Question

Question:

An illuminating alternative derivation of the WKB formula (Equation) is based on an expansion in powers of$\sout{\mathbf{h}}$. Motivated by the free particle wave function $\mathbf{\psi }\mathbf{=}\mathbf{Aexp}\mathbf{}\mathbf{(}\mathbf{\pm }\mathbf{ipx}\mathbf{/}\sout{\mathbf{h}}\mathbf{)}$,, we write

$\mathbf{\psi }\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{=}{\mathbf{e}}^{\mathbf{if}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{/}\mathbf{h}}$

Wheref(x)is some complex function. (Note that there is no loss of generality here-any nonzero function can be written in this way.)

(a) Put this into Schrödinger's equation (in the form of Equation8.1), and show that

. ${\mathbf{ihf}}^{\cancel{\mathbf{c}}\cancel{\mathbf{c}}}\mathbf{-}{\left(f^{\cancel{c}}\right)}^{\mathbf{2}}\mathbf{+}{\mathbf{p}}^{\mathbf{2}}\mathbf{+}\mathbf{0}\mathbf{}\mathbf{.}$

(b) Write f(x)as a power series in$\sout{\mathbf{h}}$:

$\mathit{f}\mathbf{\left(}\mathit{x}\mathbf{\right)}\mathbf{=}{\mathit{f}}_{\mathbf{0}}\mathbf{\left(}\mathit{x}\mathbf{\right)}\mathbf{+}\sout{\mathbf{h}}{\mathit{f}}_{\mathbf{1}}\mathbf{\left(}\mathit{x}\mathbf{\right)}\mathbf{+}{\sout{\mathbf{h}}}^{\mathbf{2}}{\mathit{f}}_{\mathbf{2}}\mathbf{\left(}\mathit{x}\mathbf{\right)}\mathbf{+}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}$And, collecting like powers of$\sout{\mathbf{h}}$, show that

${\left({\dot{o}}_0\right)}^{\mathbf{2}}\mathbf{=}{\mathbf{p}}^{\mathbf{2}}\mathbf{}\mathbf{,}\mathbf{}\mathbf{i}{\dot{\mathbf{o}}}_{\mathbf{0}}\mathbf{=}\mathbf{2}{\dot{\mathbf{o}}}_{\mathbf{0}}{\dot{\mathbf{o}}}_{\mathbf{1}}\mathbf{,}\mathbf{}\mathbf{}\mathbf{}\mathbf{i}{\dot{\mathbf{o}}}_{\mathbf{1}}\mathbf{=}\mathbf{2}{\dot{\mathbf{o}}}_{\mathbf{0}}{\dot{\mathbf{o}}}_{\mathbf{2}}\mathbf{+}{\mathbf{\left(}{\dot{\mathbf{o}}}_{\mathbf{1}}\mathbf{\right)}}^{\mathbf{2}}\mathbf{,}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}$

(c) Solve for${\mathbf{f}}_{\mathbf{0}}\left(x\right)$and${\mathbf{f}}_{\mathbf{1}}\mathbf{\left(}\mathbf{x}\mathbf{\right)}$, and show that-to first order inyou recover Equation8.10.

Short answer

(a)$i\sout{h}{f}^{"}-{\left(f^{\text{'}}\right)}^2+p^2=0$ And it’s proved.

(b)${\left(f_0^{\text{'}}\right)}^2=p^2,{if}_0^{:}=2f_0^{\text{'}}{f}_1^{\text{'}},i{f}_1^{:}=2f_0^{\text{'}}{f}_2^{\text{'}}+{\left(f_1^{\text{'}}\right)}^2$ and it’s proved.

(c)Therefore, $\mathrm{\psi }\left(\mathrm{x}\right)$ is given by

$\mathrm{\psi }\left(\mathrm{x}\right)\cong \frac{\mathrm{c}}{\sqrt{\mathrm{p}\left(\mathrm{x}\right)}}{\mathrm{e}}^{\pm \frac{\mathrm{i}}{\mathrm{h}}\int \mathrm{p}\left(\mathrm{x}\right)\mathrm{dx}}$

Step-by-step solution

Step 1:(a) Show the Schrodinger Equation.

Let, the wave equation take the form,

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

Where, f(x) is some complex function.

So that,

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

We known that Schrodinger Equation,

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

Substitute equation (a) in Schrodinger Equation we get,

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

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

(b) To write f(x) as a power series.

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.

Step 3:(c) To solve f0(x) and f1(x).

In step 2, we have to show that,

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

(or)

$f_0^{\text{'}}=\pm p$

Integrating we get,

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

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

$f_0^{"}=\pm \frac{dp}{dx}$

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" $f\left(x\right)=\pm \int p\left(x\right)dx+i\sout{h}In\sqrt{p\left(x\right)}+\mathrm{constant}+\mathrm{O}\left({\sout{\mathrm{h}}}^2\right)$

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

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

Which is exactly as equation (8.10).