A bedrock topic in quantum mechanics is the uncertainty principle. It is discussed mostly for massive objects in Chapter 4, but the idea also applies to light: Increasing certainty in knowledge of photon position implies increasing uncertainty in knowledge of its momentum, and vice versa. A single-slit pattern that is developed (like the double-slit pattern of Section 3.6) one photon at a time provides a good example. Depicted in the accompanying figure, the pattern shows that pho tons emerging from a narrow slit are spreadall-over; a photon's $\mathbf{x}$-component of momentum can be any value over a broad range and is thus uncertain. On the other hand, the $\mathbf{x}$ -coordinate of position of an emerging photon covers a fairly small range, for $\mathbf{w}$ is small. Using the single-slit diffractionformula $\mathbf{n\lambda }\mathbf{=}\mathbf{wsin\theta }$ , show that the range of likely values of ${\mathbf{p}}_{\mathbf{x}}$, which is roughly $\mathbf{psin\theta }$ , is inversely proportional to the range $\mathbf{w}$ of likely position values. Thus, an inherent wave nature implies that the precisions with which the particle properties of position and momentum can be known are inversely proportional.

Step by step solution

01

Given expressions

The single –slit diffraction formula$\mathrm{n\lambda }=\mathrm{wsin\theta }$whererole="math" localid="1660042499183" $\mathrm{n}$is the number of node and$\lambda$is the wavelength and$w$ is the width.

${\mathrm{p}}_{\mathrm{x}}=\mathrm{psin\theta }$

02

Concept de Broglie formula

The de Broglie formula for the relation between the momentum and the wavelength.

$\mathbf{wsin}\mathbf{\left(}\mathbf{\theta }\mathbf{\right)}\mathbf{=}\mathbf{n\lambda }$

Here,$\mathbf{w}$ is the width,$\mathbf{n}$ is number of node,$\mathbf{\lambda }$ is wave length and$\mathbf{sin}\mathbf{\left(}\mathbf{\theta }\mathbf{\right)}$ is the nature of wave.

03

Evaluate the relationship

Consider the de Broglie formula for the relation between the wavelength and the momentum.

$\begin{array}{c}\mathrm{wsin}\left(\mathrm{\theta }\right)=\mathrm{n\lambda }\\ \mathrm{p}=\frac{\mathrm{h}}{\mathrm{\lambda }}\\ \mathrm{wsin}\left(\mathrm{\theta }\right)=\mathrm{n}\frac{\mathrm{h}}{\mathrm{p}}\\ \mathrm{psin}\left(\mathrm{\theta }\right)={\mathrm{p}}_{\mathrm{x}}\end{array}$

Solve further as:

$\begin{array}{c}\mathrm{psin}\left(\mathrm{\theta }\right)=\frac{\mathrm{nh}}{\mathrm{w}}\\ {\mathrm{p}}_{\mathrm{x}}\propto \frac{1}{\mathrm{w}}\end{array}$

Therefore, the momentum in the $x$-direction is inversely proportional to the slit width. This is because the particle position uncertainty decreases (smaller slit width), that corresponds to the larger the uncertainty in its momentum.

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