# Asymptotic behavior of Airy functions in the WKB method

#### QuantumDuality

Problem Statement
Given a potential V(x) that varies slowly, the wkb method proposes an approximation to the solution of the Schrodinger equation for the regions where $E<V$ and $E>V$. Due to the form of the approximation, it does not work in the turning points $(E=V)$. Near this points, the Airy functions are taken as the solution, and we use the asymptotic behavior of said functions to "put together" the solutions in the different regions. What does it mean to use the asymptotic behavior instead of the function itself?
Relevant Equations
The approximated solution given by the WKB method is:

$\psi = C sin⁡( \frac{1}{\hbar} \int p dx+\delta)$ for $E > V$

$\psi= \frac{C_+}{\sqrt{|p|}} e^{-\frac{1}{\hbar}∫|p| dx}+\frac{C_-}{\sqrt{|p|}} e^{+\frac{1}{\hbar}∫|p|dx}$ for $E < V$

$p(x) = \sqrt{2 m [E - V(x)]}$

Near the turning point $x_2 >0$ , the potential is approximated to:

$V(x) \approx E + V'(x_2) (x - x_2) =>$

$p(x) \approx \sqrt{-2 m V'(x_2) (x - x_2)}$

After doing the change of variable:

$y = [\frac{2 m V'(x_2)}{\hbar^2}]^{1/3} (x - x_2)$

The solution to the Schrodinger equation is the Airy's function. If we look at the asymptotic behavior of the Airy's function and chose appropiate constants, to put together the solutions for the different regions, we get a wave function:

$\psi(y) = \begin{matrix} \frac{A'}{\sqrt{\pi }|y|^{1/4}} sin(\frac{2}{3} (-y)^{3/2} + \frac{\pi}{4}) & y << 0\\ \frac{A'}{\sqrt{\pi }y^{1/4}} e^{{\frac{2}{3} (-y)^{3/2}}} & y >> 0\\ \end{matrix}$
If it is the asymptotic behavior of the Airy's function what it's used instead of the function itself: Does it mean that the wkb method is only valid for potentials where the regions where $E<V$ and $E>V$ are "wide"?

Related Advanced Physics Homework News on Phys.org

"Asymptotic behavior of Airy functions in the WKB method"

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