Airy Functions

Airy functions are the universal local functions near a simple turning point. Whenever a one-dimensional Schrödinger equation has $E=V(x_0)$ and $V'(x_0)\ne0$, the potential is approximately linear near $x_0$, and the local equation reduces to the Airy equation.

The formulas below explain how one smooth function connects oscillations to exponential decay. For the resulting WKB matching rules, see Turning Points and Connection Formulas.

The Airy Equation

The standard Airy equation is

$$y''-xy=0.$$

Its two standard real solutions are

$$\operatorname{Ai}(x), \qquad \operatorname{Bi}(x).$$

Every solution is a linear combination

$$y(x) = A\,\operatorname{Ai}(x) +B\,\operatorname{Bi}(x).$$

The functions are entire functions of $x$. On the real axis, $\operatorname{Ai}(x)$ is the solution that decays for $x\to+\infty$, while $\operatorname{Bi}(x)$ grows for $x\to+\infty$.

Integral Representation

For real $x$, one useful oscillatory improper integral is

$$\operatorname{Ai}(x) = \frac{1}{\pi} \int_0^\infty \cos\left( \frac{t^3}{3}+xt \right)\,dt.$$

This formula already hints at the turning-point role. The phase

$$\frac{t^3}{3}+xt$$

has stationary points $t=\pm\sqrt{-x}$ on the full real line when $x<0$; they meet at $t=0$ as $x\to0$. In the half-line cosine integral, the positive saddle reaches the endpoint. Ordinary nondegenerate stationary phase then fails, and Airy scaling describes the transition. The integral is conditionally convergent, so differentiating it repeatedly requires a regulator or an equivalent contour representation.

The Wronskian normalization is

$$W\!\left[ \operatorname{Ai}, \operatorname{Bi} \right] = \operatorname{Ai}(x)\operatorname{Bi}'(x) - \operatorname{Ai}'(x)\operatorname{Bi}(x) = \frac{1}{\pi}.$$

Large Positive Argument

As $x\to+\infty$,

$$\operatorname{Ai}(x) \sim \frac{1}{2\sqrt{\pi}}\, x^{-1/4} \exp\left( - \frac{2}{3}x^{3/2} \right),$$

whereas

$$\operatorname{Bi}(x) \sim \frac{1}{\sqrt{\pi}}\, x^{-1/4} \exp\left( \frac{2}{3}x^{3/2} \right).$$

Thus $\operatorname{Ai}$ is the decaying solution on the positive real axis, and $\operatorname{Bi}$ is the growing solution. This is why bound-state tails and forbidden-region boundary conditions often select $\operatorname{Ai}$ locally.

Large Negative Argument

For $x>0$ with the Airy argument equal to $-x$,

$$\operatorname{Ai}(-x) \sim \frac{1}{\sqrt{\pi}}\, x^{-1/4} \sin\left( \frac{2}{3}x^{3/2} +\frac{\pi}{4} \right),$$

and

$$\operatorname{Bi}(-x) \sim \frac{1}{\sqrt{\pi}}\, x^{-1/4} \cos\left( \frac{2}{3}x^{3/2} +\frac{\pi}{4} \right).$$

Positive Airy argument therefore corresponds to exponential behavior, while negative Airy argument corresponds to oscillatory behavior. This is the local mathematical source of WKB connection formulas.

Linear Potential Example

Consider the one-dimensional stationary Schrödinger equation

$$- \frac{\hbar^2}{2m} \frac{d^2\psi}{dx^2} +Fx\psi = E\psi, \qquad F>0.$$

The turning point is

$$x_0=\frac{E}{F}.$$

Set

$$\ell_{\mathrm A} = \left( \frac{\hbar^2}{2mF} \right)^{1/3}, \qquad z=\frac{x-x_0}{\ell_{\mathrm A}}.$$

Then the Schrödinger equation becomes

$$\frac{d^2\psi}{dz^2} -z\psi =0.$$

Thus

$$\psi(x) = A\,\operatorname{Ai}(z) +B\,\operatorname{Bi}(z).$$

The length $\ell_{\mathrm A}$ is the Airy length. It is the scale over which the exact wavefunction rounds off the apparent WKB divergence at a simple turning point.

General Simple Turning Point

Let

$$V(x_0)=E, \qquad F=V'(x_0)\ne0.$$

Near $x_0$,

$$V(x)-E \approx F(x-x_0).$$

Define

$$\ell_{\mathrm A} = \left( \frac{\hbar^2}{2m\lvert F\rvert} \right)^{1/3}, \qquad z= \operatorname{sgn}(F) \frac{x-x_0}{\ell_{\mathrm A}}.$$

With this convention, the forbidden side has $z\gt0$ and the allowed side has $z\lt0$. To leading order in the linearization, the local equation becomes

$$\frac{d^2\psi}{dz^2} -z\psi =0.$$

The linearization must remain accurate over the region being modeled. For a twice differentiable potential, a useful local check is $|V\!\prime\prime(x_0)|\ell_{\mathrm A}\ll |F|$; higher derivatives may impose further restrictions. The decaying forbidden-side solution is locally proportional to $\operatorname{Ai}(z)$. Its negative-argument asymptotic form produces the oscillatory allowed-side WKB phase with the characteristic $\pi/4$ shift.

Relation to WKB

In a forbidden region, WKB gives a decaying form

$$\psi(x) \approx \frac{C}{\sqrt{\kappa(x)}} \exp\left[ - \frac{1}{\hbar} \int \kappa(x')\,dx' \right],$$

where

$$\kappa(x) = \sqrt{2m\left(V(x)-E\right)}.$$

Near a simple turning point this expression diverges as $\kappa^{-1/2}$, even though the exact wavefunction is finite. The Airy function supplies a uniform local replacement across the turning point.

For a decaying forbidden tail, the same $\operatorname{Ai}$ solution becomes an allowed-region oscillation with phase shift $\pi/4$. The precise orientation-dependent formulas are collected in Turning Points and Connection Formulas.

Beyond Simple Turning Points

Airy functions describe a simple, linear turning point. If the first nonzero derivative of $V(x)-E$ at the turning point is higher than first order, the local model changes. Higher-order turning points, coalescing turning points, and multidimensional caustics require different special functions or uniform approximations.

This distinction matters in applications. A smooth isolated turning point, a hard wall, a discontinuous rectangular barrier, and a barrier top are not the same local problem.

Common Mistakes

• Using $\operatorname{Ai}$ and $\operatorname{Bi}$ without stating the Airy equation convention.
• Forgetting that positive Airy argument is exponential and negative Airy argument is oscillatory.
• Choosing $\operatorname{Bi}$ for a forbidden tail that must decay at infinity.
• Dropping the Airy length $\ell_{\mathrm A}$ and treating the Airy argument as dimensionful.
• Assuming every turning point has the simple Airy form; higher-order turning points need different local models.
• Applying smooth-turning-point Airy matching directly to discontinuous potentials.
• Losing the sign of $V'(x_0)$ when deciding which side is allowed.

Related Topics

• Asymptotic Analysis
• WKB Approximation
• Bohr-Sommerfeld Quantization
• Barrier Penetration and Tunneling
• Hypergeometric Functions
• Small Parameters and Error Estimates

Exercises

1. Show that the linear potential problem with $V(x)=Fx$ and $F>0$ reduces to the Airy equation after the scaling in this page.

Solution

The Schrödinger equation is

$$- \frac{\hbar^2}{2m}\psi'' +Fx\psi = E\psi.$$

Move the potential term to the right:

$$\psi'' = \frac{2mF}{\hbar^2} (x-x_0)\psi, \qquad x_0=\frac{E}{F}.$$

With

$$z=\frac{x-x_0}{\ell_{\mathrm A}}, \qquad \ell_{\mathrm A} = \left( \frac{\hbar^2}{2mF} \right)^{1/3},$$

one has

$$\frac{d^2}{dx^2} = \frac{1}{\ell_{\mathrm A}^2} \frac{d^2}{dz^2}, \qquad x-x_0=\ell_{\mathrm A}z.$$

Substitution gives

$$\frac{1}{\ell_{\mathrm A}^2} \frac{d^2\psi}{dz^2} = \frac{2mF}{\hbar^2} \ell_{\mathrm A}z\psi.$$

Because $\ell_{\mathrm A}^3=\hbar^2/(2mF)$, this reduces to

$$\frac{d^2\psi}{dz^2} -z\psi =0.$$

2. Which Airy solution is selected by a boundary condition requiring decay as $x\to+\infty$ in the linear potential $V(x)=Fx$ with $F>0$?

Solution

For $F>0$, the Airy argument is

$$z=\frac{x-x_0}{\ell_{\mathrm A}}.$$

As $x\to+\infty$, $z\to+\infty$. The asymptotics are

$$\operatorname{Ai}(z) \sim \frac{1}{2\sqrt{\pi}}z^{-1/4} e^{-2z^{3/2}/3},$$

while $\operatorname{Bi}(z)$ grows exponentially. Therefore the decaying boundary condition selects $\operatorname{Ai}(z)$.

3. Use the negative-argument asymptotic form of $\operatorname{Ai}$ to identify the phase shift that appears when a decaying forbidden tail connects to an allowed-region oscillation.

Solution

For $x>0$,

$$\operatorname{Ai}(-x) \sim \frac{1}{\sqrt{\pi}}x^{-1/4} \sin\left( \frac{2}{3}x^{3/2} +\frac{\pi}{4} \right).$$

The oscillatory phase contains an added $\pi/4$. In WKB language this is the smooth turning-point phase shift for the standing wave connected to a decaying forbidden tail.

4. Explain why the Airy length grows when $\lvert V'(x_0)\rvert$ becomes small.

Solution

The Airy length is

$$\ell_{\mathrm A} = \left( \frac{\hbar^2}{2m\lvert V'(x_0)\rvert} \right)^{1/3}.$$

If the slope at the turning point decreases, the potential changes more slowly near $x_0$. The region over which the wavefunction transitions from oscillatory to exponential behavior becomes wider. In the limit $V'(x_0)\to0$, the turning point is no longer simple and the Airy approximation is no longer the correct local model.

References

• C. M. Bender and S. A. Orszag, Advanced Mathematical Methods for Scientists and Engineers, Springer, 1999.
• M. V. Berry and K. E. Mount, “Semiclassical approximations in wave mechanics,” Reports on Progress in Physics 35, 315-397, 1972.
• L. D. Landau and E. M. Lifshitz, Quantum Mechanics: Non-Relativistic Theory, 3rd ed., Butterworth-Heinemann, 1981.
• NIST, Digital Library of Mathematical Functions, §9.2: Definitions and Wronskians, §9.5: Integral Representations, and §9.7: Asymptotic Expansions.
• F. W. J. Olver, Asymptotics and Special Functions, A K Peters, 1997.