$ \sigma _{H}\sigma _{Q}\geqslant \frac{h}{4\pi }\frac{d\left \langle Q \right \rangle}{dt}$

$\Delta E = \sigma _{H}$

$\Delta t = \frac{\sigma _{Q}}{d\left \langle Q \right \rangle / dt}$

$\Delta E \Delta t \geq \frac{h}{4\pi }$

Q is any observable

I know that $\Delta E$ represents the standard deviation of energy distribution, but what does $\Delta t$ represent precisely?

I read an answer saying "It is the average time of the expectation value to change by one standard deviation, but I don't understand this sentence, I need some clarification.

  • $\begingroup$ Possible duplicates: physics.stackexchange.com/q/53802/2451 and links therein. $\endgroup$ – Qmechanic Sep 27 '14 at 10:08
  • $\begingroup$ @Qmechanic actually this is the same answer found in my book. $\endgroup$ – user3613971 Sep 27 '14 at 10:15
  • $\begingroup$ @Qmechanic I don't understand what does it mean to say"average time for the expectation value to change by one standard deviation" $\endgroup$ – user3613971 Sep 27 '14 at 10:16
  • $\begingroup$ @user3613971 joshphysics' answer gives a short mathematical explanation. $\endgroup$ – Danu Sep 27 '14 at 10:38

As an example, http://en.wikipedia.org/wiki/Particle_decay, you can regard $\Delta t$ as the lifetime of the particle decayed.


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