In the book on QM by D.J. Griffiths, the time-energy uncertainty relation is in general proved for any observable $Q$ whose operator is not a function of time.

  • Now it has been derived that if uncertainty in energy is $\Delta E$ and the time required for an observable's expectation value to change by its standard deviation be $\Delta T$, then $$\Delta E \Delta T\geq h/4\pi$$

  • Now, my question is,say, for the time evolution of a free particle wavefunction, when the standard deviation itself is a function of time, there how can we interpret this result? Because then we cannot define a time in which the expectation changed by an s.d as the s.d is also a function of time.

  • $\begingroup$ Related: physics.stackexchange.com/q/53802/2451 and links therein. $\endgroup$
    – Qmechanic
    Commented Oct 30, 2017 at 7:34
  • $\begingroup$ What do you mean when you say for a free particle "the standard deviation itself is a function of time", are you modeling a free particle as as a wave packet in QFT ? I would think that the uncertainty volume defined by the relation would be a function of time. Whats the problem? $\endgroup$
    – anna v
    Commented Oct 30, 2017 at 11:40
  • $\begingroup$ Actually i am asking about the time evolution of some free particle wavefunction,where s.d is a function of time $\endgroup$
    – user157588
    Commented Oct 30, 2017 at 11:41
  • $\begingroup$ Do you have a link for the wave function of this free particle? $\endgroup$
    – anna v
    Commented Oct 30, 2017 at 12:05
  • $\begingroup$ I can give pics from griffiths $\endgroup$
    – user157588
    Commented Oct 30, 2017 at 12:06

1 Answer 1


Quoting Griffith,

$\Delta t$ represents the amount of time it takes the expectation value of $Q$ to change by one standard deviation.

Your questions stems from the fact that Griffith did not spell out the times at which each quantity is evaluated in that phrase.

First, the standard deviation always depend on time in a demonstration such as this one since $\sigma_Q^2 = \langle Q^2 \rangle - \langle Q \rangle^2$ and $\langle Q \rangle$ depends on time, and so does $\langle Q^2 \rangle$, a priori, and there is no reason to postulate that those dependencies cancel.

Then in the phrase I quoted above, you have to read:

  • the change of the expectation value from time $t$ to time $t+\Delta t$; and
  • the standard deviation at time $t$.

It is quite clear if we look at the formula this quote comments:

$$\sigma_Q = \left|\frac{d\langle Q\rangle}{dt}\right|\Delta t$$

From the time $t$ to the time $t+\Delta t$, $\langle Q\rangle$ changes by $\left|\frac{d\langle Q\rangle}{dt}\right|\Delta t$ if $\Delta t$ is small enough: this is a classic linear approximation. And then we compare that to $\sigma_Q$ at time $t$, since this formula is correct only if $\sigma_Q$ is that at time $t$, as should be clear from the demonstration in Griffith, or from this answer to the question quoted by @Qmechanic in a comment to your question.


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