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Just a short question regarding an interpretation of the Heisenberg uncertainty principle $$\sigma_x \sigma_p \geq \frac{\hbar}{2}$$.

Question:

  • The uncertainty principle is sometimes written as $\displaystyle{\Delta x \Delta p \geq \frac{\hbar}{2}}$, where to calculate $\Delta x$ (the position uncertainity), we could consider $\Delta p \approx mv$. Why is this a valid approximation for the standard deviation of momentum? An example of this is in Zettili's book "Quantum Mechanics" where an example states:

Estimate the uncertainty in the position of a neutron moving at $5 \times 10^6 m s^-1$. The solution is given by first stating $$\Delta x \geq \frac{\hbar}{2 \Delta p} \approx \frac{\hbar}{2 m_n v}.$$

Why is this a valid approximation? Thanks for any assistance.

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For a neutron of that speed, the uncertainty in the momentum is expected to be less than the momentum magnitude. Using the actual momentum will be an upper bound on the momentum uncertainty. That correlates to a lower bound on the position uncertainty. So, $\Delta x$ is lower bounded by $\hbar/(2p)$: $$\Delta x \ge \frac{\hbar}{2m_nv}.$$

$\Delta x$ could have an even larger lower bound if one knows more about the momentum uncertainty. This is simply a first estimate.

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  • $\begingroup$ Are you Bill Nye (the science guy)? Thanks that makes sense. Would have probably been clearer if the author just used "$\geq$" instead of "$\approx$". $\endgroup$ – user100411 May 2 '16 at 16:22
  • $\begingroup$ No, not Nye. Yeah, another $\ge$ would have been better. $\endgroup$ – Bill N May 2 '16 at 17:30
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Bill has given very nice answer. I will just add some little formalism. We know that neutron speed as well as it's momentum is already measured. This means that momentum uncertainty must be small, on the order $ \frac{\Delta p}{p} \ll 1 $.
Thus it will not make any harm to assume that upper bound is $\frac{\Delta p}{p} \approx 1$, so $\Delta p \approx p = mv$ for upper momentum uncertainty bound. This gives neutron momentum uncertainty bounds as : $$ \left[\Delta p \ll mv; mv \right] $$

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Is it about if is $ Δp≈mv $ a good aproximattion, because is it a no relativistic one? It is, because really $ mv_x $ is a least-upper-bound for the real expressión $ \frac{mv_x}{1-\sqrt{v^2/c^2}} $

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  • $\begingroup$ Heisenberg uncertainty principle has nothing to do with relativity, just with measurement(s) of particle properties, like speed, position, etc. $\endgroup$ – Agnius Vasiliauskas Jan 8 '20 at 7:40

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