How does this formula relate to particle creation?

AbsoluteZer0
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Hi,

ΔEΔt ≥\frac{h}{4\pi}

How does this formula relate to particle creation? I understand that it is relevant to the uncertainty principle, but that is essentially all that I am aware of. Does this formula indicate that when a particle with energy 'E' exists for time 't' it then decays?

Thanks,
 
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Both are related to quantum physics, but I don't see a direct relation between particle creation and this uncertainty relation.
For short-living particles, the uncertainty relation gives them a natural width in their mass.
 
What this formula means is that if a particle only exists for a short time ##\Delta t##, then its energy is necessarily uncertain by an amount ##\Delta E = h / 4 \pi \Delta t##. For example, the mass of the rho meson is nominally 770 MeV. However, the rho meson is extremely short-lived: ##\Delta t \approx 4 \times 10^{-24}## seconds, so we actually observe rho mesons with a range of masses, with the extent of that range being about ##\Delta E = 145## MeV. So it wouldn't be uncommon to observe a rho meson with mass 700 MeV, for instance.See http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/parlif.html
 
Insights auto threads is broken atm, so I'm manually creating these for new Insight articles. Towards the end of the first lecture for the Qiskit Global Summer School 2025, Foundations of Quantum Mechanics, Olivia Lanes (Global Lead, Content and Education IBM) stated... Source: https://www.physicsforums.com/insights/quantum-entanglement-is-a-kinematic-fact-not-a-dynamical-effect/ by @RUTA
If we release an electron around a positively charged sphere, the initial state of electron is a linear combination of Hydrogen-like states. According to quantum mechanics, evolution of time would not change this initial state because the potential is time independent. However, classically we expect the electron to collide with the sphere. So, it seems that the quantum and classics predict different behaviours!
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