= Higgs resonance resolution and background model
{c}
{title2=$m_{\gamma\gamma}^2=2E_1E_2(1-\cos\alpha)$}
The bridge from detector readout to a <particle resonance> is <four-momentum> conservation. In units $c=1$, sum the measured final-state <four-momenta>:
$$
m^2=\left(\sum_i E_i\right)^2-\left|\sum_i\mathbf p_i\right|^2.
$$
For two <photons> separated by angle $\alpha$, $|\mathbf p_i|=E_i$ gives the displayed diphoton formula. Two back-to-back <photons> of $60\,\mathrm{GeV}$ each have $m_{\gamma\gamma}=120\,\mathrm{GeV}$. A four-lepton candidate instead uses four <four-momenta> in the same sum. Collision debris outside the selected decay is not included. Since the <partons> carry unknown fractions of the incoming <proton> momenta, reconstructing the decay products is more informative than assigning the full beam <energy> to each candidate.
\Image[/higgs-detector-readout-and-resonance-resolution.png]
{title=Detector functions, reconstructed mass and a chosen resolution comparison}
{description=Functional schematic followed by an original expectation model: sixty selected signal events at $120\,\mathrm{GeV}$, a flat background of eight events per $10^9$ <electronvolts>, and chosen <Gaussian> mass widths $1.5$ and $3\,\mathrm{GeV}$. No historical event counts or copied discovery histograms are shown.}
Suppose the intrinsic <decay width> is negligible compared with the stipulated detector spread. Let the reconstructed signal mass follow a <normal distribution> with mean $m_0=120\,\mathrm{GeV}$ and <standard deviation> $\sigma_m$. For a total expected selected signal $S=60$ and flat background density $b=8$ events per $10^9$ <electronvolts>, the expected count density is
$$
\frac{dN}{dm}=b+\frac{S}{\sqrt{2\pi}\sigma_m}\exp\!\left[-\frac{(m-m_0)^2}{2\sigma_m^2}\right].
$$
This is a deliberately chosen response model, not a fit to either experiment. In the fixed window $m_0-a<m<m_0+a$ with $a=3\,\mathrm{GeV}$, direct integration gives
$$
B_w=2ab=48,\qquad S_w=S\operatorname{erf}\!\left(\frac{a}{\sqrt2\sigma_m}\right).
$$
The expected signal counts are $57.2700$ for $\sigma_m=1.5\,\mathrm{GeV}$ and $40.9614$ for $\sigma_m=3\,\mathrm{GeV}$. Doubling the spread lowers the fraction in the same window without changing the total signal. Detector resolution can therefore change the visibility of a peak even with identical particle production.
As a separate counting example, stipulate an observed integer count $n=70$ and an exactly known background mean $B_w=48$. The <Poisson distribution> gives a <likelihood> $L(s)=e^{-(48+s)}(48+s)^{70}/70!$ for nonnegative signal mean $s$. Its <maximum-likelihood estimator> is $\widehat s=22$, and the <likelihood ratio> statistic against $s=0$ is
$$
q_0=-2\log\frac{L(0)}{L(22)}=2\left[70\log\frac{70}{48}-22\right]\simeq8.82119.
$$
This tests a specified window with a specified background, and is not either experiment's discovery significance. If the background mean is allowed to vary freely, background alone can choose $70$ and fit that count exactly. Control samples, a defensible mass-shape model, <nuisance parameters> and corrections for searching many masses are therefore essential to the experimental inference.
Back to article page
