Let us consider the following potential $V(x)$ such that

$$ V(x) = \begin{cases} 0&0<x<b \\ V_0&b<x<a \\ \infty&x>a \\ \end{cases}$$

And we have $V(x)=V(-x)$. We are considering states with energy $E \lt V_0$

If we had $x\rightarrow \infty$, we would have our finite square well. In the region between $-b$ and $b$, we would have a sinusoidal wavefunction which would be :$$\psi(x)=A\sin(kx)+B\cos(kx)$$

In the region $[-a,-b] \space\cup\space[b,a]$, we would have exponentially decaying wavefunctions of the form :

$$\psi(x\in[-a,-b])=Ce^{\alpha x} $$

$$\psi(x\in[b,a])=Ce^{-\alpha x} $$

This is derived by solving the Schrodinger equation, and setting the boundary conditions $\psi(x)=0$ for $x\rightarrow \pm\infty$.

Using the Dirichlet and Neumann boundary conditions, we could derive transcendental equations, and solve them to get our bound states and the energies.

However, all this was true when $a$ was infinite. Now we have a case where $a$ is finite. Is there any way to solve this analytically, and find the wavefunction ? I have solved this problem using perturbation theory, but I can't seem to be able to use the boundary conditions properly.

The solution in the region $[-b,b]$ remains the same, with different amplitudes ofcourse. Let us consider this :


In the region $[b,a]$ however, the TISE yields the solution :

$$\psi(x\in[b,a])=C'e^{-\alpha x} +D'e^{\alpha x}$$

I don't think we can set $D=0$ by claiming that wavefunction vanishes at infinity. Moreover, this new wavefunction must vanish at $a$. Hence $\psi(a)=0$.

I have no clue how to apply the boundary conditions in this case, at $x=a$, and get the correct relations that would tell us about the energy levels and the wavefunction.

Furthermore, I want to set $b\rightarrow a$ and show that this solution becomes the solution for the infinite well potential.

Any help in solving this problem would be highly appreciated.

  • $\begingroup$ I think you mean “a goes to inf” instead of x, for the case of the finite well $\endgroup$ May 31 at 5:30

2 Answers 2


The boundary conditions at $x=\pm a$ are $\psi(x)=0$, just like in the usual particle in an infinite potential well problem, i.e. the wavefunction must vanish at the edges of the impenetrable potential barrier. In total these provide 2 boundary conditions, one for each infinite barrier.

At $x = \pm b$, the boundary conditions are the continuity of $\psi(x)$ and $\psi'(x)$. These provide 4 boundary conditions: two equations for each of the two interfaces.

In total there are 6 unknown coefficients (two for each region with finite potential) and 6 boundary conditions. I'm not sure an analytical solution is possible, but you can certainly write down the equations and solve them numerically.


In addition to what Puk pointed out, your potential is symmetric, so you can solve for the even/odd parity solutions separately (which usually simplifies the algebra considerably). I also see that you have chosen a wavefunction for $[b,a]$ which implies that you have $E \leq V_0$. For a complete set of solutions you should also consider $E>V_0$ as these states will also be bound. As you take your limit $a \to \infty$, you can consider only the first set of results and see if you get the established results for a square well.

Addendum: I beleive your problem has already been discussed here: Finite Square Well Inside an Infinite Square Well


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