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How uncertainty principle can be used to calculate the range of actual variables

Physics Asked on April 19, 2021

I know the uncertainty relation $Delta x Delta p ge frac{hbar}{2}$ tells the uncertainty in position and momentum, or energy and time should be greater than or equal to $frac{hbar}{2}$, but I don’t understand how this relation can be used to give an estimation on size of system or energy of system.

i.e. When we want to prove that an electron cannot remain inside the nucleus, we use this formula and put the size of nucleus (femtometers) in it and get energy(using momentum calculated from the relation) of electron which contradicts the actual value measured in the experiment, so we conclude that electron cannot reside in the nucleus.

Also we use energy-time uncertainty relation to calculate the time range of exchange particles using their energy values, but I don’t understand how we can do that since uncertainty relation only tells about the uncertainty in variables and not anything about their typical values?

A similar type of question is asked here Why the uncertainty principle can be used for estimation?
but no satisfactory answer is provided

One Answer

The uncertainties in the Heisenberg uncertainty principle are actually standard deviation from the mean. The actual value of a measured quantity is usually probabilistic and so the measured quantity has an associated distribution with it. That distribution would have a mean and standard deviation.

Let's say that the electron is confined to the nucleus. And the boundaries of the nucleus have x coordinates $x_{min}$ and $x_{max}$. Then the probability of measuring the electron at position $x<x_{min}$ and $x>x_{max}$ is zero. So for this electron confined to the nucleus, the minimum value of the distribution is $x_{min}$ and the maximum value is $x_{max}$. Since the standard deviation can never exceed the range of a distribution, a reasonable upper bound on it would be the range itself $x_{max} - x_{min}$. So you can treat the size of the nucleus as an upper bound to $Delta x$. A similar argument could be applied to your other example.

Correct answer by Brain Stroke Patient on April 19, 2021

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