Minimum-uncertainty wave function -- contradiction?

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SUMMARY

The discussion centers on the minimum-uncertainty wave function represented by equation (5.122), which is identified as a Gaussian wave packet. The relationship between position uncertainty (Δx) and momentum uncertainty (Δpx) is established through the equation ΔxΔpx = ħ/√2, which contradicts the established minimum uncertainty principle stating ΔxΔpx = ħ/2. The correct approach to derive Δx involves calculating the probability density |ψ(x)|² from the Gaussian distribution rather than directly from the wave function ψ(x).

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The following equation (5.122) is the minimum-uncertainty wave function, which is a Gaussian wave packet. Since it is Gaussian in ##x##, we may get ##\Delta x## directly from the ##\sigma## of the Gaussian distribution: ##(\Delta x)^2=\frac{\hbar^2}{2(\Delta p_x)^2}##. We have ##\Delta x\Delta p_x=\frac{\hbar}{\sqrt2}##, which contradicts the fact that the wave function has minimum uncertainty, ie., ##\Delta x\Delta p_x=\frac{\hbar}{2}##.

Derivation of (5.122):
Screen Shot 2015-12-23 at 5.08.19 am.png

Screen Shot 2015-12-23 at 5.08.42 am.png


Gaussian distribution:
Screen Shot 2015-12-23 at 5.09.04 am.png
 
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You should first calculate ##|\psi(x)|^2## and then extract ##\Delta x## from there, not from ##\psi(x)##
 
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