# Finite Potential well inside an infinite potential well

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

$$V(x) = \begin{cases} 0&0a \\ \end{cases}$$

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 :

$$\psi(x)=A'\sin(kx)+B'\cos(kx)$$

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.

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

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 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.