Eigenvalue and Eigenfunction for a particle trapped in a 1D infinite asymmetric potential well As we're barely scratching the surface of Quantum Physics in class, we haven't been taught about asymmetric potential wells. However, I find it fascinating, moreover difficult, to find the eigenvalues and eigenfunction for a particle trapped in a 1D infinite asymmetric potential well.
I don't know how to do this, so here's what I thought:
I came up with the region-wise potentials, i.e.
$$V=+\infty , x<-l_1$$
$$V=V_1 , -l_1<x<0$$
$$V=V_2 , 0<x<l_2$$
$$V=+\infty , x> l_2$$
Now, we need to write four Schrodinger Equations, two for V1 and two for V2.
Then we need to find respective solutions to the four equations, apply boundary conditions and solve it.
I have a gut feeling I've gone wrong somewhere in my thought process. As I've just started learning about Quantum Mechanics, I request you to help me out.
I'd really appreciate if someone could find the eigenvalues and eigenfunctions, and post the step-by-step solution, as I feel I won't be able to solve this on my own(and I'm a bit tired from trying to solve this...)
Any help would be much appreciated!!Thanks, in advance!!
 A: So here's how you'd go about solving a question like this. However, keep in mind that you will not be able to get a simple analytic answer: you will eventually need to resort to numerical methods to solve for the explicit values of certain quantities.
The first step is to realise that the potential is only ever defined up to some global constant. As a result, you don't need two potentials, you could just as well set one of them (the "lower" one) to $0$. This will simplify the calculations slightly, and is perfectly justified physically. So now our potential well looks like this:
$$V(x) = \begin{cases}0, \quad &-1\leq x<0 \texttt{  (Region I)}\\ V_0, \quad &{\,\,\,\,\,0}\leq x\leq 1 \texttt{  (Region II)}\\ \infty \quad &\,\,\,\text{ otherwise.}\end{cases}$$
Notice that I've defined the "center" of the well at $x=0$. This is not essential, but greatly simplifies the calculations as well. I've also chosen the length to be $2$ in some units. It's not hard to generalise this.
Now, let's look for solutions in the different regions. Obviously, for $|x|>1$, $\psi(x) =0$, so I'm not going to worry about those regions. I am only interested in the regions I and II I've defined above. Furthermore, the wavefunctions should clearly go to zero at $x=-1$ and $x=1$. (Why?) Using this fact, we can guess the following general solutions in the two regions:
$\texttt{Region I}$
$$\psi_I(x) = A \sin\Big(k (x+1)\Big), \quad \quad \quad k = \sqrt{\frac{2mE}{\hbar^2}}$$
$\texttt{Region II}$
$$\psi_{II}(x) = \begin{cases}B \sin\Big(q\, (x-1)\Big), \quad \quad \quad q = \sqrt{\frac{2m(E-V_0)}{\hbar^2}} \quad \quad (E>V_0)\\~\\ C \sinh\Big(\kappa \, (x-1)\Big), \quad \quad \quad \kappa = \sqrt{\frac{2m(V_0-E)}{\hbar^2}} \quad \quad (E\leq V_0)\end{cases}$$
At first glance you might be confused as to how we've guessed these solutions, but if you look at it carefully, you should understand: in Region I, the solution would always be some combination of sines and cosines. The most general combination that satisfies the boundary condition that $\psi_I(-1)=0$ is the one provided. The same holds true for Region II, when $E>V_0$. When $E<V_0$ in Region II, the solution is some combination of (real) exponentials. Again, by choosing the hyperbolic sine function below, we have incorporated the fact that this combination go to zero at $x=1$. If you are still skeptical, I invite you to start off with the general solution with real or complex exponentials, and impose the boundary conditions. I guarantee that -- up to some overall multiplicative constant -- you'll get the same result.
The next step is to make sure that the wavefunction and its derivative match when $x=0$ (at the interface of the two regions). You will have to do this for both cases, $E<V_0$ and $E>V_0$. This will give you equations that define $k$, $\kappa$, and $q$ in terms of the parameters in this problem.
If you work out the complete problem, you should be able to see that when $E>V_0$, you require: $$\frac{\tan{k}}{k} = -\frac{\tan{q}}{q}.$$
Similarly, when $E<V_0$, you require: $$\frac{\tan{k}}{k} = -\frac{\tanh{\kappa}}{\kappa}.$$
(It's really easy to show the above relations, I'm not explicitly showing it here, since it's a good exercise for someone just beginning a course in QM.) To explictly solve these relations, you'll need to use numerical method. Once you do, you will have the allowed values of $k$, $\kappa$, and $q$. There will also be an overall multiplicative constant $A$ that you can set by requiring that the wavefunction be normalised. And that's it! Problem solved.
If you're interested, there is a nice animation of this exact problem, showing what the wavefunctions look like in the different regions here. However, I should point out that the "derivation" given on that site is filled with typos and other small errors that make it a little annoying. The animation is great, though.
