## Delta x

$\Delta p \Delta x\geq \frac{h}{4\pi }$

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Ruiting Jia 4D
Posts: 65
Joined: Fri Sep 28, 2018 12:27 am

### Delta x

If the uncertainty equation is delta p * delta x is greater than or equal to 1/2 h/2pi, then can we combine m*delta x (delta p) with delta x so that it's m*delta x^2?

caseygilles 1E
Posts: 73
Joined: Fri Sep 28, 2018 12:18 am

### Re: Delta x

Well, the change in momentum can be replaced with mass times the change in velocity. Momentum is not mass times position (m*x) as you are referring to it. So, the proper way to substitute for uncertainty in momentum would be m*delta v, not m*delta x. Therefore, the total equation would be (m*delta V)(delta X) >= h/4pi

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