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According to what I know is that a classical turning point in Newtonian Mechanics is a point where a particle has a zero kinetic energy (Total energy is equal to potential energy) and must be instantaneously at rest. This means it stop its motion and reverse direction similar to harmonic motion oscillating back and forth between points $x=-A$ and $x=+A$. In the equation given below doesn't the wavelength $$ \lambda(x) = \frac{h}{\sqrt{2m(E-U(x))}}. $$ tend to infinity when $E=U(x)$?

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closed as unclear what you're asking by Aaron Stevens, John Rennie, eranreches, Jon Custer, Cosmas Zachos May 25 at 20:03

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    $\begingroup$ What is unclear? $\endgroup$ – Qmechanic May 28 at 11:00
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  1. The WKB approximation is valid in regions where $$\left| \frac{d\lambda(x)}{dx} \right|~\ll~2\pi, \tag{46.6}$$ i.e. the WKB approximation breaks down near a turning point $x$, cf. Ref. 1. Generically, the velocity/momentum behaves as $\sim |x-x_0|^{1/2}$, so that the de Broglie wavelength has a singularity $\sim |x-x_0|^{-1/2}$

  2. Recall that the wavefunction solution $\psi$ to the 2nd-order TISE has 2 integration constants. Typically the solutions in two semiclassical regions [satisfying (46.6)] on each side of a turning point are matched via WKB connection formulas.

References:

  • [LL] L.D. Landau & E.M. Lifshitz, QM, Vol. 3, 2nd & 3rd ed, 1981; $\S46$.
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