91Ó°ÊÓ

Let, the wave equation take the form,

$\mathrm{ψ}\left(\mathrm{x}\right)={\mathrm{e}}^{\mathrm{if}\left(\mathrm{x}\right)/\sout{\mathrm{h}}}$

Where, f(x) is some complex function.

So that,

$$\begin{gathered} \frac{d\mathrm{ψ}}{dx}=\frac{i}{\sout{h}}{f}^{\text{'}}\left(x\right)\mathrm{ψ}\left(\mathrm{x}\right) \\ \frac{{\mathrm{d}}^2\mathrm{ψ}}{{\mathrm{dx}}^2}=\left[\frac{\mathrm{i}}{\sout{\mathrm{h}}}{\mathrm{f}}^{"}{\left(\mathrm{x}\right)-\frac{1}{{\sout{\mathrm{h}}}^2}\left({\mathrm{f}}^{\text{'}}(\mathrm{x}\right))}^2\right]\mathrm{ψ}\left(\mathrm{x}\right)...\left(\mathrm{a}\right) \end{gathered}$$

We known that Schrodinger Equation,

$\frac{{\mathrm{d}}^2\mathrm{ψ}}{{\mathrm{dx}}^2}=\frac{p^2}{{\sout{h}}^2}\mathrm{ψ}$

Substitute equation (a) in Schrodinger Equation we get,

$\left[i\frac{f^{"}}{\sout{h}}-\frac{{\left(f^{\text{'}}\right)}^2}{{\sout{h}}^2}\right]\mathrm{ψ}\left(x\right)=-\frac{p^2}{{\sout{h}}^2}\mathrm{ψ}\left(x\right)$This leads to

$i\sout{h}{f}^{"}-{\left(f^{\text{'}}\right)}^2+p^2=0$ ............(1)

Writeas a power series in:

$f\left(x\right)=f_0\left(x\right)+\sout{h}{f}_1\left(x\right)+{\sout{h}}^2{f}_2\left(x\right)+...,$

So that,

$$\begin{gathered} f^{\text{'}}\left(x\right)=f_0^{\text{'}}x)+\sout{h}{f}_1^{\text{'}}(x)+{\sout{h}}^2{f}_2^{\text{'}}(x)+..., \\ {f}^{"}(x)=f_0^{"}x)+\sout{h}{f}_1^{"}\left(x\right)+{\sout{h}}^2{f}_2^{"}\left(x\right)+..., \end{gathered}$$

Putting this into equation (1) we have,

$i\sout{h}\left[f_0^{"}+\sout{h}{f}_1^{"}+{\sout{h}}^2{f}_2^{"}+...,\right]-\left[f_0^{\text{'}}+\sout{h}{f}_1^{\text{'}}+{\sout{h}}^2{f}_2^{\text{'}}+...,\right]+p^2=0$

(Or)

$p^2-{\left(f_0^{\text{'}}\right)}^2+\sout{h}\left[i{f}_0^{"}-2f_0^{\text{'}}{f}_1^{\text{'}}\right]+{\sout{h}}^2\left[i{f}_0^{"}-2f_0^{\text{'}}{f}_2^{\text{'}}-{\left(f_1^{\text{'}}\right)}^2\right]+\mathrm{O}\left(\sout{h^3}\right)$

Finally, power of yields,

${\left(f_0^{\text{'}}\right)}^2=p^2,i{f}_0^{"}-2f_0^{\text{'}}{f}_1^{\text{'}},i{f}_0^{"}-2f_0^{\text{'}}{f}_2^{\text{'}}-{\left(f_1^{\text{'}}\right)}^2$

........etc.

In step 2, we have to show that,

${\left(f_0^{\text{'}}\right)}^2=p^2$

(or)

$f_0^{\text{'}}=Â\pm p$

Integrating we get,

$f_0\left(x\right)=Â\pm ∫p\left(x\right)dx+C_1$(C1 is constant)

Also, it follows (by taking the x-derivative) that,

$f_0^{"}=Â\pm \frac{dp}{dx}$

Again from step 2, we have shown that

$i{f}_0^{"}=2f_0^{\text{'}}{f}_1^{\text{'}}$

Solving for $f_1^{\text{'}}$an substituting for $f_0^{\text{'}}$and $f_0^{\text{'}}$then,

$f_1^{\text{'}}=\frac{i}{2p}\frac{dp}{dx}=\frac{i}{2}\frac{dinp\left(x\right)}{dx}$

By integrating, we get

$f_1^{\text{'}}=\frac{i}{2}inp\left(x\right)+C_2$($C_2$ is constant)

Hence, we can write f(x) into first order in $\sout{h}$as,

role="math" localid="1658465427689" $f\left(x\right)=Â\pm ∫p\left(x\right)dx+i\sout{h}In\sqrt{p\left(x\right)}+\mathrm{constant}+\mathrm{O}\left({\sout{\mathrm{h}}}^2\right)$

Hence $\mathrm{ψ}\left(\mathrm{x}\right)$is given by,

$\mathrm{ψ}\left(\mathrm{x}\right)≈\exp \left[Â\pm \frac{i}{\sout{h}}∫p\left(x\right)dx-i\sout{h}In\sqrt{p\left(x\right)}+\mathrm{constant}\right]=\frac{C}{\sqrt{p\left(x\right)}}{e}^{Â\pm \frac{i}{\sout{h}}∫p\left(x\right)dx}$

Which is exactly as equation (8.10).