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Question: The uncertainty in position is equal to the uncertainty in momentum. What is the uncertainty in velocity?

What I did: I know that the uncertainty in position multiplied by uncertainty in momentum is equal to $\frac{h}{2\pi}$ As momentum is mass times velocity, and mass is constant and definite, the uncertainty in momentum is equal to the uncertainty in velocity. Thus the answer should be $$\sqrt{\frac{h}{m2\pi}}$$ However, this is incorrect. Where am I wrong?

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  • $\begingroup$ Comment to the question (v1): Consider rephrasing the sentence Uncertainty in position and momentum are equal, since they are two quantities of different dimensions/units. $\endgroup$
    – Qmechanic
    Commented Jul 13, 2014 at 19:54
  • $\begingroup$ The new version "The uncertainty in position is equal to the uncertainty in momentum" is no different, it is still talking about equality of two quantities of different units. Either you have misunderstood the question or something is missing here. $\endgroup$
    – JiK
    Commented Jul 14, 2014 at 14:11
  • $\begingroup$ But that's what the question says. @JiK $\endgroup$ Commented Jul 14, 2014 at 14:46

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The Heisenberg inequalities reads :

$$\Delta x \Delta p_x \geq \frac{\hbar}{2},$$ where $$\hbar = \frac{h}{2\pi}. $$

Therefore, for a free particle, your add the expressions $$ p_x=mv_x$$ so you get : $$\Delta x \Delta v_x \geq \frac {\hbar}{2m}$$ you have the uncertainty on the velocity : $\Delta v_x \geq \frac{\hbar}{2m\Delta x}$

the Heisenberg inequalities relies two observables. here, $x$ and $p_x$ or $x$ and $v_x$. Be carefull, if the particule is not free, there might be some changes.

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  • $\begingroup$ I don't understand. I have read that the formula is $\frac{h}{2\pi}$?? Is this wrong? And your answer isn't correct either. The correct answer is $\frac{1}{2m}\sqrt{\frac{h}{\pi}}$ $\endgroup$ Commented Jul 13, 2014 at 13:08
  • $\begingroup$ you don't say what is the full formula... So we cannot check. But I can tell you that my formula is correct. $\endgroup$
    – sailx
    Commented Jul 13, 2014 at 13:13
  • $\begingroup$ ?????? What do you mean? $\endgroup$ Commented Jul 13, 2014 at 13:16
  • $\begingroup$ give the full equation. not juste one side. For instance, my formulae is $\Delta x \Delta p_x \geq \frac{\hbar}{2m}$ there is the two sides of the equation $\endgroup$
    – sailx
    Commented Jul 13, 2014 at 13:19
  • $\begingroup$ @salix Same LHS but for the RHS its just $\frac{h}{2\pi}$ $\endgroup$ Commented Jul 13, 2014 at 15:11

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