Question

The "inversion theorem" for Fourier transforms states that

$\cancel{\mathbf{\varphi }}\left(Z\right)\mathbf{=}{\mathbf{\int }}_{\mathbf{-}\mathbf{\infty }}^{\mathbf{\infty }}\cancel{\mathbf{\varphi }}\left(k\right){\mathbf{e}}^{\mathbf{ikz}}\mathbf{dk}\mathbf{\iff }\cancel{\mathbf{\varphi }}\left(k\right)\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}\mathbf{\pi }}{\mathbf{\int }}_{\mathbf{-}\mathbf{\infty }}^{\mathbf{\infty }}\cancel{\mathbf{\varphi }}\left(z\right){\mathbf{e}}^{\mathbf{-}\mathbf{ikz}}\mathbf{dz}$

Use this to determine $\cancel{\mathbf{A}\mathbf{}}\left(k\right)$, in Eq. 9.20, in terms of $\mathbf{f}\left(z,0\right)$and$\stackrel{\mathbf{*}}{\mathbf{f}}\left(z,0\right)$

Short answer

The expression for$\cancel{\mathrm{A}}\left(\mathrm{k}\right)$ is$\frac{1}{2\mathrm{\pi }}{\int }_{-\infty }^{\infty }\left[\mathrm{f}\left(\mathrm{z},0\right)+\frac{\mathrm{i}}{\mathrm{\omega }}\stackrel{\mathrm{g}}{\mathrm{f}}\left(\mathrm{z},0\right)\right]{\mathrm{e}}^{-\mathrm{ikz}}\mathrm{dz}$

Step-by-step solution

Expression for the linear combination of sinusoidal wave:

Write the expression for the linear combination of a sinusoidal wave.

$\cancel{\stackrel{\mathbf{0}}{\mathbf{f}}}\left(z,t\right)\mathbf{=}{\mathbf{\int }}_{\mathbf{-}\mathbf{\infty }}^{\mathbf{\infty }}\cancel{\stackrel{\mathbf{0}}{\mathbf{A}}}\left(k\right){\mathbf{e}}^{\mathbf{i}\left(\mathrm{kz}-\mathrm{\omega t}\right)}\mathbf{dk}$.............(1)

Here, $\mathrm{k}$is the propogation vector and $\mathrm{\omega }$is the angular frequency.

Determine A0(k) in term of data-custom-editor="chemistry" f (z,0) and data-custom-editor="chemistry" f *(z,0)

Substitute $\mathrm{t}=0$in equation (1).

$$\begin{gathered} \cancel{\stackrel{0}{\mathrm{f}}}\left(\mathrm{z},0\right)={\int }_{-\infty }^{\infty }\cancel{\stackrel{0}{\mathrm{A}}}\left(\mathrm{k}\right){\mathrm{e}}^{\mathrm{i}\left(\mathrm{kz}-\mathrm{\omega }\left(0\right)\right)}\mathrm{dk} \\ ={\int }_{-\infty }^{\infty }\cancel{\stackrel{0}{\mathrm{A}}}\left(\mathrm{k}\right){\mathrm{e}}^{\mathrm{ikz}}\mathrm{dk} \end{gathered}$$

Write the conjugate of the above expression.

$$\begin{gathered} \cancel{\stackrel{0}{\mathrm{f}}}{\left(\mathrm{z},0\right)}^{*}={\int }_{-\infty }^{\infty }\cancel{\stackrel{0}{\mathrm{A}}}{\left(-\mathrm{k}\right)}^{*}{\mathrm{e}}^{-\mathrm{ikz}}\mathrm{dk} \\ \cancel{\stackrel{0}{\mathrm{f}}}{\left(\mathrm{z},0\right)}^{*}={\int }_{-\infty }^{\infty }\cancel{\stackrel{0}{\mathrm{A}}}{\left(\mathrm{k}\right)}^{*}{\mathrm{e}}^{-\mathrm{ikz}}\left(-\mathrm{dk}\right) \end{gathered}$$

It is known that,

$$\begin{gathered} \mathrm{f}\left(\mathrm{z},0\right)=\mathrm{Re}\left[\cancel{\stackrel{0}{\mathrm{f}}}\left(\mathrm{z},0\right)\right] \\ \mathrm{f}\left(\mathrm{z},0\right)=\frac{1}{2}\left[\cancel{\stackrel{0}{\mathrm{f}}}\left(\mathrm{z},0\right)+\cancel{\stackrel{0}{\mathrm{f}}}{\left(\mathrm{z},0\right)}^{*}\right]....\left(2\right) \end{gathered}$$

Substitute $\frac{1}{2}\cancel{\stackrel{0}{\mathrm{A}}}\left(\mathrm{k}\right){\mathrm{e}}^{\mathrm{ikz}}\mathrm{dk}$for $\cancel{\stackrel{0}{\mathrm{f}}}\left(z,0\right)$and $\frac{1}{2}\cancel{\stackrel{0}{\mathrm{A}}}{\left(-\mathrm{k}\right)}^{*}{\mathrm{e}}^{\mathrm{ikz}}\mathrm{dk}$for $\cancel{\stackrel{0}{\mathrm{f}}}{\left(z,0\right)}^{*}$in equation (2)

$\mathrm{f}\left(\mathrm{z},0\right)={\int }_{-\infty }^{\infty }\frac{1}{2}\left[\cancel{\stackrel{0}{\mathrm{A}}}\left(\mathrm{k}\right)+\cancel{\stackrel{0}{\mathrm{A}}}{\left(-\mathrm{k}\right)}^{*}\right]{\mathrm{e}}^{\mathrm{ikz}}\mathrm{dk}$

Substitute $\mathrm{f}\left(\mathrm{z},0\right)={\int }_{-\infty }^{\infty }\frac{1}{2}\left[\cancel{\stackrel{0}{\mathrm{A}}}\left(\mathrm{k}\right)+\cancel{\stackrel{0}{\mathrm{A}}}{\left(-\mathrm{k}\right)}^{*}\right]{\mathrm{e}}^{\mathrm{ikz}}\mathrm{dk}$in equation (2).

$\frac{1}{2}\left[\cancel{\stackrel{0}{\mathrm{A}}}\left(\mathrm{k}\right)+\cancel{\stackrel{0}{\mathrm{A}}}{\left(-\mathrm{k}\right)}^{*}\right]=\frac{1}{2\mathrm{\pi }}{\int }_{-\infty }^{\infty }\mathrm{f}\left(\mathrm{z},0\right){\mathrm{e}}^{-\mathrm{ikz}}\mathrm{dk}$ ......(3)

Solve the conjugate for $\cancel{\stackrel{0}{\mathrm{f}}}\left(z,0\right)$.

$$\begin{gathered} \cancel{\stackrel{0}{\mathrm{f}}}\left(\mathrm{z},\mathrm{t}\right)={\int }_{-\infty }^{\infty }\cancel{\stackrel{0}{\mathrm{A}}}\left(\mathrm{k}\right)\left(-\mathrm{i\omega }\right){\mathrm{e}}^{\mathrm{i}\left(\mathrm{kz}-\mathrm{\omega t}\right)}\mathrm{dk} \\ \cancel{\stackrel{0}{\mathrm{f}}}\left(\mathrm{z},\mathrm{t}\right)={\int }_{-\infty }^{\infty }\left[-\mathrm{i\omega }\cancel{\stackrel{0}{\mathrm{A}}}\left(\mathrm{k}\right)\right]{\mathrm{e}}^{\mathrm{ikz}}\mathrm{dk} \\ \cancel{\stackrel{0}{\mathrm{f}}}\left(\mathrm{z},\mathrm{t}\right)={\int }_{-\infty }^{\infty }\left[\mathrm{i\omega }\cancel{\stackrel{0}{\mathrm{A}}}\left(\mathrm{k}\right)\right]{\mathrm{e}}^{\mathrm{ikz}}\left(-\mathrm{dk}\right) \end{gathered}$$

On further solving, the above equation becomes,

$$\begin{gathered} \cancel{\stackrel{\mathrm{g}}{\mathrm{f}}}{\left(\mathrm{z},0\right)}^{*}={\int }_{-\infty }^{\infty }\left[\mathrm{i\omega }\cancel{\stackrel{\mathrm{g}}{\mathrm{A}}}{\left(-\mathrm{k}\right)}^{*}\right]{\mathrm{e}}^{-\mathrm{ikz}}\left(\mathrm{dk}\right) \\ \stackrel{\mathrm{g}}{\mathrm{f}}\left(\mathrm{z},0\right)=\mathrm{Re}\left[\cancel{\stackrel{\mathrm{g}}{\mathrm{f}}}\left(\mathrm{z},0\right)\right] \\ \stackrel{\mathrm{g}}{\mathrm{f}}\left(\mathrm{z},0\right)=\frac{1}{2}\left[\cancel{\stackrel{\mathrm{g}}{\mathrm{f}}}\left(\mathrm{z},0\right)+\cancel{\stackrel{\mathrm{g}}{\mathrm{f}}}{\left(\mathrm{z},0\right)}^{*}\right].......\left(4\right) \end{gathered}$$

Substitute $\frac{1}{2}-\mathrm{i\omega }\cancel{\stackrel{0}{\mathrm{A}}}\left(\mathrm{k}\right){\mathrm{e}}^{\mathrm{ikz}}\mathrm{dk}$for $\cancel{\stackrel{\mathrm{g}}{\mathrm{f}}}\left(\mathrm{z},0\right)$and $\frac{1}{2}-\mathrm{i\omega }\cancel{\stackrel{0}{\mathrm{A}}}{\left(\mathrm{k}\right)}^{*}{\mathrm{e}}^{\mathrm{ikz}}\mathrm{dk}$for $\cancel{\stackrel{\mathrm{g}}{\mathrm{f}}}{\left(\mathrm{z},0\right)}^{*}$in equation (4).

$$\begin{gathered} \stackrel{\mathrm{g}}{\mathrm{f}}\left(\mathrm{z},0\right)={\int }_{-\infty }^{\infty }\frac{1}{2}\left[-\mathrm{i\omega }\cancel{\stackrel{0}{\mathrm{A}}}\left(\mathrm{k}\right)+\mathrm{i\omega }\cancel{\stackrel{0}{\mathrm{A}}}\left(-\mathrm{k}\right)\right]{\mathrm{e}}^{\mathrm{ikz}}\mathrm{dk} \\ \stackrel{\mathrm{g}}{\mathrm{f}}\left(\mathrm{z},0\right)=\frac{-\mathrm{i\omega }}{2}\left[\cancel{\stackrel{0}{\mathrm{A}}}\left(\mathrm{k}\right)-\cancel{\stackrel{0}{\mathrm{A}}}{\left(-\mathrm{k}\right)}^{*}\right] \end{gathered}$$

The inversion theorem for Fourier transformation states that,

$\stackrel{\mathrm{g}}{\mathrm{f}}\left(\mathrm{z},0\right)=\frac{1}{2\mathrm{\pi }}{\int }_{-\infty }^{\infty }\stackrel{\mathrm{g}}{\mathrm{f}}\left(\mathrm{z},0\right){\mathrm{e}}^{\mathrm{ikz}}\mathrm{dz}$