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SM Higgs boson phenomenology

17. About the delta function influence SOLVED - WE DECIDED TO USE DIFFERENT APPROACH, SEE LOGBOOK 7.2

We are interested in \sigma (pp \rightarrow Zh)(s) = \int_0^1 dx_1 dx_2\ f_1(x_1) f_2 (x_2)\ \delta(\hat{s}x_1 x_2 - s)\ \sigma (b\bar{b} \rightarrow Zh)(\hat{s}x_1 x_2), which can be rewritten:

\sigma (pp \rightarrow Zh)(s) = \sigma (b\bar{b} \rightarrow Zh)(s) \int_0^1 \frac{dx_1}{\hat{s}x_1} f_1(x_1) f_2 (\frac{s}{\hat{s}}x_1)\ \theta(\hat{s}x_1 - s)

This is probably also wrong, since this integration is giving result \sim 10^{-3}\ \text{pb}


Mathematica notebook - int.nb

Notebook by Stefan:int_SL.nb

2HDMC & SusHi out files - 2HDMC_250.out & sushi_250.out & sushi_400.out & sushi_1000.out

The employed b-quark masses are

for mA=250GeV: 2.8831183569691126 GeV

for mA=400GeV: 2.7980373758125761 GeV

for mA=1000GeV: 2.6512880401491685 GeV

Results for cross section differ \sim 9 times:

\sigma_{\text{SusHi}}(250\ \text{GeV}) \times \text{BR}

2\int dx_1 dx_2 f_1(x_1) f_2(x_2) \sigma_{\text{anal}} (x_1 x_2 \hat{s})

0.167587\ \text{pb}

1.52302\ \text{pb}

But I am not sure whatever should I trust it or not since NIntegrate is giving the following warnings:

Also I am not sure that I understand why those two answers for cross section should be the same. The first one is just hadronic cross section for one particular invariant mass 250 Gev, while the second one is something like a total cross section for given hadron CM energy \hat{s} (it includes all kinetically allowed s).

Here is the table for all masses:

M_A, \text{GeV}
\sigma_{\text{SusHi}}(M_A) \times \text{BR}

2\int dx_1 dx_2 f_1(x_1) f_2(x_2) \sigma_{\text{anal}} (x_1 x_2 \hat{s})














new notebook - int.nb



  • Does \frac{d\sigma}{d\sqrt{s}} in the BW approximation formula really mean the cross section derivative? I am wondering since I have got the following plots, where the pattern of the analytical cross section fits the \frac{d\sigma}{d\sqrt{s}} pattern from BW.

The result of BW formula and the integrated over \sqrt{s} result

Bare analytical result and after the PDF's integration

\frac{d\sigma}{d\sqrt{s}}  units are \text{pb} \times \text{GeV}^{-1}, \sigma units are \text{pb}

\sigma units are \text{GeV}^{-2} \approx 0.389379 \times 10^{9}\ \text{pb}

  • Is it OK that cross section for pp \rightarrow Zh is higher than b\bar{b} \rightarrow Zh? - not clear
  • There is a problem with units

10. Question on FormCalc SOLVED

  • When the sum over polarizations is done is the numerical factor also taken into account by FormCalc? - seems like not, it calculates just bare polarization sum
  • Since Z boson has 3 polarization states, should I multiply the polarization sum by \frac{1}{9}?

14. Bad behavior of the cross section. Integrals do not coverage. SOLVED

At small s \rightarrow 0 the analytical expression for the cross section behaves like \sigma \sim \frac{1}{s} = \frac{1}{x_1 x_2 \hat{s}}, the PDFs are also rise at small x, that is why integrals do not coverage. May be we got this problem because of the our assumption of zero mass for quarks?


13. What are we going to compare? SOLVED

The SusHi+2HDMC result for the BW is function of s, while analytical answer will be the function of \hat{s}. How to compare this two results? or should they be the same?

12. Cross section formula SOLVED

According to Peskin's book the cross section is given with

d \sigma = \frac{1}{2s}\times \frac{|\bar{k}_1|}{16 \pi^2 \sqrt{s}}\times |M(s)|^2d \Omega,

where s = (p_1 + p_2)^2 is the CM Energy squared and |\bar{k}_1| is the magnitude of the momentum of one of the outgoing particles. In our case we do not have an angle dependence, so after angular integration

\sigma = \frac{1}{s}\times \frac{|\bar{k}_1|}{8 \pi \sqrt{s}}\times |M(s)|^2.


  • What to do with outgoing momentum in the formula?
s = \left[ \begin{pmatrix} E_Z\\ \bar{k} \end{pmatrix} + \begin{pmatrix} E_h\\ -\bar{k} \end{pmatrix} \right]^2 = (E_Z + E_h)^2 = m_Z^2 + m_h^2 + 2|\bar{k}|^2 + 2\sqrt{m_Z^2 + |\bar{k}|^2}\sqrt{m_h^2 + |\bar{k}|^2}

for |\bar{k}| we can obtain

|\bar{k}| \to \frac{\sqrt{-2 s m_h^2-2 m_h^2 m_Z^2+m_h^4-2 s m_Z^2+m_Z^4+s^2}}{2 \sqrt{s}}
  • Should we now substitute this expression in \sigma(b\bar{b} \rightarrow Zh)(s) and differentiate with respect to \sqrt{s}, since we are interested in \frac{d \sigma (b\bar{b} \rightarrow Zh)}{d \sqrt{s}}?
  • Why \frac{d \sigma (b\bar{b} \rightarrow Zh)}{d \sqrt{s}} not just \sigma(b\bar{b} \rightarrow Zh)(s)?

11. Difficulties with the BW formula (yes, again (sad)) SOLVED

BW formula is given by

\frac{d\sigma(b\bar b\to ZH_{125})}{d\sqrt{s}} = \sigma(b\bar b\to A)(s)\frac{2\sqrt{s}}{(s-m_A^2)^2+m_A^2\Gamma_A^2}\frac{\sqrt{s}\Gamma(A\to ZH_{125})(s)}{\pi}

When I applied it previously I used SusHi results and, since SusHi integrates cross section with parton distribution functions, I actually used \sigma \left(pp \rightarrow A \right). That is why probably it is better to say that I have calculated \frac{d \sigma (pp \rightarrow Zh)}{d \sqrt{s}}, where s is still the CM energy of b\bar{b} squared s = (p_1 + p_2)^2.

Now I am trying to understand what is \sigma \left(pp \rightarrow A \right):

\sigma(pp \rightarrow A) = \int_0^1 dx_1 dx_2 f_b(x_1) f_{\bar{b}}(x_2) \sigma(b\bar{b} \rightarrow A)(s),

where f(x) - PDF, this function gives the probability for quark of a certain type to carry the fraction x of initial proton momentum. To perform the integration we should rewrite s in terms of fractions x_1 and x_2:

s = (p_1 + p_2)^2 = \left( \begin{pmatrix} \frac{1}{2}E_{CM} \\ x_1\bar{k}_1 \end{pmatrix} + \begin{pmatrix} \frac{1}{2}E_{CM} \\ x_2\bar{k}_2 \end{pmatrix} \right)^2 = E_{CM}^2 - (x_1 \bar{k}_1 + x_2 \bar{k}_2)^2

here \bar{k}_1 and \bar{k}_2 are the proton momenta in quark CM frame \Rightarrow x_1 \bar{k}_1 = -x_2 \bar{k}_2 \Rightarrowx_1 |\bar{k}_1| = x_2 |\bar{k}_2| and we are obtaining the following result:

\sigma (pp \rightarrow A) = \sigma(b\bar{b} \rightarrow A)(s) \times \int_0^1 d x f_b(x) f_\bar{b} \left(x\frac{|\bar{k}_1|}{|\bar{k}_2|}\right)


  • Is it correct?
  • I've got the dependence on \frac{|\bar{k}_1|}{|\bar{k}_2|}, how to get rid of it?
  • I was assuming that quark from the first proton and antiquark from the second proton participate in the reaction, but the opposite situation is also possible how to take this into account correctly? (multiply by factor 2?)
  • Here I was assuming that PDFs do not depend on s but only on factorization scale which is constant \sim \frac{1}{4}M_A, is it right?

9. Failed with FormCalc compilation SOLVED

./compile: 1: ./compile: : Permission denied

mkdir: cannot create directory ‘’: No such file or directory

Cannot create directory

8. Questions about FeynArts: SOLVED

  • What is G^0 (scalar line)?
  • What is the logic in particles numeration? Is there any table of particles and codes?
  • How to extract amplitude from the FeynAmpList?
  • How to make FeynArts output amplitudes readable? {Tried to use FeynCalc - FCFAConver[] }
  • What are EL, MW, CW, SW?


7. V-diagram in FeynArts output - how to get rid of it? SOLVED


6. BW for M_A = 1000\ \text{GeV}: the nlo-curve below while the nnlo-curve is above lo-curve (?) SOLVED

5. How to understand the K-factor structure? SOLVED

4. Discussion of the graphs for heavy Higgses and graphs for M_A = 1000\ \text{GeV}  SOLVED

  • g_{AH^{\pm}W^{\mp}} \sim 1, is it true?
  • How to understand the onion-like structure of A \rightarrow h^{\pm}W^{\mp}?
  • Is it right that  t\bar{t} and b\bar{b} decay channels are cancel each other when we are looking at the picture of Zh channel? (they have different magnitude)
  • How to understand the heavy Higgs graphs for BR? Why Higgses masses changed them in this way?
  • What other channels might have a significant contribution?

3. Total decay width for M_A = 1000\ \text{GeV} SOLVED - SEE THE LOGBOOK SECTION 4

Find attached sushi and 2HDMC output files for M_A = 1000\ \text{GeV}. According to those outputs total width has value in order of 7 \times 10^2 GeV, which is suspicious (?)


1. I've a problem when making a transition from the Lagrangian SOLVED

(1) \mathcal{L}^{hadr}_Y = - \left(\bar{Q}^\prime \phi {h_D}^\prime D^\prime + \bar{D}^\prime \phi_c^\dagger {{h_D}^\prime}^\dagger Q^\prime\right) - \left(\bar{Q}^\prime \phi {h_U}^\prime U^\prime + \bar{U}^\prime \phi_c^\dagger {{h_U}^\prime}^\dagger Q^\prime\right)

to the Lagrangian (see lecture notes by G. Ridolfi pp. 18-19)

(2) \mathcal{L}^{hadr}_Y = - \frac{1}{\sqrt{2}} \left(v + H \right) \sum_{f = 1}^n \left( h_D^f \bar{d}^f d^f + h_U^f \bar{u}^f u^f \right).

To do this transition I use eq. 2.2.42, 44-48 and definition 2.1.39 from G. Ridolfi lecture notes.

For example, in the initial Lagrangian there is a term

\bar{Q}^\prime \phi {h_D}^\prime D^\prime = \left( \bar{u}_L^\prime \bar{d}_L^\prime\right) \frac{1}{\sqrt{2}} \begin{pmatrix}0 \\ v + H(x)\end{pmatrix} V_L^D h_D {V_R^D}^\dagger d_R^\prime = \frac{1}{\sqrt{2}} (v + H) \bar{d}_L^\prime V_L^D h_D {V_R^D}^\dagger d_R^\prime = \frac{1}{\sqrt{2}} (v + H) \color{Orange}{\bar{d}_L^\prime V_L^D} h_D d_R\\ d_L^\prime = V_L^D d_L \rightarrow \bar{d}_L^\prime = d_L^\dagger {V_L^D}^\dagger \gamma_0 \rightarrow \color{Orange}{\bar{d}_L^\prime V_L^D} = d_L^\dagger {V_L^D}^\dagger \gamma_0 V_L^D.

2. Problem with the script SOLVED - FOR CORRECTED FILES SEE LOGBOOK 3.2

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