What is meant by a classical turning point in quantum mechanics and why does the WKB approximation fail at that point?


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A classical turning point is a point at which the system's total energy $E$ equals the potential energy $V.$ Past this point, i.e. for $E<V$ the potential is greater than the total energy, such cases we denote as classically forbidden regions, because from a purely classical point of view, the system has 0 chance of being in a state where its potential energy is larger than its total energy or in other words having a negative kinetic energy $E - V < 0.$

In order to see why at this point the WKB approximation fails, recall the time independent form of Schrödinger equation (let's work at 1D to simplify):

$$ \begin{align*} \frac{d^2\psi }{dx^2}+k^2(x)\psi(x)&= 0 \\ k^2(x) = \frac{2m}{\hbar^2}(E-V(x))&=\frac{p^2(x)}{\hbar^2} \end{align*} $$ With $k(x)$ the wavevector, and $p(x)$ is the local classical momentum corresponding to a classical decomposition of energy $E.$ Now it suffices to look at the 0-th order WKB solutions, which are planewaves with amplitudes independent of $x$, as follows:

$$ \psi(x)=Ae^{\pm iu(x)} $$ Which if you substitute into TISE, and assume a slowly varying $k(x)$ then one can neglect the second derivative of $u(x)$ and solve for $u(x)$, which then is simply the classical action integral: $$ s(x)=u(x)\hbar = \pm \int^x p(x')dx' $$ This brings us to our 0th order WKB solution:

$$ \psi_0 (x) = \exp \left[\pm i \int^x k(x')dx'\right] $$

Now we need to check the validity of this solution, for this we insert $\psi_0(x)$ back into our first TISE equation and obtain:

$$ \frac{d^2\psi_0}{dx^2}+ \left[k^2(x) \mp i\frac{dk}{dx}\right] \psi_0(x)=0 $$ From the above result, it is now clear that for $\psi_0(x)$ to be a rigorous solution (i.e. to correspond to a valid solution of our original TISE), then:

$$ \begin{align*} \left\lvert \frac{dk}{dx} \right\rvert &<< k^2(x) \\ \left\lvert \frac{1}{k}\frac{dk}{dx} \right\rvert &<< k(x) \end{align*} $$ A condition which is never fulfilled at the turning point ($k \to 0$) because $1/k$ diverges. This validity condition propagates through higher order solutions of WKB, e.g. the 1st order WKB solution is of type:

$$ \psi_1(x) = \frac{C}{\sqrt{k(x)}}\exp\left[\pm i \int^x k(x')dx'\right] $$ Again at the classical turning point, when $k \to 0$ we see that the amplitude explodes and the wavefunction $\psi_1(x)$ is not a valid one anymore (for more on this matter, you will find useful answers here)

Now when you try to apply WKB, the intuition is that, in the classically allowed regions ($E>V$), the trigonometric WKB solutions are used and in the classically forbidden regions, exponentially decaying solutions are taken (and not just zero, because quantum mechanics still allows a non-zero probability for the system to be in a classically forbidden region), all of which, so that the approximation remains physically sensible.


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