# Triangular barrier in infinite potential well

Suppose I am looking to solve the wavefunction for the following 1D potential: $$U(x) = \begin{cases}V_0\frac{a-|x|}{a}&\quad\text{for}\quad|x|<a \\\infty&\quad\text{for}\quad|x|>a\end{cases} \tag{1}$$ Since our potential is symmetric, we have even and odd solutions and so we can solve the system for $x\ge 0$ and afterward construct the full solution for either $\psi_{even}$ or $\psi_{odd}$. For the region inside the well, we can cast the time-independent Schrodinger eq. into a form that is the Airy differential equation (note that it is a linear potential), namely $$\frac{d^2\psi}{dz^2}-z\psi = 0, \tag{2}$$ where $$z\equiv Q(1 - \eta), \tag{3}$$

where

$$\eta = \frac{x}{a_0}, \ \ \ Q = \frac{2mV_0a_0}{\hbar^2a}, \ \ \ a_0 \equiv a-\frac{a}{V_0}E. \tag{4}$$

Therefore we can express $\psi$ as

$$\psi(z) = C_1Ai(z)+C_2Bi(z). \tag{5}$$

Our boundary conditions tell us: $$\psi(\zeta_a) = 0 \tag{6}$$ $$\psi(\zeta_0) = 0\quad\text{for}\quad\psi_{odd} \tag{7}$$ $$\psi'(\zeta_0) = 0\quad\text{for}\quad\psi_{even} \tag{8}.$$

where I am denoting $\zeta_0= z|_{x=0}$ and $\zeta_a= z|_{x=a}$.

Question: Is this system exactly solvable?

Typically we only have one zero boundary condition in these types of problems which allows us to quantize the energy levels through the zero's of the Airy function. But here we must simultaneously satisfy two zero boundary conditions. To clarify, here is our system of linear equations.

$\psi_{odd}$: $$C_1Ai(\zeta_0)+C_2Bi(\zeta_0) = 0 \tag{9}$$ $$C_1Ai(\zeta_a)+C_2Bi(\zeta_a) = 0 \tag{10}$$

$\psi_{even}$: $$C_1Ai'(\zeta_0)+C_2Bi'(\zeta_0) = 0, \tag{11}$$ $$C_1Ai(\zeta_a)+C_2Bi(\zeta_a) = 0. \tag{12}$$

My proposed idea would be to solve for the zero determinant of the homogeneous linear equations, but this would force me to a numerical solution. I would like to find the energies in terms of the zeros of the Airy functions. Ideas?

• I trust you know the article Airy equation in Wikipedia. The point $x=a$ where you want $\psi=0$, falls at $z=Q(1-a/a_0)$. It will give you the ratio $C_2/C_1$. But, pay attention that as $x$ decreases, $z$ increases. You can see how looks like the region $z>0$ in the graphs in the article. You'll be interested to require also that $\psi$ or its derivative, be zero at $x=0$ i.e. $z=Q$, which is to the left of $z = Q(1-a/a_0)$. It doesn't seem that in this region you'll get another zero or a maximum (or minimum) for $\psi(z)$.(I continue) – Sofia Mar 3 '15 at 3:15
• In general, a system of 2 homogeneous eqs. with 2 variables, may be incompatible. What could save the situation were the value of $E$, which determines the value of $a_0$. But, again I see a problem in the fact that the values of $z$ fall in the region $z>0$. It seems to me that you could have obtained solutions of your equation including the condition of symmetry or antisymmetry, if $z$ took values in the region $z<0$ where the functions $Ai$ and $Bi$ oscillate. But with $z>0$ I am quite skeptical. – Sofia Mar 3 '15 at 3:15

## 2 Answers

Indeed, there is no getting around having to solve this problem numerically. Departing from the determinant method originally posted, perhaps an alternative approach here would be to solve the system using only the energy. We start with the full form of our points of interest taking the odd wavefunctions as an example: $$z(0) = x_0a\left(-\frac{E}{V_0}+1\right)$$ $$z(a) = x_0a\left(-\frac{E}{V_0}\right).$$ I denote $\alpha_i$ and $\beta_i$ as the $i$th zeros of $Ai(z)$ and $Bi(z)$ respectively. Likewise $\alpha'_i$ and $\beta'_i$ the zeros of the derivatives. Also $x_0 \equiv \left[\frac{2mV_0}{\hbar^2a}\right]^{1/3}$ - notation of original post was edited :(

Firstly, we know that $E>V_{min}$ and thus $E>0$. Therefore, we see that $z(0)>z(a)$. Now, the first zero of either $Ai(z)$ or $Bi(z)$ must occur at $z<0$. For an appropriate value of $E$, we can see that if the zeros of the Airy functions differed by $x_0a$, then our condition would be satisfied for a single energy. Working with the Airy functions of the first kind, we can impose the condition $$\alpha_i = x_0a\left(-\frac{E}{V_0}\right)$$ or $$E = -\frac{V_0}{x_0a}\alpha_i.$$ The criteria on the second zero is then $$\alpha_j = x_0a+\alpha_i.$$ We could numerically cycle through all the zeros of the Airy function and for those with a separation of $$\alpha_j-\alpha_i=x_0a$$ we would then have a corresponding energy of $$E = -\frac{V_0}{x_0a}\alpha_i.$$ Applying the same concept to both even and odd wavefucntions we have: $$\psi_{even}(z) = C_1Ai(z);\quad E = -\frac{V_0}{x_0a}\alpha_i;\quad\alpha'_j -\alpha_i= x_0a$$ $$\psi_{odd}(z) = C_2Ai(z);\quad E = -\frac{V_0}{x_0a}\alpha_i;\quad\alpha_j -\alpha_i= x_0a$$ $$\psi_{even}(z) = C_3Bi(z);\quad E = -\frac{V_0}{x_0a}\beta_i;\quad\beta'_j -\beta_i= x_0a$$ $$\psi_{odd}(z) = C_4Bi(z);\quad E = -\frac{V_0}{x_0a}\beta_i;\quad\beta_j -\beta_i= x_0a$$ where the third column denotes the condition that must be satisfied by our zeros, $C_i$ is the appropriate normalization constant for each wavefunction, and $$z\equiv \left[\frac{2mV_0}{\hbar^2a}\right]^{1/3}(a_0-x)=x_0(a_0-x);\quad a_0 \equiv \left[a-\frac{a}{V_0}E\right].$$ In looking at a graph of the Airy function, we can see that at $z\ll 0$ the wavelength'' is very small and so there should be a great number of possible zero's that will satisfy our criteria. In the same light, at this range of $z$, our energies will be very large and we have the semblance of the infinite square well. Thank you Sofia and Ali Moh for the helpful comments.

What you're asking for is impossible. The energy eigenvalues are given by the zeros of the determinant (as you surmised) and cannot be related to the zeros of the airy function. $$Ai(\zeta_a)Bi(\zeta_0)-Ai(\zeta_0)Bi(\zeta_a)=0$$ For the odd solution (and with derivatives on $Ai$ for the even solution).

Also be careful, your $z$ is not dimensionless as it should be, so your change of variables is not quite right...

Please see comments for physical arguments that guarantee infinitely many solutions to this condition!

• I regret that you didn't read my comments. Through the determinant the OP tries to find the constants $C$ that allow fulfilling the boundary conditions. It is not sure whether the determinant can be made zero. However, if the determinant can be made zero, from the value $a_0$ for which the determinant becomes zero, one can find the energy $E$. – Sofia Mar 3 '15 at 18:25
• I read your comment but I don't agree with it. The determinant can always be made zero, the energy can be changed continuously to satisfy this condition. One variable with one condition. But my guarantee mostly stems from physics, it is impossible that there aren't many energy eigenvalues! because this is a potential well with infinite height, so for high enough energies the wave function will be actually blind to the triangular bump and we know in that case that there is an infinite number of discrete energy eigenvalues – Ali Moh Mar 3 '15 at 18:34
• So I cannot tell for sure whether these eigenvalues will comparable to $V_0$ (the height of the bump), but I can guarantee from the physical argument above that there exists at least eigenstates with $E>>V_0$.. More over it seems very unlikely to me that low lying energy eigenvalues don'e exist – Ali Moh Mar 3 '15 at 18:36
• An equivalent way is to note the asymptotic trigonometric behavior of airy functions, which validates the physical intuition – Ali Moh Mar 3 '15 at 18:41
• There is no need of $E >> V_0$, it seems sufficient $E>V_0$. The problem is for $E < V_0$. You are looking at the determinant and I am looking at how look the wave-functions in the region $z > 0$. – Sofia Mar 3 '15 at 18:42