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Q13E

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Found in: Page 224

### Modern Physics

Book edition 2nd Edition
Author(s) Randy Harris
Pages 633 pages
ISBN 9780805303087

# Show that ${\mathbit{\psi }}\left(x\right){\mathbf{=}}{\mathbit{A}}{\mathbf{\text{'}}}{{\mathbit{e}}}^{\mathbf{i}\mathbf{k}\mathbf{x}}{\mathbf{+}}{\mathbit{B}}{\mathbf{\text{'}}}{{\mathbit{e}}}^{\mathbf{-}\mathbf{i}\mathbf{k}\mathbf{x}}$ is equivalent to ${\mathbit{\psi }}{\mathbf{\left(}}{\mathbit{x}}{\mathbf{\right)}}{\mathbf{=}}{\mathbit{A}}{\mathbit{s}}{\mathbit{i}}{\mathbit{n}}{\mathbit{k}}{\mathbit{x}}{\mathbf{+}}{\mathbit{B}}{\mathbit{c}}{\mathbit{o}}{\mathbit{s}}{\mathbit{k}}{\mathbit{x}}$, provided that ${\mathbit{A}}{\mathbf{\text{'}}}{\mathbf{=}}\frac{\mathbf{1}}{\mathbf{2}}\left(B-iA\right)$ ${\mathbit{B}}{\mathbf{\text{'}}}{\mathbf{=}}\frac{\mathbf{1}}{\mathbf{2}}\left(B+iA\right)$.

Hence, the proof for the equation is obtained.

See the step by step solution

## Step 1: Concept involved

According to the Euler’s formula in complex numbers, it can be written that:

${{\mathbit{e}}}^{\mathbf{i}\mathbf{\phi }}{\mathbf{=}}{\mathbit{c}}{\mathbit{o}}{\mathbit{s}}{\mathbit{\phi }}{\mathbf{+}}{\mathbit{i}}{\mathbit{s}}{\mathbit{i}}{\mathbit{n}}{\mathbit{\phi }}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{\dots }}{\mathbf{\dots }}{\mathbf{}}{\mathbf{\left(}}{\mathbf{1}}{\mathbf{\right)}}$

## Step 2: Given/known parameters

Consider the given function:

$\psi \left(x\right)=A\text{'}{e}^{ikx}+B\text{'}{e}^{-ikx}$

Consider the equations:

$A\text{'}=\frac{1}{2}\left(B-iA\right)\dots \dots \left(2\right)$

$B\text{'}=\frac{1}{2}\left(B+iA\right)\dots \dots \left(3\right)$

## Step 3: Solution

Apply Euler’s formula from equation (1) and solve:

$\psi \left(x\right)=A\text{'}\left(\mathrm{cos}\left(kx\right)+i\mathrm{sin}\left(kx\right)\right)+B\text{'}\left(\mathrm{cos}\left(kx\right)-i\mathrm{sin}\left(kx\right)\right)$

Rewrite the above equation as,

$\psi \left(x\right)=\left(A\text{'}+B\text{'}\right)\mathrm{cos}kx+\left(A\text{'}-B\text{'}\right)\mathrm{sin}\left(kx\right)\dots ..\left(4\right)$

Now, by using equation (2) and (3) in equation (4) solve as:

$\psi \left(x\right)=\left(\frac{1}{2}\left[B-iA\right]+\frac{1}{2}\left[B+iA\right]\right)\mathrm{cos}\left(kx\right)-\left(\frac{1}{2}\left[B-iA\right]-\frac{1}{2}\left[B+iA\right]\right)\mathrm{sin}\left(kx\right)\phantom{\rule{0ex}{0ex}}\psi \left(x\right)=B\mathrm{cos}\left(kx\right)+A\mathrm{sin}\left(kx\right)$

Thus, you can say that: $\psi \left(x\right)=A\text{'}{e}^{ikx}+B\text{'}{e}^{-ikx}$ is equivalent to

$\psi \left(x\right)=B\mathrm{cos}\left(kx\right)+A\mathrm{sin}\left(kx\right)$ provided that $A\text{'}=\frac{1}{2}\left(B-iA\right)$ and $B\text{'}=\frac{1}{2}\left(B+iA\right)$