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I am working on an LHCb experiment, in particular the $B^0 \rightarrow K^{*0} \gamma$ decay.

The $K^{*0}$ decays into $K^+$ and $\pi^-$. So the decay products of the decay are $\gamma, K^+ $and $ \pi^-$ (and their antiparticles in case of an $\overline{B^0}$ decay).

The mass is reconstructed in the following way: the 4-momenta of the decay products are summed together to get the total 4-momentum $p^{\mu}_{B^0} $ : $p^{\mu}p_{\mu}$ then gives the invariant mass squared, $(mc^2)^2$.

From the uncertainties of the detectors, I can propagate the uncertainty on the reconstructed mass of every single event (i.e. every single collection of $\gamma, K^{+} $ and $\pi^-$).

When we make a histogram of all the events (5,000,000+), I get this:

enter image description here

And I want to estimate, beforehand, what the spread in the signal peak should be. How can I do this?

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  • $\begingroup$ What does it mean "spread in the signal peak" ? You mean RMS? I'd be glad if you be more clear. $\endgroup$ – aQuestion Mar 15 '15 at 20:30
  • $\begingroup$ the width of the Gaussian-like signal $\endgroup$ – SuperCiocia Mar 15 '15 at 22:21
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There are two methods for making predictions for LHC searches:

  1. Data-driven. If you are searching for a signal, extrapolate the background prediction from a "sideband" region to a "signal" region of phase space.
  2. Monte Carlo (MC). Generate events with in pseudo-experiments (event generation from matrix elements, followed showering, hadronization, and detector simulation). Apply the relevant selections to the MC events. Calculate the width the resulting invariant mass histogram.

For predicting the width of a distribution, I think only the latter option is possible. There are two contributions to the width of that distribution:

  1. The intrinsic width of the of the $B_0$, $\Gamma$. You can calculate this.
  2. The experimental resolution, resulting from experiment errors etc.

I think in your case the experimental resolution dominates, so you probably cannot make an approximate guess from the intrinsic width. In summary: do the MC.

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  • $\begingroup$ yes. The spread of the signal is due to the detectors' resolution. But how can I calculate it in advance, knowing the errors in each individual detector (obtained from previous calibrations)? $\endgroup$ – SuperCiocia Mar 15 '15 at 22:26
  • $\begingroup$ With a MC. For very simple detectors you can sometimes add the "width" of each bin in quadrature with the intrinsic width, but LHCb is not "very simple" by any stretch of the imagination. $\endgroup$ – dmckee Mar 15 '15 at 23:14

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