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I have seen that the more a particle has a high energy, i.e $E$, the more its lifetime is short, respecting so the uncertainty principle.

But by the definition of this uncertainty principle :

$E\,\Delta t \geq \dfrac{\hbar}{2}$, I can write :

$\Delta t \geq \dfrac{\hbar}{2E}$, then $\Delta t$ has a lower limit and not an upper limit.

If this would be an upper limit, this would mean that $\Delta t$, i.e. the apparition time, should be observed in a time interval lower than $\dfrac{\hbar}{2E}$ : for example, if a detector had a time resolution greater than $\dfrac{\hbar}{2E}$, the particle could not be detected, could it ?

So in which case can we write : $\Delta t \leq \dfrac{\hbar}{2E}$ ??

It seems that I have confusions with this principle. Any clarification is welcome

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    Related and also. – Cosmas Zachos Jul 20 '18 at 00:46
  • there is a basic misrepresentation here " high energy, i.e E, the more its lifetime is short" .lifetimes are defined at the rest system of the particle, and the E is the mass of the particle/resonance in that rest frame. In a different frame with a different E it is the lorentz transformation that will define the lifetime in that frame. The HUP is more general than decays, see links given by Zachos – anna v Jul 20 '18 at 15:06
  • Your formula is wrong. E is irrelevant, but the width $\Delta E \sim \Gamma$ is not. Fast-decaying (unstable) particles have a broad width, while slow-decaying (more stable) particles have a narrow width, so $\Gamma \sim 1/\tau$. The faster the particle decays (the shorter its lifetime), the less certain is its mass (the larger its width), as dictated by the Breit-Wigner distribution. – Cosmas Zachos Jul 20 '18 at 16:35
  • Related. – Cosmas Zachos Jul 20 '18 at 18:49
  • @Cosmas Zachos thanks for your answer. Just a last question, when you talk about "fast particle decays (the shorter its lifetime)", you talk about the lifetime of atom excited or lifetime of particles emitted by excited atom (i.e for example gamma rays) ? if it is about lifetime of atom excited, its mass decreases and this variation of mass is equal to $\Delta m = E/c^2$ with E the energy of gamma ray emitted, isn't it ? –  Jul 21 '18 at 11:21
  • "Fast" means fast decay, short lifetime. $\Delta E$ is not the energy of the products!! It is the "slop" in the reconstructed energy of the parent atom or decaying particle. That is the whole point of the discussion: that the parent atom/particle lacks a well-defined energy; it is a wave packet of a range of energies, and $\Delta E$, or $\Gamma$, etc... quantifies this range/spread. Read up on the UP wikipedia article. – Cosmas Zachos Jul 21 '18 at 13:18
  • If the 3 linked answers and the WP article are not enough, you could edit your question to list correct statements and focus on the parts you do not understand. My sense is the middle one really describes decay for you, and the uncertainty principle is mere icing on the cake. – Cosmas Zachos Jul 21 '18 at 15:09

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