Question

Equation (4-21) expresses a function $\mathbf{}\mathit{\psi }\mathbf{\left(}\mathit{x}\mathbf{\right)}$as a sum of plane waves, each with a coefficient $\mathit{A}\left(k\right)$. Equation (4-22) finds the coefficients from the given function$\mathbf{}\mathit{\psi }\mathbf{\left(}\mathit{x}\mathbf{\right)}$ . The equations aren't independent statements; in fact, one is the inverse of the other. Equation (4-22) gives$\mathit{A}\left(k\right)$when $\mathbf{}\mathit{\psi }\mathbf{\left(}\mathit{x}\mathbf{\right)}$is known, and (4-21) does the reverse. Example 4.7calculates $\mathit{A}\left(k\right)$from a specific$\mathbf{}\mathit{\psi }\mathbf{\left(}\mathit{x}\mathbf{\right)}$. Show that when this $\mathit{A}\left(k\right)$ is inserted into (4-21) , the original $\mathit{\psi }\left(x\right)$is returned. Use the Euler formula and the symmetry properties of odd and even functions to simplify your work.

Short answer

Using the result of example 4.7, we can use the inverse relation given in equation 4-21 to recover the original expression of $\psi \left(x\right)$.

Step-by-step solution

Concept of the euler formula and the symmetry properties of odd and even function

$$\begin{gathered} \because A\left(k\right)=\frac{c}{\mathrm{\pi }}\frac{\sin \left({\displaystyle \frac{k\omega }{2}}\right)}{k},e^{ikx}=\cos (kx)+i\sin (kx) \\ \therefore \psi \left(x\right)={\int }_{-\infty }^{\infty }\frac{C}{\mathrm{\pi }}\frac{\sin \left({\displaystyle \frac{k\omega }{2}}\right)}{k}\cos (kx)+i\sin (kx)dk \end{gathered}$$

Determine the final expression of ψ(x) 

As noted in the problem statement, equations 4-21 and 4-22 are inverses of each other, hence, given the expression of $A\left(k\right)$, example 4.7, we should recover the original expression for $\psi \left(x\right)$using equation 4-21. We are given the direction on how to proceed, so we will start by plugin the expression for $A\left(k\right)$into 4-21 and with the help of a table of integration, we can solve for $\psi \left(x\right)$

$$\begin{gathered} \because \psi \left(x\right)={\int }_{-\infty }^{\infty }A\left(k\right)e^{ikx}dk \\ \because A\left(k\right)=\frac{c\sin \left({\displaystyle \frac{k\omega }{2}}\right)}{k},e^{ikx}=\cos \left(kx\right)+i\sin \left(kx\right) \\ \therefore \psi \left(x\right)={\int }_{-\infty }^{\infty }\frac{c}{\mathrm{\pi }}\frac{\sin \left({\displaystyle \frac{k\omega }{2}}\right)}{k}\left(\cos \left(kx\right)+i\sin \left(kx\right)\right)dk \end{gathered}$$

Noting that $A\left(k\right)$is an even function of$k\left(A\left(k\right)=A\left(-k\right)\right)$

$\because \psi \left(x\right)=\frac{c\sin \left({\displaystyle \frac{k\omega }{2}}\right)}{k}\cos \left(kx\right)dk$

From symmetry, the sine term (odd) cancel out and the cosine term (even) doubles

$\therefore \psi \left(x\right)=\frac{2c}{\mathrm{\pi }}{\int }_0^{\infty }\frac{\sin \left({\displaystyle \frac{k\omega }{2}}\right)\cos \left(kx\right)}{k}dk$

Well, a table of integration will be helpful at this point, performing this integral, will give us $0,\frac{\mathrm{\pi }}{2}$for x more than $\frac{\omega }{2}$and less than $\frac{\omega }{2}$respectively. The final expression for $\psi \left(x\right)$will be as follows,

$\therefore \psi \left(x\right)=\left\{\begin{array}{ll}0& \left|x\right|>\frac{\omega }{2}\\ c& \left|x\right|<\frac{\omega }{2}\end{array}\right.$