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I have read the following sentence in a section which is explaining why particle accelerators need such high energy:

As we require high precision position measurement, the Heisenberg uncertainty principle tells us that the standard deviation of momentum $\sigma_p$ must be large. The standard deviation of momentum of a particle can never be greater than its momentum $p$. Therefore, high momentum is required.

Could someone explain this relationship (bold sentence)? I know that the Heisenberg uncertainty principle makes a statement about the fundamental knowability of position and momentum of a particle, but I don't know the relationship between its actual momentum and the standard deviation of momentum.

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For any measurement there is no general relationship between the measurement (average) itself and the uncertainty (standard deviation) of that measurement. The bold statement is about measurements in general.

We usually report a measurement as $$p\pm\Delta p$$ where $p$ represents an average of our measurements, and $\Delta p$ is related to the standard deviation of our measurements. This expression is essentially saying "I am fairly confident that the value we are trying to measure is somewhere within this range from $p-\Delta p$ to $p+\Delta p$.

So, if $\Delta p>p$, then we are saying that our actual value could be in a range that will extend below $0$ and more than double what our average is. This is not good.

You can also look at the fractional uncertainty $\frac{\Delta p}{p}$. If $\Delta p>p$, then the fractional uncertainty is larger than $1$. Once again, this is an issue, as we are putting our actual measurement in a range that covers a large interval compared to our average.

So really if your uncertainty is larger than the average, then you basically can't say anything useful about your measurement.

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The misconception here is that there is an actual momentum, and not a superposition of momenta. Moreover, the average momentum (or expectation value) is not relevant.

What the book is saying is that if $\Delta p$ is large, then there must be high momentum states in the actual state. In fact, the lowest momentum wave function with $\Delta p$ is:

$$ \psi(p) = \frac{1}{\sqrt 2}[\delta(\Delta p) \pm \delta(-\Delta p)] $$

which is a superposition of $p = \pm\Delta p$ eigenstates, which exactly confirms what the book says. Any other state with a standard deviation of $\Delta p$ (and mean $\bar p = 0$) will have even higher momentum components.

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