In HEP experiments, a particle of interest (e.g. Higgs) is reconstructed using information about decay products, momentum, energy deposits, angle of those particles, etc. The invariant mass of the said particle is calculated using relativistic kinematics. We end up with a histogram of mass vs. number of events, such as the one shown below.

My question is, if the invariant mass of a particle is its unique property, why does it have a "width" of a noticeable $\Delta m$ in the histogram?

(I guess $\pm$few GeVs comes from various sources of uncertainties. But some reconstructed mass distributions have $\pm$tens of GeVs, which is what I do not quite understand.)

enter image description here


3 Answers 3


There are two contributions to the width of a measured peak:

Detector resolution: like you hint in your question, measured quantities come with uncertainties. The momentum resolution of the trackers and energy resolution of the calorimeters will combine to "smear out" an invariant mass distribution.

Momentum resolution is strongly related to the spatial resolution of the tracking system, since momentum is inferred from the curvature of a charged particle's trajectory in a magnetic field. More closely-spaced sensors will typically give better resolution.

Energy resolution is typically parametrised as:

$$\frac{\sigma_E}{E} = \frac{A}{\sqrt{E}} + B + \frac{C}{E}$$

where $A$ is due to random fluctuations in the shower and sampling, $B$ is due to inhomogeneities, imperfections etc, and $C$ is due to electronic noise, radioactivity etc.

Natural width of the particle: even with zero resolution, all unstable particles have some width in invariant mass, which is inversely proportional to its lifetime: $$\Gamma = \frac{\hbar}{\tau}$$ you can think of this in terms of the time-energy uncertainty principle: $\Delta E \Delta t > \frac{\hbar}{2}$. The invariant-mass distribution of decaying particles is referred to as its resonance lineshape (since you see the same shape in scattering experiments). A typical function to parametrise a lineshape is a relativistic Breit-Wigner function: $$ T (m) \propto \frac{1}{m_0^2 - m^2 - im_0\Gamma}, $$ The width can itself have some dependence on mass and angular momentum. A typical function is: $$ \Gamma(m) = \Gamma_0 \left(\frac{q}{q_0}\right)^{2J+1} \left(\frac{m_0}{m}\right) B^\prime_J(q)^2 $$ where $B^\prime$ is a Blatt-Weisskopf barrier factor.

The way to model a peak in a fit depends on how large the natural width is compared to the resolution. If the width is much larger than the resolution, then you can just use a function to describe the resonance lineshape (e.g. a Breit-Wigner). If the resolution is much larger than the width (as with Higgs measurements at the LHC) then you just use a function that models the resolution, typically a Gaussian distribution (or sum of several Gaussians) or some variation thereon to account for e.g. radiative tails. If the resolution is similar to the width, then you have to convolve the lineshape with the resolution function. The convolution of a Breit-Wigner function with a Gaussian function is called a Voigtian function. Sometimes it's too complicated to obtain the analytical convolution, so numerical methods are used, e.g. FFTs.


There are two factors:

  1. The Higgs boson has a small intrinsic width, $\Gamma \approx 4 \, \text{MeV}$. This occurs because the Higgs need not be on mass-shell $p^2 = m_h^2$ when it is a virtual particle
  2. We cannot resolve the invariant mass of the decay products that well, resulting in a smearing of the resonance feature

The intrinsic width is substantially smaller than the resolution at the moment, which as about $1.5\, \text{GeV}$ in the diphoton channel. So predominantly the latter one results in the width of the resonance feature you see plotted.


(I guess ±few GeVs comes from various sources of uncertainties. But some reconstructed mass distributions have ±tens of GeVs, which is what I do not quite understand.)

In the plot you give, the experimental resolution is not +/- tens of GeV's, as you state. The experimental resolution is the width at half maximum of the gaussian ( classical statistics). In the figure it is by eye about 2 GeV.

It is a mistake to consider the whole spread of the red curve on the left as the experimental width/resolution. The curve is also a classical probability distribution, and it is improbable that a Higgs with a standard model decay width of 4 MeV will have such measurement errors in the decay products as to end up into the tails of the distribution in GeV.

Some decay channels have very large errors and the narrow width resonance has a probability of statistical error much larger than cleaner channels experimentally and theoretically ( extra interactions in decay products destroy resolution, jets have large errors associated with them etc.)

From the particle data group:

For the case of the Standard Modelthe prediction for the total width is about 4 MeV, which is three orders of magnitude smaller than the experimental mass resolution.

The combined error on the measured Higgs mass in the tables is given as $+/- 0.16$ GeV

Because of the complexity of the errors entering into the statistical analysis Monte Carlo simulated events are used to fit the measured curves, the MC has in all the known errors , measurement and theoretical, in fitting the mass curves.


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