivier Mattelaer, that solved my question.