Question: Figure 38-13 shows a case in which the momentum component

${\mathbf{p}}_{\mathbf{x}}$of a particle is fixed so that $\mathbf{∆}{\mathbf{p}}_{\mathbf{x}}\mathbf{=}\mathbf{0}$ ; then, from Heisenberg’s uncertainty principle (Eq. 38-28), the position x of the particle is completely unknown. From the same principle it follows that the opposite is also true; that is, if the position of a particle is exactly known $\left(∆x=0\right)$, the uncertainty in its momentum is infinite.Consider an intermediate case, in which the position of aparticle is measured, not to infinite precision, but to within a distanceof $\frac{\mathbf{\lambda }}{\mathbf{2}\mathbf{\pi }}$, where $\mathbf{\lambda }$is the particle’s de Broglie wavelength.Show that the uncertainty in the (simultaneously measured) momentumcomponent is then equal to the component itself; that is,$\mathbf{∆}{\mathbf{p}}_{\mathbf{x}}\mathbf{=}\mathbf{p}$. Under these circumstances, would a measured momentumof zero surprise you? What about a measured momentum of $\mathbf{0}\mathbf{.}\mathbf{5}\mathbf{p}$? Of $\mathbf{0}\mathbf{.}\mathbf{2}\mathbf{p}$? Of $\mathbf{12}\mathbf{p}$?

Step by step solution

01

Concept used to solve the question.

Heisenberg’s Uncertainty Principle:

According to Heisenberg’s Uncertainty Principle, the uncertainty in position and momentum are related as,

$∆x∆p_x\ge \frac{h}{2\mathrm{\pi }}$

Where;

$∆x$is uncertainty in position along the axis.

$∆p_x$is uncertainty in momentum along the axis.

$h$is plank’s constant.

02

Finding The uncertainty in momentum 

From Heisenberg’s uncertainty Principle,

$∆x∆p_x\ge \frac{h}{2\mathrm{\pi }}$

For minimum uncertainty

$∆x∆p_x=\frac{h}{2\mathrm{\pi }}$

Let, $∆x=\frac{\lambda }{2\mathrm{\pi }}$

$$\begin{gathered} \frac{\lambda }{2\mathrm{\pi }}∆p_x=\frac{h}{2\mathrm{\pi }} \\ ∆p_x=\frac{h}{\lambda } \end{gathered}$$

From de Broglie relation we know

$p=\frac{h}{\lambda }$

Therefore,

$∆p_x=p$

Hence it is proved that $∆p_x=p$

Yes, measure momentum of zero, $0.5p,2p$and $12p$would all be surprising.

Unlock Step-by-Step Solutions & Ace Your Exams!

• Full Textbook Solutions

  Get detailed explanations and key concepts

• Unlimited Al creation

  Al flashcards, explanations, exams and more...

• Ads-free access

  To over 500 millions flashcards

• Money-back guarantee

  We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!