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Why don’t we notice Heisenberg’s uncertainty principle in everyday life? The answer is that Planck’s constant is very small. Thus the lower limit in the uncertainty of measuring the position and momentum of large objects is negligible. We can detect sunlight reflected from Jupiter and follow the planet in its orbit around the Sun. The reflected sunlight alters the momentum of Jupiter and creates an uncertainty in its momentum, but this is totally negligible compared with Jupiter’s huge momentum. The correspondence principle tells us that the predictions of quantum mechanics become indistinguishable from classical physics for large objects, which is the case here.
There is another form of Heisenberg’s uncertainty principle for simultaneous measurements of energy and time . In equation form,
where $\mathrm{\Delta}E$ is the uncertainty in energy and $\mathrm{\Delta}t$ is the uncertainty in time . This means that within a time interval $\mathrm{\Delta}t$ , it is not possible to measure energy precisely—there will be an uncertainty $\mathrm{\Delta}E$ in the measurement. In order to measure energy more precisely (to make $\mathrm{\Delta}E$ smaller), we must increase $\mathrm{\Delta}t$ . This time interval may be the amount of time we take to make the measurement, or it could be the amount of time a particular state exists, as in the next [link] .
An atom in an excited state temporarily stores energy. If the lifetime of this excited state is measured to be ${\text{1.0\xd710}}^{-\text{10}}\phantom{\rule{0.25em}{0ex}}\mathrm{s}$ , what is the minimum uncertainty in the energy of the state in eV?
Strategy
The minimum uncertainty in energy $\mathrm{\Delta}E$ is found by using the equals sign in $\mathrm{\Delta}E\mathrm{\Delta}t\ge h\text{/4}\pi $ and corresponds to a reasonable choice for the uncertainty in time. The largest the uncertainty in time can be is the full lifetime of the excited state, or $\mathrm{\Delta}t={\text{1.0\xd710}}^{-\text{10}}\phantom{\rule{0.25em}{0ex}}\mathrm{s}$ .
Solution
Solving the uncertainty principle for $\mathrm{\Delta}E$ and substituting known values gives
Now converting to eV yields
Discussion
The lifetime of ${\text{10}}^{-\text{10}}\phantom{\rule{0.25em}{0ex}}\text{s}$ is typical of excited states in atoms—on human time scales, they quickly emit their stored energy. An uncertainty in energy of only a few millionths of an eV results. This uncertainty is small compared with typical excitation energies in atoms, which are on the order of 1 eV. So here the uncertainty principle limits the accuracy with which we can measure the lifetime and energy of such states, but not very significantly.
The uncertainty principle for energy and time can be of great significance if the lifetime of a system is very short. Then $\mathrm{\Delta}t$ is very small, and $\mathrm{\Delta}E$ is consequently very large. Some nuclei and exotic particles have extremely short lifetimes (as small as ${\text{10}}^{-\text{25}}\phantom{\rule{0.25em}{0ex}}\text{s}$ ), causing uncertainties in energy as great as many GeV ( ${\text{10}}^{9}\phantom{\rule{0.25em}{0ex}}\text{eV}$ ). Stored energy appears as increased rest mass, and so this means that there is significant uncertainty in the rest mass of short-lived particles. When measured repeatedly, a spread of masses or decay energies are obtained. The spread is $\mathrm{\Delta}E$ . You might ask whether this uncertainty in energy could be avoided by not measuring the lifetime. The answer is no. Nature knows the lifetime, and so its brevity affects the energy of the particle. This is so well established experimentally that the uncertainty in decay energy is used to calculate the lifetime of short-lived states. Some nuclei and particles are so short-lived that it is difficult to measure their lifetime. But if their decay energy can be measured, its spread is $\mathrm{\Delta}E$ , and this is used in the uncertainty principle ( $\mathrm{\Delta}E\mathrm{\Delta}t\ge h\text{/4}\pi $ ) to calculate the lifetime $\mathrm{\Delta}t$ .
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