The Heisenberg uncertainty principle: Interpreting $\Delta p$, $\Delta t$, etc

(1) I have a textbook question that states the following:

An electron has a speed of 500 m/s with an accuracy of 0.004%. Calculate the certainty with which we can locate the position of the electron.

The solution starts with $$\frac{\Delta p}{p}=\frac{0.004}{100}$$

I don't understand how the question is being interpreted to give the above expression.

(2) The very next question says:

The average lifetime of an excited atomic state is $10^{-9}$s. If the spectral line associated with the decay of this state is 6000 Angstrom, estimate the width of the line.

The solution begins with $\Delta t = 10^{-9}$s.

Again, I don't understand how this is arrived at. I'd expect it to be $t=10^{-9}$s.

Can somebody give me an intuitive understanding of uncertainty (perhaps in words) that I can apply to these two problems? I know the statistical definition, but I haven't found it very helpful in this context.

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Usually "The uncertainty principle" refers to the "Heisenberg Uncertainty Principle" en.wikipedia.org/wiki/Heisenberg_uncertainty_principle . The HUP is a fundamental uncertainty imposed by the behavior of quantum mechanical operators on certain pairs of variables. Maybe you should edit the title. – anna v Oct 19 '12 at 4:07
I think I understood my problem with the 1st part. The word "accuracy" is used in a counter-intuitive way such that 0% accuracy would mean zero uncertainty in the momentum. Ordinarily, one would expect this to be called 100% accuracy. – Joebevo Oct 19 '12 at 5:08

Very roughly speaking, the uncertainty $\Delta p$ can be interpreted as follows: Make a billion particles with a momentum p and an uncertainty $\Delta p$. Measure the momentum of each particle. More times than not, the particles will be observed to have a momentum between $p-\Delta p$ and $p+\Delta p$.

In fact if p has a normal distribution, you will find that 68% of the time you will find it in this interval.

You can think of $\Delta t$ as the charachteristic time scale in which the excited atom will stay excited.

Dividing an uncertainty by its variable just gives us a way to quote a dimensionless uncertainty.

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What confuses me is that $\Delta t$ ordinarily refers to a spread (from the mean) in the measured values. However, the characteristic timescale is something like the mean value itself, right? – Joebevo Oct 19 '12 at 5:16
Yes, I think you have the right intuition. The difference between delta t and delta x or delta p is that time is not an observable in quantum mechanics. In other words, there is no hermitian operator that gives you time. – hwlin Oct 19 '12 at 13:41

You are discussing the meaning of statistical uncertainty, something that can be calculated or measured in an experiment.

When given a delta, usually called measurement error, of a quantity you are given a measure on how sure you can be on your measurement.

In the case of the momentum whenever you measure p you can only be sure that the value you found is true within +/- one sigma which will be four percent of the value , as this is what you have been given as the error behavior.

In the case of the atomic state you are given the width of the decay line. The width of the decay line depends on the properties of the decaying atomic configuration and will not change according to the clock time you are using. This is within the given of the problem.

So if you decide to measure the exact time an atom has decayed your measurement will be precise to 1 sigma of the width you have been given for the line, +/- 10^-9 seconds.

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