0
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ 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 Jul 13 '14 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 Jul 14 '14 at 14:11
  • $\begingroup$ But that's what the question says. @JiK $\endgroup$ – Gummy bears Jul 14 '14 at 14:46
1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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 $\frac{1}{2m}\sqrt{\frac{h}{\pi}}$ $\endgroup$ – Gummy bears Jul 13 '14 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 Jul 13 '14 at 13:13
  • $\begingroup$ ?????? What do you mean? $\endgroup$ – Gummy bears Jul 13 '14 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 Jul 13 '14 at 13:19
  • $\begingroup$ @salix Same LHS but for the RHS its just $\frac{h}{2\pi}$ $\endgroup$ – Gummy bears Jul 13 '14 at 15:11

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.