Question

Consider the wave function$\mathbf{\Psi }\left(x,0\right)\mathbf{=}\{\begin{array}{l}\frac{1}{\sqrt{2\mathrm{v\lambda }}}{e}^{i2\mathrm{\pi x}/\lambda },-\mathrm{n\lambda }<x<\mathrm{n\lambda }\\ 0,\end{array}$

where$\mathbf{n}$is some positive integer. This function is purely sinusoidal (with wavelength$\mathbf{\lambda }$)on the interval$\mathbf{-}\mathbf{n\lambda }\mathbf{<}\mathbf{x}\mathbf{<}\mathbf{n\lambda }$, but it still carries a range of momenta, because the oscillations do not continue out to infinity. Find the momentum space wave function$\mathbf{\Phi }\left(p,0\right)$. Sketch the graphs of${\left|\Psi (x,0)\right|}^2$and${\left|\Phi (p,0)\right|}^2$, and determine their widths,${\mathbf{w}}_{\mathbf{x}\mathbf{}}$and${\mathbf{w}}_{\mathbf{p}}$(the distance between zeros on either side of the main peak). Note what happens to each width as$\mathbf{n}\mathbf{\to }\mathbf{\infty }$. Using${\mathbf{w}}_x$and${\mathbf{w}}_{\mathbf{p}}$as estimates of$\mathbf{∆}\mathbf{x}$and$\mathbf{∆}\mathbf{p}$, check that the uncertainty principle is satisfied. Warning: If you try calculating${\mathbf{\sigma }}_{\mathbf{p}}$, you're in for a rude surprise. Can you diagnose the problem?

Short answer

The widths are ${\mathrm{w}}_{\mathrm{x}}=2\mathrm{\lambda }$and ${\mathrm{w}}_{\mathrm{p}}=2\mathrm{\pi }\sout{\mathrm{h}}/\mathrm{n\lambda }$.

Step-by-step solution

The momentum-space representation and the probability densities in the coordinate space and momentum.

The momentum-space representation is connected to the coordinate representation by Fourier transform:

$\mathbf{\Phi }\mathbf{\left(}\mathbf{p}\mathbf{\right)}\mathbf{=}\frac{\mathbf{1}}{\sqrt{\mathbf{2}\mathbf{\pi }\sout{\mathbf{h}}}}{\mathbf{\int }}_{\mathbf{-}\mathbf{\infty }}^{\mathbf{\infty }}\mathbf{\psi }\mathbf{\left(}\mathbf{x}\mathbf{\right)}{\mathbf{e}}^{\mathbf{-}\mathbf{ipx}\mathbf{/}\sout{\mathbf{h}}}\mathbf{dx}$

In this case, the limits of the integration shrink since the wave function in the coordinate representation vanishes outside the certain region. Therefore:

$\mathrm{\Phi }\left(\mathrm{p}\right)=\frac{1}{\sqrt{2\mathrm{\pi }\sout{\mathrm{h}}}}\frac{1}{\sqrt{2\mathrm{\pi \lambda }}}{\int }_{-n\lambda }^{n\lambda }{\mathrm{e}}^{-\mathrm{i}2\mathrm{\pi x}/\mathrm{\lambda }}{\mathrm{e}}^{-\mathrm{ipx}/\sout{\mathrm{h}}}\mathrm{dx}$

Now try to evaluate the given integral.

$$\begin{gathered} \mathrm{\Phi }\left(\mathrm{p}\right)=\frac{1}{\sqrt{2\mathrm{\pi }\sout{\mathrm{h}}}}\frac{1}{\sqrt{2\mathrm{\pi \lambda }}}{\int }_{-n\lambda }^{n\lambda }{{\mathrm{e}}^{\mathrm{i}2\mathrm{\pi x}/\mathrm{\lambda }}}^{-\mathrm{ipx}/\sout{\mathrm{h}}}\mathrm{dx} \\ =\frac{1}{\sqrt{2\mathrm{\pi }\sout{\mathrm{h}}}}{\overline{)\frac{{{\mathrm{e}}^{\mathrm{i}2\mathrm{\pi x}/\mathrm{\lambda }}}^{-\mathrm{ipx}/\sout{\mathrm{h}}}}{{{}^{\mathrm{i}2\mathrm{\pi x}/\mathrm{\lambda }}}^{-\mathrm{ipx}/\sout{\mathrm{h}}}}}}_{-n\lambda }^{n\lambda } \end{gathered}$$

The latter expression after inserting the limits of integration and by the use the exponential form of the sine function can be simplified to:

$\Phi \left(p\right)=\sqrt{\frac{\sout{h\lambda }}{n\pi }}\frac{\sin np\lambda \left(np\lambda /\sout{h}\right)}{p\lambda -2\pi \sout{h}}$

Therefore, the probability densities in the coordinate space and momentum space are:

${\left|\psi \left(x\right)\right|}^2=\frac{1}{2n\lambda }$

And outside the given region

${\left|\Phi \left(p\right)\right|}^2=\frac{\sout{h\lambda }}{n\pi }\frac{\sin np\lambda \left(np\lambda /\sout{h}\right)}{\left(p\lambda -2\pi \sout{h}\right)}$

And the corresponding graph for the coordinate space density:

Plot the graph for momentum space density.

This graph is for momentum space density.

Calculate the widths.

Now, calculate the widths. For the coordinate space representation, the job is trivial, since the function spreads only in the region $\left[-n\lambda ,n\lambda \right]$therefore the total width is $w_x=2n\lambda$. As for the momentum space function, the peak is centered at $p=2\pi \sout{h}/\lambda$. Find the global maximum by the use of derivatives. The zeroes of this function can be found by hand, assuming the function vanishes, or simply by the use of Mathematics, which has been done. The zeroes are found to be:

$p_0=\frac{2\pi }{\lambda }\left(1\pm 1/2n\right)$. Therefore, the total width (from first zero on the left of the peak to the first zero right of the peak) since the zeroes are symmetric is$w_p=2\pi \sout{h}/n\lambda$. When $n\to \infty$, the uncertainty in position is infinity and uncertainty in position is zero; the particle tends to have a well-defined momentum while being completely delocalized. The product of widths is:

$$\begin{gathered} w_x.w_p=2n\lambda .\frac{2\pi \sout{h}}{n\lambda } \\ =4\pi \sout{h}>\sout{h}/2 \end{gathered}$$

So the uncertainty relation holds. If onetries to correctly calculate the uncertainty in momentum, its mandatory to find the expectation value , for which the integral diverges. The reason behind it is because the function in the position space is discontinuous (the action of the momentum operator in the coordinate basis yields a derivative, and since the function is discontinuous, the derivative provides a delta function, therefore the squared operator provides a squared delta function, for which the integral blows up).