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The derivation is obtained from Introduction to Quantum Mechanics by Griffiths

Let assume the turning point occurs at $x=0$, then the WKB solutions right and left to the turning point are:

$$ \psi=\dfrac{C}{\sqrt{|p|}}\exp{\left(-\dfrac{1}{\hbar}\left|\int_0^x pdx\right|\right)} \quad \text{for} \quad x>0, \tag{8.31a} $$

$$ \psi=\dfrac{C_1}{\sqrt{p}}\exp{\left(\dfrac{i}{\hbar}\int_0^x pdx\right)}+\dfrac{C_2}{\sqrt{p}}\exp{\left(-\dfrac{i}{\hbar}\int_0^x p dx\right)}\quad \text{for} \quad x<0.\tag{8.31b} $$

By assuming the potential near $x=0$ to be $$ V(x) \cong E + V'(0)x \tag{8.32} $$

and solving the Schrödinger equation, he takes the patching wave function as $$ \psi_p(x) = aAi(\alpha x) + bBi(\alpha x).\tag{8.37} $$

Then for $x > 0$, he uses the asymptotic forms of the Airy functions to obtain $$ \psi_p(x) \cong \dfrac{a}{2\sqrt{\pi}(\alpha x)^{1/4}}\exp(-\dfrac{2}{3}(\alpha x)^{3/2}) + \dfrac{b}{\sqrt{\pi}(\alpha x)^{1/4}}\exp(\dfrac{2}{3}(\alpha x)^{3/2}),\tag{8.40} $$

and use this to obtain the constant $C$,

My question is, why can we assume $x \gg 0$? Shouldn't this wave function stay in the neighborhood of the origin?

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Actually, Griffiths does not assume that the position coordinate $x\to \infty$ is unbounded, cf. footnote 8 on p. 288. To the contrary, he assumes that $x$ does not leave the patching region so that the linear approximation (8.32) to the potential is valid, cf. Fig. 8.9. Instead he assumes that the argument (to the Airy function) $$ z~\equiv~\alpha x~\gg~1 \tag{8.35}$$ is large, where $$ \alpha~~\equiv~\left[\frac{2m}{\hbar^2}V^{\prime}(0)\right]^{1/3}. \tag{8.34}$$ Semiclassically, $\alpha\to \infty$ for $\hbar\to 0$. Griffiths makes the assumption $z\gg 1$, since he is only after the leading semiclassical behavior.

For alternative derivations of the WKB connection formulas, see e.g. this related Phys.SE post.

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  • $\begingroup$ This is the key point, often overlooked in textbook discussions. $\alpha$ scales like $(1/\hbar^{2/3})$ and so in the limit $\hbar\to 0$ $z=\alpha x >> 1$ for any $x\ne 0$. For the same reason one cannot simply expand the exponentials near $x=0$ even if $x$ is small as there's no reason to believe that $\alpha x$ is itself small. $\endgroup$ Nov 11, 2021 at 14:36

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