In this Coursera lecture by Mercedes Paniccia on inelastic scattering of electrons off hadrons, at 1:45 she says that

At moderate $q^2$, inelastic processes produce excited states of the nucleon, also called resonances, such as $\Delta^+$, which have the same quantum numbers as the proton. These resonances have an extremely short lifetime and therefore a broad mass distribution.

I assume that the broad mass distribution means that the invariant mass of the produced resonance can assume values over a broad range. Why does the resonance lifetime being short imply that its mass distribution is broad?


A typical propagator in QFT at tree level looks like $(p^2-M^2)^{-1}$. Naively, this tells you that if the four-momentum of your virtual particle becomes on shell, $p^2=M^2$, then the amplitude for this process diverges. However, if you take into account loops in your scattering processes, you can arrive at the famous Breit-Wigner distribution, which states the the propagator gets corrected to


where $\Gamma$ is the total decay rate of your virtual particle. Squaring this gives you cross section which will look like


Thus, our original divergent peak at $p^2=M^2$ has been broadened. The full-width the distribution of $\sigma$ against $\sqrt{p^2}$ at half-maximum is given by $\Gamma$. Thus, if the decay rate is very high (the lifetime is very short), then the width is very wide, and the distribution of the cross section against $\sqrt{p^2}$ has a very broad peak (equivalently, a very broad mass distribution).

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  • $\begingroup$ I suspect the OP is only asking about the connection of τ to this... he is asking about the connection to the rate he probably skipped in his particle physics course. $\endgroup$ – Cosmas Zachos Jul 4 '17 at 13:29

By the generalization of Heisenberg's uncertainty principle, $\Delta E \Delta t \geq \frac{\hbar}{2}$ and so if the lifetime is very short and well defined, the energy distribution is badly defined (the masses are broad).

  • 2
    $\begingroup$ I am not convinced this really explains anything (though I may well be mistaken): since time (nor lifetime, which is not the same thing of course) is not an observable, this is not really an instance of the Heisenberg uncertainty principle. There are somewhat similar results involving energy and time that are often cast in the same form, but their meaning is not exactly the same. $\endgroup$ – doetoe Jul 3 '17 at 23:53
  • $\begingroup$ It's not a very technical explanation but I think the main concept emerges from this fundamental property, because the squeezing of an excited state in the time dimension must make it energy distribution very broad. And I think the existence of this particle is an observable which gives pretty much the lifetime and so it is related to the energy of the state. Please tell me what we don't agree on! $\endgroup$ – gingras.ol Jul 4 '17 at 4:32
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    $\begingroup$ It is not an observable in the technical sense that it doesn't have an associated set of eigenstates that form a basis for the full state space, and which is the context in which the general uncertainty principle applies. The notation of the energy-time uncertainty principle is suggestive, but since ∆t doesn't have the same kind of meaning as the corresponding quantity for an observable, i don't feel comfortable to draw quick conclusions from it. That doesn't mean it cannot be done, or that it cannot serve as a heuristic aid. $\endgroup$ – doetoe Jul 4 '17 at 7:16

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