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This is a problem in quantum mechanics when one considers a linear potential; in physics-speak the equation would be written as

$$\frac{d^2\psi}{dx^2} + \frac{2m}{\hbar^2}(E-ax)\psi = 0,$$ with $V(x) = ax$.

I've been looking at it for a while and can't find a solution, and I thought I'd ask here for a hint before I go running to W|A for help. Disclaimer: I have no idea if a closed form solution exists.

The first thing I thought is to try a series expansion, but the recurrence relation between the coefficients is pretty ugly. Then I tried a Fourier transform. I have no reason to think the solution will be integrable in $\mathbb{R}$, but as a physicist I've been trained not to care about such things.


$$ \phi(k) = \frac1{\sqrt{2\pi}}\int_{-\infty}^\infty \psi(x)e^{-ikx}\ dx $$

we get the equation

$$\frac{d\phi}{dk} - \frac{i}{a}\left(\frac{\hbar^2 k^2}{2m}-E\right)\phi = 0$$

which is easily solved to get

$$\large \phi(k) = Ae^{-\frac{ik}{a}(\frac{\hbar^2 k^2}{6m}-E)}$$

which not only can't be Fourier-transformed since it doesn't go to zero at infinity, I also have no idea how to integrate.

So this is as far as I got. Does anyone know how to solve this equation, and if there's no closed form (which I suspect), how to get a nice series or something?

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Side note; how do I make the expression for $\phi(k)$ look nice? The symbols look extremely small to me. – Javier Nov 15 '13 at 19:51
"but as a physicist I've been trained not to care about such things." : +1000000 :-) That cracked me up :D – Avitus Nov 15 '13 at 19:52
Look at Airy functions: – Felix Marin Nov 15 '13 at 19:57
You could certainly give it a try with a numerical solution assuming you have initial conditions – user88595 Nov 15 '13 at 19:57
@JavierBadia You can try \large and/or \dfrac (instead of \frac), but\dfrac in exponents looks ugly. Usually it looks better to write a fraction $\dfrac a b$ in an exponent as $a/b$. I'd also use\left( and \right) for the parentheses. – Git Gud Nov 15 '13 at 19:57

1 Answer 1

up vote 4 down vote accepted

The solution is in term of the Airy functions, $\text{Ai}(x)$, $\text{Bi}(x)$ (which are well defined, have asymptotic formulas and series representation, etc. see for example Abramowitz & Stegun or Szegö).

These functions are the linearly indepedent solutions of $$\frac{d^2y(x)}{dx^2}-xy(x)=0 $$ With the change of variable $$\tilde{x} = -\frac{b+ax}{(-a)^{2/3}}:=\mu x +\nu$$ (where $\mu,\nu$ are just used to abreviate the longer formula), you will be able to transform the initial form to a Airy form. The solution is then directly

$$y(x)=\alpha \text{Ai}(\mu x + \nu)+\beta \text{Bi}(\mu x + \nu) $$

where $\alpha,\beta$ are the integration constants.

Source: I spent the last 3 days playing around with these functions..

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