Turning points and connection formulas · Quantum Mechanics

Intuition

Near a classical turning point the WKB formulas blow up, but there the potential is nearly a straight line, and the Schrödinger equation in a linear potential is solved exactly by the Airy function. Far from the turning point on either side, the Airy function turns into a WKB wave: an oscillation on the allowed side and a decay on the forbidden side, joined with a definite phase of $\pi/4$ . These connection formulas let WKB solutions be carried across turning points, and so give energy levels and tunnelling rates.

A road crossing a river needs a bridge built for the crossing itself; on each bank the ordinary road continues. The Airy function is the bridge at a turning point, and the connection formula says how the roads on the two banks line up.

Near a turning point, at $x=0$ , the exact solution in a linear potential is the Airy function (solid). Away from it on either side the WKB forms (dashed) match it — an oscillation with phase $+\pi/4$ on the allowed side, a decay on the forbidden side — and both blow up near the turning point itself.

Connection formulas

For a turning point at $x=a$ with the allowed region on the left, the decaying WKB solution on the right continues into the oscillating one on the left with a phase of $\pi/4$ .

\begin{aligned}&\frac{2C}{\sqrt{p}}\cos\Bigl(\frac1\hbar\int_{x}^{a}p\,dx'-\frac{\pi}{4}\Bigr)\\ &\longleftrightarrow\ \frac{C}{\sqrt{|p|}}\exp\Bigl(-\frac1\hbar\int_{a}^{x}|p|\,dx'\Bigr)\end{aligned}

Properties

Near $x=a$ the potential is $V\approx E+V'(a)(x-a)$ , and the Schrödinger equation becomes the Airy equation in a scaled variable .

Where the connection formula comes from

\begin{aligned}&V(x)\approx E+F\,(x-a)\\ &\Longrightarrow\ \psi\propto\text{Ai}\bigl(\beta(x-a)\bigr),\quad \beta=\Bigl(\frac{2mF}{\hbar^{2}}\Bigr)^{1/3}\end{aligned}

Straighten the potential near the turning point and rescale the variable: the equation becomes Airy’s. The solution that decays into the forbidden region is $\text{Ai}$ . Its known asymptotic forms on both sides are exactly the two WKB expressions of the connection formula.

Proof steps

-\frac{\hbar^{2}}{2m}\psi''+F(x-a)\,\psi=0

With $V\approx E+F(x-a)$ , $F>0$ , the equation near the turning point.

z=\beta(x-a),\quad \beta^{3}=\frac{2mF}{\hbar^{2}}\ \Rightarrow\ \frac{d^{2}\psi}{dz^{2}}=z\,\psi

Rescaling removes every constant: Airy’s equation.

\psi=C'\,\text{Ai}(z)

Of the two solutions, $\text{Ai}$ is the one that decays for $z\to+\infty$ , into the forbidden region.

\begin{aligned}\text{Ai}(z)&\approx\frac{e^{-\frac23z^{3/2}}}{2\sqrt\pi\,z^{1/4}}\quad(z\gg1)\\ \text{Ai}(z)&\approx\frac{\sin\bigl(\frac23|z|^{3/2}+\frac\pi4\bigr)}{\sqrt\pi\,|z|^{1/4}}\quad(z\ll-1)\end{aligned}

The asymptotic forms of the Airy function, quoted without proof.

\begin{aligned}&\tfrac23|z|^{3/2}=\frac1\hbar\int_{x}^{a}p\,dx'\\ &\sin\bigl(\theta+\tfrac\pi4\bigr)=\cos\bigl(\theta-\tfrac\pi4\bigr)\end{aligned}

In the linear region the Airy phases are the WKB integrals, which gives the connection formula.

Applications

Practice

A Straight Line Near the Turn

Close to a turning point the potential can be replaced by its tangent line. The Schrödinger equation in a linear potential is solved exactly by the Airy function.

Try it

Why does the Airy function appear at turning points?