# TISE for a triangular potential

I want to solve the TISE of a particle of charge $q$ and mass $m$ in a one dimensional triangular potential, with an infinitely high potential wall at $x = 0$, i.e.

$$\hat{H}=-\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2}+V(x),$$ with

$$V\left(x\right)=\begin{cases}qEx&{\rm if}\,x>0 \\ \infty &{\rm if}\,x\leq0\end{cases}$$

For $x>0$, this gives the TISE $$-\frac{\hbar^2}{2m}\frac{\partial^2\Psi_n}{\partial x^2}+qEx\cdot\Psi_n = E_n\Psi_n.$$ I know that the solutions of $$\frac{d^2y}{dx^2} - xy = 0$$ are given by the Airy functions, however I don't know how I can transform the TISE to this form. I know it's possible, because I have the solution to the problem, but I don't know how I can myself come up with this, I don't see the aim which has to be reached in order for a substitution to be successful. Also, I don't know how the differential will transform under a substitution including $x$. Any help?

The substitution used in the solution is $$u=\left(\frac{2mqE}{\hbar^2}\right)^\frac{1}{3}\left(x-\frac{E_n}{qE}\right).$$

• possible duplicate of Wave function of a particle in a gravitational field – Kyle Kanos Jan 18 '14 at 15:34
• It's not a duplicate, I'm not asking for the method described there. I need an explanation of the method using change of variable. I understand that both yield the same result, but the result is of minor importance for me. – Pascal Engeler Jan 18 '14 at 18:26
• @KyleKanos nope, it is not a duplicate, what does this question have to do with the gravitational potential? BTW it would be great if people could refrain from voting to close such and similar perfectly legitimate questions generally ... – Dilaton Jan 19 '14 at 6:47
• @Dilaton KyleKanos is right in that the differential equations are essentially the same (both have potential $V(z)=Az$), however in the other post another method is (thoroughly) explained to solve it. – Pascal Engeler Jan 19 '14 at 13:21
• @Dilaton Last I checked both problems involve linear potentials. You don't need to cry over it because the problem was solved within a few hours of posting and 11 hours before you complained about it. I doubt at this point it will get closed. – Kyle Kanos Jan 19 '14 at 14:54

Well the first step is to rearrange the equation to take the form $$\frac{d^2\psi}{dx^2}=\frac{2mqE}{\hbar^2}\left(x-\frac{E_n}{qE}\right)\psi$$
Since we are free to choose any substitution we like, we let the term in the parenthesis, $x-E_n/qE$, be equal to the new variables, $z$, times some normalizing factor $\beta$ (so that we end up with $\psi''=z\psi$): $$z=\beta\left(x-\frac{E_n}{qE}\right) \\ dz=\beta\,dx$$ Using the above two equations, $$\beta^2\frac{d^2\psi}{dz^2}=\frac{2mqE}{\hbar^2}\beta^{-1} z\psi$$ Moving the $\beta^2$ term to the right, we have $$\frac{d^2\psi}{dz^2}=\frac{2mqE}{\hbar^2}\beta^{-3} z\psi$$ Since $\beta$ is there to make the whole co-factor be 1, we get $$\frac{2mqE}{\hbar^2}\beta^{-3}=1\to\beta=\left(\frac{2mqE}{\hbar^2}\right)^{1/3}$$ Such that our substitute variable is now defined as $$z=\left(\frac{2mqE}{\hbar^2}\right)^{1/3}\left(x-\frac{E_n}{qE}\right)$$ which is what you get.