Determine the momenta that can never be measured when a particle has a wave function.

$\mathit{\psi }\mathbf{\left(}\mathit{x}\mathbf{\right)}\mathbf{=}\mathbf{\left\{}\begin{array}{c}\mathbf{C}\mathbf{;}\left|x\right|\mathbf{\le }\mathbf{+}\frac{\mathbf{1}}{\mathbf{2}}\mathbf{w}\\ \mathbf{0}\mathbf{;}\left|x\right|\mathbf{>}\frac{\mathbf{1}}{\mathbf{2}}\mathbf{w}\end{array}\mathbf{\right\}}$

The generalization of the Fourier series is known as Fourier transform and it can also refer to both the frequency domain representation and the mathematical function used. The Fourier transform facilitates the application of the Fourier series to non-periodic functions, allowing every function to be viewed as a sum of simple sinusoids.

The equation of the Fourier transform as,

$\mathit{A}\mathbf{\left(}\mathit{k}\mathbf{\right)}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}\mathbf{\pi }}{\mathbf{\int }}_{\mathbf{-}\mathbf{\infty }}^{\mathbf{+}\mathbf{\infty }}\mathit{\psi }\mathbf{\left(}\mathit{x}\mathbf{\right)}{\mathit{e}}^{\mathbf{i}\mathbf{k}\mathbf{x}}\mathit{d}\mathit{x}$

As given particle has a wave function $\psi \left(x\right)=\left\{\begin{array}{c}\mathrm{C};\left|\mathrm{x}\right|\le +\frac{1}{2}\mathrm{w}\\ 0;\left|\mathrm{x}\right|>\frac{1}{2}\mathrm{w}\end{array}\right\}$, the Fourier transform A(k) is needed.

The equation for momentum p

$\mathit{p}\mathbf{=}\sout{\mathbf{h}}\mathit{k}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{\left(}\mathbf{1}\mathbf{\right)}$

Here, $\sout{\mathbf{h}}$is reduced Planck's constant, and k wave number.

Euler formula will also be used for exponential form to trigonometry form

${\mathit{e}}^{\mathbf{i}\mathbf{\theta }}\mathbf{=}\mathit{c}\mathit{o}\mathit{s}\mathbf{}\mathit{\theta }\mathbf{+}\mathit{s}\mathit{i}\mathit{n}\mathbf{}\mathit{\theta }$

The wave function for the particle into the Fourier equation

$$\begin{gathered} A\left(k\right)=\frac{1}{2\mathrm{\pi }}{\int }_{-\infty }^{+\infty }\psi \left(x\right)e^{\mathrm{ikx}}dx \\ A\left(k\right)=\frac{1}{2\mathrm{\pi }}{\int }_{-\frac{W}{2}}^{+\frac{W}{2}}\left(C\right)e^{-\mathrm{ikx}}dx \\ A\left(k\right)=\frac{C}{2\mathrm{\pi }}\left[{\int }_{-\frac{W}{2}}^{+\frac{W}{2}}\cos \left(kx\right)dx-0\right] \\ A\left(k\right)=\frac{1}{2\mathrm{\pi }}{\int }_{-\frac{W}{2}}^{+\frac{W}{2}}\cos \left(kx\right)dx \end{gathered}$$

Integrate the above expression from the limits using - w / 2 to + w / 2

$$\begin{gathered} A\left(k\right)=\frac{C}{2\mathrm{\pi }}{\int }_{-\frac{W}{2}}^{+\frac{W}{2}}\cos \left(kx\right)dx \\ A\left(k\right)=\frac{C}{2\mathrm{\pi }}{\left[\frac{\sin \left(kx\right)}{k}\right]}_{-\frac{W}{2}}^{+\frac{W}{2}} \\ A\left(k\right)=\frac{C}{2\mathrm{\pi }}\left[\frac{\sin \left({\displaystyle \frac{kw}{2}}\right)}{k}-\frac{\sin \left({\displaystyle -\frac{kw}{2}}\right)}{k}\right] \end{gathered}$$

sin (-ve) Is an odd function, the negative can be pulled out of the $\sin \left(-\frac{kw}{2}\right)$, and simplified

$$\begin{gathered} A\left(k\right)=\frac{C}{2\mathrm{\pi }}\left[\frac{\sin \left({\displaystyle \frac{kw}{2}}\right)}{k}+\frac{\sin \left({\displaystyle \frac{kw}{2}}\right)}{k}\right] \\ A\left(k\right)=\frac{C}{2\mathrm{\pi }}\left(\frac{\sin \left({\displaystyle \frac{kw}{2}}\right)}{k}\right) \end{gathered}$$

Since that is the function for the amplitude in terms of the wave numbers, there are certain values of k for which A(k) will be zero. Those are the wave numbers, for which no particles will be observed.

$\begin{array}{rcl}A\left(k\right)& =& \frac{C}{\mathrm{\pi }}\frac{\sin \left({\displaystyle \frac{kw}{2}}\right)}{k}\\ 0& =& \frac{C}{\mathrm{\pi }}\frac{\sin \left({\displaystyle \frac{kw}{2}}\right)}{k}\\ 0& =& \sin \left(\frac{kw}{2}\right)\end{array}$

The only time that the sine is equal to zero is when the argument of the sine is equal to n times $\mathrm{\pi }$, with n being some integer:

$\begin{array}{rcl}0& =& \sin \left(\frac{kw}{2}\right)\\ n\mathrm{\pi }& =& \frac{\mathrm{kw}}{2}\end{array}$

Solve for k

$\begin{array}{rcl}n\mathrm{\pi }& =& \frac{\mathrm{kw}}{2}\\ \frac{2\mathrm{n\pi }}{\mathrm{w}}& =& \mathrm{k}\\ & & \end{array}$

Substitute $\frac{2\mathrm{n\pi }}{\mathrm{w}}$for k in the equation (1)

role="math" localid="1658424411417" $$\begin{gathered} p=\sout{h}k \\ p=\sout{h}\left(\frac{2n\mathrm{\pi }}{w}\right) \\ p=\frac{2n\mathrm{\pi h}}{w} \end{gathered}$$

Conclusion

Momenta that can never be observed which obey the relation$p=\frac{2n\pi \sout{h}}{w}$