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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 :

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

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.

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  • $\begingroup$ 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

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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.

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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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