# Return matrix element squared of all terms

Asked by Ana Luisa Moreira de Carvalho

Dear experts,

I am generating the following process: g g > h1 h1 [QCD] using the 2HDM_NLO model from the feynrules database. It consists of three types of diagrams, let's call them M1, M2 and M3. These diagrams include one loop of fermions. We want to have separately the contribution for each diagram when only top or bottom quarks are allowed to run in this loop. Therefore, each diagram can be seen as a sum of two: M1= M1t+ M1b, M2=M2t+ M2b,M3= M3t+ M3b, where t and b indicate the fermion contributing to the loop. The total matrix element is then given by:

|M|² =|M1t+M1b+M2t+M2b+M3t+M3b|²= |M1t|²+|M1b|²+ |M2t|²+|M2b|²+ |M2t|²+|M2b|² + 15 interference terms.

My question is the following: is there a way of generating the entire process at once and then access the value of the matrix element squared for each of the terms described above?

Thank you,
Ana Luisa

## Question information

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Valentin Hirschi Edit question
Last query:
 Revision history for this message Olivier Mattelaer (olivier-mattelaer) said on 2018-09-28: #1

Hi,

This would require 21 different call to the loop reduction algorithm and therefore slow down the computation by a factor 21. So no this is not possible in one go.

Cheers,

Olivier

> On 27 Sep 2018, at 13:57, Ana Luisa Moreira de Carvalho <email address hidden> wrote:
>
> New question #674456 on MadGraph5_aMC@NLO:
>
> Dear experts,
>
> I am generating the following process: g g > h1 h1 [QCD] using the 2HDM_NLO model from the feynrules database. It consists of three types of diagrams, let's call them M1, M2 and M3. These diagrams include one loop of fermions. We want to have separately the contribution for each diagram when only top or bottom quarks are allowed to run in this loop. Therefore, each diagram can be seen as a sum of two: M1= M1t+ M1b, M2=M2t+ M2b,M3= M3t+ M3b, where t and b indicate the fermion contributing to the loop. The total matrix element is then given by:
>
> |M|² =|M1t+M1b+M2t+M2b+M3t+M3b|²= |M1t|²+|M1b|²+ |M2t|²+|M2b|²+ |M2t|²+|M2b|² + 15 interference terms.
>
> My question is the following: is there a way of generating the entire process at once and then access the value of the matrix element squared for each of the terms described above?
>
> Thank you,
> Ana Luisa
>
> --

 Revision history for this message Ana Luisa Moreira de Carvalho (ana-luisa-carvalho) said on 2018-09-28: #2

Hello,

That means that if I want to have access to the individual matrix elements I need to generate the corresponding diagrams separately right? That I know how to do so it should be ok. But then how can I access the matrix element instead of the cross section?

Cheers,
Ana Luísa

-------- Mensagem original --------
De:Olivier Mattelaer
Enviado:Fri, 28 Sep 2018 08:13:44 +0100
Para:Ana Carvalho
Assunto:Re: [Question #674456]: Return matrix element squared of all terms

Olivier Mattelaer proposed the following answer:
Hi,

This would require 21 different call to the loop reduction algorithm and
therefore slow down the computation by a factor 21. So no this is not
possible in one go.

Cheers,

Olivier

> On 27 Sep 2018, at 13:57, Ana Luisa Moreira de Carvalho <email address hidden> wrote:
>
> New question #674456 on MadGraph5_aMC@NLO:
>
> Dear experts,
>
> I am generating the following process: g g > h1 h1 [QCD] using the 2HDM_NLO model from the feynrules database. It consists of three types of diagrams, let's call them M1, M2 and M3. These diagrams include one loop of fermions. We want to have separately the contribution for each diagram when only top or bottom quarks are allowed to run in this loop. Therefore, each diagram can be seen as a sum of two: M1= M1t+ M1b, M2=M2t+ M2b,M3= M3t+ M3b, where t and b indicate the fermion contributing to the loop. The total matrix element is then given by:
>
> |M|² =|M1t+M1b+M2t+M2b+M3t+M3b|²= |M1t|²+|M1b|²+ |M2t|²+|M2b|²+ |M2t|²+|M2b|² + 15 interference terms.
>
> My question is the following: is there a way of generating the entire process at once and then access the value of the matrix element squared for each of the terms described above?
>
> Thank you,
> Ana Luisa
>
> --

--
know that it is solved:

If you still need help, you can reply to this email or go to the
following page to enter your feedback:

 Revision history for this message Olivier Mattelaer (olivier-mattelaer) said on 2018-09-28: #3

Hi,

You have to generate the process like this:
generate g g > h1 h1 [virt=QCD]
output

Then you have a standalone code for the evaluation of the matrix-element and nothing else.
Cheers,

Olivier

> On 28 Sep 2018, at 10:43, Ana Luisa Moreira de Carvalho <email address hidden> wrote:
>
> Question #674456 on MadGraph5_aMC@NLO changed:
>
>
> Ana Luisa Moreira de Carvalho is still having a problem:
> Hello,
>
> Thank you for the reply.
>
> That means that if I want to have access to the individual matrix
> elements I need to generate the corresponding diagrams separately right?
> That I know how to do so it should be ok. But then how can I access the
> matrix element instead of the cross section?
>
> Cheers,
> Ana Luísa
>
> -------- Mensagem original --------
> De:Olivier Mattelaer
> Enviado:Fri, 28 Sep 2018 08:13:44 +0100
> Para:Ana Carvalho
> Assunto:Re: [Question #674456]: Return matrix element squared of all terms
>
>
>
> Olivier Mattelaer proposed the following answer:
> Hi,
>
> This would require 21 different call to the loop reduction algorithm and
> therefore slow down the computation by a factor 21. So no this is not
> possible in one go.
>
> Cheers,
>
> Olivier
>
>> On 27 Sep 2018, at 13:57, Ana Luisa Moreira de Carvalho <email address hidden> wrote:
>>
>> New question #674456 on MadGraph5_aMC@NLO:
>>
>> Dear experts,
>>
>> I am generating the following process: g g > h1 h1 [QCD] using the 2HDM_NLO model from the feynrules database. It consists of three types of diagrams, let's call them M1, M2 and M3. These diagrams include one loop of fermions. We want to have separately the contribution for each diagram when only top or bottom quarks are allowed to run in this loop. Therefore, each diagram can be seen as a sum of two: M1= M1t+ M1b, M2=M2t+ M2b,M3= M3t+ M3b, where t and b indicate the fermion contributing to the loop. The total matrix element is then given by:
>>
>> |M|² =|M1t+M1b+M2t+M2b+M3t+M3b|²= |M1t|²+|M1b|²+ |M2t|²+|M2b|²+ |M2t|²+|M2b|² + 15 interference terms.
>>
>> My question is the following: is there a way of generating the entire process at once and then access the value of the matrix element squared for each of the terms described above?
>>
>> Thank you,
>> Ana Luisa
>>
>> --
>
> --
> know that it is solved:
>
> If you still need help, you can reply to this email or go to the
> following page to enter your feedback:
>
>
> --

 Revision history for this message Ana Luisa Moreira de Carvalho (ana-luisa-carvalho) said on 2018-09-28: #4

Hello,

Thank you! Just a final question: I generated just like you suggested and it worked. However, I get the following output:

Unknown numerical stability because MadLoop is in the initialization stage.
---------------------------------
---------------------------------
This is a loop induced process, so only the
unnormalized finite part is output here. Be aware
that all loops are expected to beUV-finite as no
renormalization prescription is considered.

I am not sure what this MadLoop return code means. Does this message mean that some terms are not being taken into account ?