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?
Best Answer
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.