Higgs production and PDF

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1. Aug 1, 2013

Safinaz

Hi all,

I try to calculate the Higgs production cross section by Mathematica using
for example the formula in [arXiv:hep-ph/0503172].

The problem comes when I calculate " σ_0 " it's of order 10^-10 pb at mh = 125 GeV ( 3.57), then when I convolute with the parton distribution function to get the pp -> H production according to equ.( 3.58) it gives me of order 10^-7 pb !! while in the plot (3.18) pp -> H is around 50 pb.

I don't know what is the problem I have, any help please.

Best,
Safinaz

2. Aug 1, 2013

Staff: Mentor

It is hard to impossible to tell what is wrong if you do not show your calculations.

10-7 pb is too small.

3. Aug 2, 2013

Safinaz

for σ0:

σ0[mh_] : = ( ( c^2)/ (256 *Pi)) *(1/2)^2* ( ((Yht)/(mT))* At [mh])^2
where
c = 0.117, Yht= mt/v, v= 174 and At is the fermion loop function.

This gives 10^-10 pb .

For σ(pp -> h) I used:

LogPlot[ {σ0[mh]*(q)* NIntegrate[(1/x)* pdf[iset, iparton, x, 14000] * pdf[iset, iparton, (q/x), 14000], {x, q, 1}], {mh, 100, 200}]

With

iset = 1; (* CTEQ5M*)
iparton = 0;
q := mh^2/14000^2;

4. Aug 2, 2013

Bill_K

This is way too small also.

What you gave doesn't correspond very well to the equation Eq (3.57) But after all, finding a missing factor of so many orders of magnitude should not require the use of Mathematica!

If you take a look at the equation as given, the Fermi coupling constant Gμ from Eq (1.76) is 10-5 GeV-2, the strong coupling constant from Eq (1.78) is αs = 0.1, and there is a numerical factor of about 1000 in the denominator. The form factor A is order of magnitude 1.

Apparently a dimensionful factor has been suppressed. To get something with the right dimensions, i.e. a cross-section, we need to reintroduce (ħc)2 where ħc = 200 MeV-f. So we have

10-5 GeV-2 (200 MeV-f)2 (0.1)2/1000 ≈ 10-10 f2 = 10-36 cm2 = 10-12 b = 1 pb

5. Aug 2, 2013

Safinaz

So the problem is only the dimensions. Now σ0[125] of order 10^-5 pb .. but what about
σ(pp -> h) , at page 115 it's mentioned that $\int L dt = 100 pb^{-1}$ so that should
we multiply by some factor here because I still have σ(pp -> h) [100] = 0.006.

s.

6. Aug 3, 2013

Staff: Mentor

I don't see how the integrated luminosity at hadron colliders would be relevant for the cross-section. If you care about the total number of Higgs bosons produced, multiply the cross-section with the integrated luminosity. With 1pb (see post 4) and 100/pb this gives 1pb*100/pb = 100 Higgs bosons.

7. Aug 3, 2013

Safinaz

So what about equ.(3.58) in [arXiv:hep-ph/0503172], they gave the total cross section by multiplying
$\sigma0$ by $\tau_h ( dL/d\tau_h)$ I don't understand this expression.

They define $dL/d\tau_h$ as the parton distribution function and this is what I used in my calculation
to get(pp->h) as in post 3 after getting $\sigma0$ (pb).

S

Last edited: Aug 3, 2013