Question

Question: Figure 38-13 shows that because of Heisenberg’s uncertainty

principle, it is not possible to assign an $\mathit{x}$coordinate to the positionof a free electron moving along an $\mathit{x}$axis. (a) Can you assign a $\mathit{y}$or a $\mathit{z}$coordinate? (Hint:The momentum of the electron has no $\mathit{y}$or $\mathit{z}$component.) (b) Describe the extent of the matter wave in threedimensions.

Short answer

(a) It is not possible to assign $y$and $z$coordinates.

(b) It describes the extent of the matter wave in three dimensions.

Step-by-step solution

Concept used to solve the question.

Heisenberg’s Uncertainty Principle

According to Heisenberg’s uncertainty principle, it is not possible to measure the position and the momentum of a particle simultaneously with

unlimited precision.

$\mathbf{∆}\mathbf{r}\mathbf{∆}\mathbf{p}\mathbf{=}\frac{\mathbf{h}}{\mathbf{2}\mathbf{\pi }}$

Where, $\mathit{r}$is the position and $\mathit{p}$is momentum

Finding weather we can assign a y or a z coordinate or not

From the figure, we can see that the probability density function does not have any $y$and $z$components

So, from the figure, we can see that

$p_y=p_z=0$

This implies,

$∆p_y=∆p_z=0$

Therefore, from Heisenberg’s Uncertainty principle $∆y$and $∆z$both are infinite.

Therefore, it is not possible to assign $y$and $z$coordinates.

Finding the extent of matter wave.

Since probability density ${\left|\psi \left(x\right)\right|}^2$is independent from $y$and $z$,therefore the wave function $\psi \left(x\right)$describes a wave extending infinitely along $y$axis and $z$ axis it also has $x$component, therefore describes the extent of the matter wave in three dimensions.

Hence the density function describes the extent of the matter wave in three dimensions.