# Connection formulas: Why do we assume asymptotic behavior of the Airy functions?

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?

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.