In most of examples, I notice that uncertainty principle for time & energy is given between mass & lifetime. The UP for time and energy is $$ \Delta t\,\Delta E\geq\frac h{4π} $$ where $$Δt =σ_A \frac{dt}{d\langle \hat A\rangle}$$

with $σ_A$ being the standard deviation of operator $\hat{A}$. But what operator should one define so that $Δt$ becomes the lifetime of a particle?


1 Answer 1


Usually, when you want to probe the lifetime $\tau$ of a particle (or a quasi-particle), what you basically do is looking for the way the associated wave-function $\psi$ is significantly decreasing :


Let us consider a free particle with a given energy $E$. As $\langle E\rangle=E$ is fixed, then the standard deviation of energy falls to zero : $\sigma_H=\Delta E=0$ . According to the uncertainty principle, you have to verify :

$$\Delta E \Delta t \sim \hbar$$

where $\Delta t$ is the caracteristic time it takes to the energy to significantly change.

In our case, it follows that $\Delta t\rightarrow\infty$. Then here you can directly read $\Delta t$ as the lifetime of your particle, which tends to be infinite because your system is in a stationary state (i.e. its energy does not change).

In fact, the "proper" definition of $\Delta t$ :

$$\Delta t = \frac{\sigma_\hat{O}}{|\frac{d\langle\hat{O}\rangle}{dt}|}$$

will give you the lifetime of your particle as the observable $\hat{O}$ allows you to follow the dynamics of the wave function of your system.

But generally, this definition is not very usefull. Usually, what you do is determine the standard deviation in energy $\Delta E$ of your system and then simply say that the associated lifetime is qualitatively : $$\tau\sim\frac{\hbar}{\Delta E}$$ Another possibility is to use the Fermi's Golden Rule that gives you the transition rate $\Gamma$ associated to the particle state you are probing. Then, the lifetime of your particle will simply follow as :


  • $\begingroup$ so by definition there is no operator which can make dt as standard deviation in lifetime $\endgroup$
    – teja4477
    Jul 14, 2014 at 9:30
  • $\begingroup$ No, there is no particular operator for which the definition of a lifetime is exclusive. $\endgroup$
    – dolun
    Nov 10, 2014 at 12:16

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