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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

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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.

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  • $\begingroup$ I don't understand the line "a reasonable upper bound on it would be the range itself $x_max - x_min$" you mean to say that range will have same order as of the variable? $\endgroup$
    – smallest quanta
    Commented Jul 30, 2020 at 10:30
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    $\begingroup$ I meant to say standard deviation $\le$ range. $\endgroup$
    – Brain Stroke Patient
    Commented Jul 30, 2020 at 10:43
  • $\begingroup$ I got your point, but in the uncertainty relation we use lower bound rather than upper bound i.e. $\Delta x \Delta p = \frac{\hbar}{2}$ $\endgroup$
    – smallest quanta
    Commented Jul 30, 2020 at 11:00
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    $\begingroup$ Because you don't know the probability distribution of x inside the nucleus, the standard deviation can be anything between 0 and the range, i.e., $0 \le \Delta x \le x_{max} - x_{min}$. To find the lowest possible value of $\Delta p$ in a state of minimum uncertainty (when $\Delta x \Delta p = \frac{\hbar}{2}$), you want the highest possible value of $\Delta x$ which in this case would be the range. But even the lowest value of $\Delta p$ gives results far greater than experimental results so that you can conclude that $\Delta x$ must be much larger than the size of the nucleus. $\endgroup$
    – Brain Stroke Patient
    Commented Jul 30, 2020 at 11:21
  • $\begingroup$ Thankyou I completely understand it now. $\endgroup$
    – smallest quanta
    Commented Jul 30, 2020 at 12:06

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