Chapter 5, Problem 35P

(a)

To determine

The sketch of the real part of the matter wave pulse shape $f\left(x\right)$ for a Gaussian amplitude distribution where $a\left(k\right)=A{e}^{-{\alpha }^2{\left(k-k_0\right)}^2}$ .

Answer to Problem 35P

The real part of the matter wave pulse shape $f\left(x\right)$ for a Gaussian amplitude distribution where $a\left(k\right)=A{e}^{-{\alpha }^2{\left(k-k_0\right)}^2}$ is $\frac{A}{\alpha \sqrt{2}}{e}^{-4x^2{\alpha }^2}{\mathrm{cosk}}_{0}x$ and its sketch is

Explanation of Solution

Write the expression for the matter wave pulse shape.

  $f\left(x\right)=\frac{1}{\sqrt{2\pi }}{\displaystyle \underset{-\infty }{\overset{+\infty }{\int }}a\left(k\right)e^{ikx}dk}$        (I)

Here, $f\left(x\right)$ is the matter wave pulse shape, $a\left(k\right)$ is the amplitude distribution, $k$ is the wavenumber.

Write the expression for $a\left(k\right)$ .

  $a\left(k\right)=A{e}^{-{\alpha }^2{\left(k-k_0\right)}^2}$

Here, $k_0$ is the position of the peak of $a\left(k\right)$ .

Put the above equation in equation (I).

  $\begin{array}{c}f\left(x\right)=\frac{1}{\sqrt{2\pi }}{\displaystyle \underset{-\infty }{\overset{+\infty }{\int }}\left(A{e}^{-{\alpha }^2{\left(k-k_0\right)}^2}\right)e^{ikx}dk}\\ =\frac{A}{\sqrt{2\pi }}{\displaystyle \underset{-\infty }{\overset{+\infty }{\int }}{e}^{-{\alpha }^2{\left(k-k_0\right)}^2}{e}^{ikx}dk}\\ =\frac{A}{\sqrt{2\pi }}{e}^{-{\alpha }^2{k}_0^2}{\displaystyle \underset{-\infty }{\overset{+\infty }{\int }}{e}^{-{\alpha }^2\left(k^2-\left(2k_0+ix/{\alpha }^2\right)k\right)}dk}\\ =\frac{A}{\sqrt{2\pi }}{e}^{-{\alpha }^2{k}_0^2}{\displaystyle \underset{-\infty }{\overset{+\infty }{\int }}{e}^{{\alpha }^2{\left(k_0+ix/2{\alpha }^2\right)}^2}{e}^{-{\alpha }^2{\left(k-\left(k_0+ix/2{\alpha }^2\right)\right)}^2}dk}\end{array}$

Simplify the above equation.

  $\begin{array}{c}f\left(x\right)=\frac{A}{\sqrt{2\pi }}{e}^{-{\alpha }^2{k}_0^2}{e}^{{\alpha }^2{\left(k_0+ix/2{\alpha }^2\right)}^2}{\displaystyle \underset{-\infty }{\overset{+\infty }{\int }}{e}^{-{\alpha }^2{\left(k-\left(k_0+ix/2{\alpha }^2\right)\right)}^2}dk}\\ =\frac{A}{\sqrt{2\pi }}{e}^{-x^2/4{\alpha }^2}{e}^{i{k}_{0}x}{\displaystyle \underset{-\infty }{\overset{+\infty }{\int }}{e}^{-{\alpha }^2{\left(k-\left(k_0+ix/2{\alpha }^2\right)\right)}^2}dk}\end{array}$        (II)

Take $z=k-\left(k_0+\frac{ix}{2{\alpha }^2}\right)$ .

Take the derivative of $z$ .

  $\begin{array}{c}dz=dk-0\\ dk=dz\end{array}$

Put the above two equations in equation (II).

  $\begin{array}{c}f\left(x\right)=\frac{A}{\sqrt{2\pi }}{e}^{-x^2/4{\alpha }^2}{e}^{i{k}_{0}x}{\displaystyle \underset{-\infty }{\overset{+\infty }{\int }}{e}^{-{\alpha }^2{z}^2}dz}\\ =\frac{A}{\sqrt{2\pi }}{e}^{-x^2/4{\alpha }^2}{e}^{i{k}_{0}x}\left(\frac{{\pi }^{1/2}}{\alpha }\right)\\ =\frac{A}{\alpha \sqrt{2}}{e}^{-x^2/4{\alpha }^2}{e}^{i{k}_{0}x}\\ =\frac{A}{\alpha \sqrt{2}}{e}^{-x^2/4{\alpha }^2}\left(\cos k_{0}x+i\sin k_{0}x\right)\end{array}$

Write the real part of $f\left(x\right)$ from the above equation.

  $\text{Re }f\left(x\right)=\frac{A}{\alpha \sqrt{2}}{e}^{-x^2/4{\alpha }^2}\cos k_{0}x$        (III)

Here, $\text{Re }f\left(x\right)$ is the real part of $f\left(x\right)$ .

The plot of $\text{Re }f\left(x\right)$ is shown in figure 1.

Conclusion:

Therefore, the real part of the matter wave pulse shape $f\left(x\right)$ for a Gaussian amplitude distribution where $a\left(k\right)=A{e}^{-{\alpha }^2{\left(k-k_0\right)}^2}$ is $\frac{A}{\alpha \sqrt{2}}{e}^{-4x^2{\alpha }^2}{\mathrm{cosk}}_{0}x$ and its sketch is shown in figure 1.

(b)

To determine

To show that the width of the matter wave pulse is $\Delta x=\alpha$ .

Answer to Problem 35P

It is showed that width of the matter wave pulse is $\Delta x=\alpha$ .

Explanation of Solution

Write the standard form of the Gaussian function with width $\Delta x$ .

  $f\left(x\right)\propto A{e}^{-{\left(x/2\Delta x\right)}^2}$        (IV)

Comparison of equation (III) with the above expression shows that

  $\Delta x=\alpha$        (V)

Conclusion:

Therefore, it is showed that width of the matter wave pulse is $\Delta x=\alpha$ .

(c)

To determine

The width $\Delta k$ of $a\left(k\right)$ and to show that $\Delta x\Delta k=\frac{1}{2},\text{ independent of }\alpha$ .

Answer to Problem 35P

The width $\Delta k$ of $a\left(k\right)$ is $\frac{1}{2\alpha }$ and it is showed that $\Delta x\Delta k=\frac{1}{2},\text{ independent of }\alpha$ .

Explanation of Solution

Compare equations (III) and (IV) to write the expression for ${\alpha }^2$ .

  ${\alpha }^2=\frac{1}{4{\left(\Delta k\right)}^2}$

Rewrite the above equation for $\Delta k$ .

  $\begin{array}{c}{\left(\Delta k\right)}^2=\frac{1}{4{\alpha }^2}\\ \Delta k=\sqrt{\frac{1}{4{\alpha }^2}}\\ =\frac{1}{2\alpha }\end{array}$        (VI)

Use equations (V) and (VI) to find the value of $\Delta x\Delta k$ .

  $\begin{array}{c}\Delta x\Delta k=\alpha \left(\frac{1}{2\alpha }\right)\\ =\frac{1}{2}\end{array}$

Conclusion:

Thus, the width $\Delta k$ of $a\left(k\right)$ is $\frac{1}{2\alpha }$ and it is showed that $\Delta x\Delta k=\frac{1}{2},\text{ independent of }\alpha$ .