# [Physics] Quantum Mechanics and the Airy Function, the physics of the turning point

homework-and-exercisespotentialquantum mechanicsschroedinger equation

I'm working with the Airy function, and the book states at $x=0$ is a turning point, and there are two very different behaviors on either side. In general what a does a turning point mean in differential equations? A point where the highest order derivative in the equation equals zero? The book says this makes sense, oscillatory and decaying on either side of the turning point. But if $E>$ turning point it decays, that's the opposite case of the finite square well right? if $E<$ potential then it decays. What does it physically mean to have a turning point there, and why is it oscillatory when $E$ is greater than the turning point? This is so unintuitive, these potentials confuse me.

Consider first the classical theory. Turning points are these points $x$ where the total energy is completely stored in the potential energy, $E = V(x)$. Hence the kinetic energy is zero at these points. For example consider the potential of the Harmonic Oscillator $V(x) = x^2 / 2$. Then the turning points are $x_t = \pm \sqrt{2 E}$. The pendulum oscillates between $x_{-} = - \sqrt{2 E}$ and $x_{+} = - \sqrt{2 E}$, and hence makes a turn at these maximal points.
Now turn to the quantum mechanic picture, or to be more exact to the semiclassical WKB method. One tries to approximate the quantum mechanic solution by some other functions. In the case of the Harmonic Oscillator these approximating functions are exactly the Airy functions (with some prefactors). So looking at the solution from wikipedia, one sees that the behavior of the wavefunction changes at the turning points (the potential is drawn in blue, so the turning points are approximately where the wavefunctions intersects the blue graph). For the region between the $x_t$ one has an oscillation whereas for greater $x$ the wavefunction is exponentially damped. This is just as for the Airy function.