Heisenberg
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Another consequence of unmitigated quantum mechanics is that there are lower limits for the certainty of some measurements. For example, consider a particle P described by a historically ordered repetitive chain of events

$\Psi ^{\sf{P}} = \left( \sf{\Omega}_{1}, \sf{\Omega}_{2} \ \ldots \ \sf{\Omega}_{\it{i}} \ \ldots \ \sf{\Omega}_{\it{f}} \ \ldots \ \right)$

If P is isolated then the elapsed time between events $i$ and $f$ is

$\Delta t = \left( \, f-i \right) \hat{\tau}$

where $\hat{\tau}$ is the period. Consider measuring the elapsed time from observations of event indices and periods. The experimental uncertainty in repeated measurments of $\Delta t$ is written as $\delta t$. By the usual rules for assessing the propagation of this uncertainty is given by

$\delta t = \left( \delta f + \delta i \right) \hat{\tau} + \left( \, f-i \right) \delta \hat{\tau}$

As discussed [ where? ] the uncertainty in the period is bounded by

$\delta \hat{\tau} \ge \hat{\tau} k_{S}$

And some unavoidable uncertainty is also associated with event indices. They are required to be integers, so rounding-off errors are $\delta i \ge 1/2$ and $\delta f \ge 1/2$. Then

$\delta t \ge \hat{\tau} + k_{S} \left(\, f-i \right) \hat{\tau}$

And since $i < f$ we know that $\, f - i \ge 1$ so

$\delta t \ge \left( 1 + k_{S} \right) \hat{\tau}$

or in terms of the energy

$\delta t \ge h \left( 1 + k_{S} \right) /E$

Thus the uncertainty in a time measurement can be decreased by working with high energy particles. In contrast, if $\delta E \ge k_{S} E$ where

$k_{S} = \frac{ 1}{2} \sqrt{ 1 + 1 / \pi \ \vphantom{\Sigma^{2}} } - \frac{ 1}{2}$

then the uncertainty in an energy measurement is increased for larger particles. The two effects cancel for the product of the uncertainties

$\delta E \, \delta t \ge h k_{S} \left( 1 + k_{S} \right)$

leaving a constant

$\delta E \, \delta t \ge h /4\pi$

This is one of uncertainty relationships.

page revision: 348, last edited: 25 Aug 2018 15:32
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