This is my first question on here. I'm trying to numerically solve the Schrödinger equation for the Woods-Saxon Potential and find the energy eigenvalues and eigenfunctions but I am confused about how exactly this should be done.
I've solved some initial value problems in the past using iterative methods such as Runge–Kutta. I've read that Numerov's method is the way to solve Schrödinger's equation but Wikipedia also describes it as an iterative method for initial value problems.
How do I use it to solve an eigenvalue problem?
This confuses me for the following reasons:
- Wouldn't iteratively solving the DE require knowledge of the energy eigenvalues to use as input to the calculation? I don't know the eigenvalues yet; they're precisely what I'm trying to calculate.
- If I did that, wouldn't I simply get a unique solution, instead of a family of eigenfunctions and eigenvalues?
I've seen some mention of "tridiagonal matrices" being generated somehow, but am not sure what the elements of that matrix would be or how that applies to the problem. Leandro M. mentioned that "the discretization defines a finite dimensional (matrix) eigenvalue problem". This seems like the correct road I should be going down, but I haven't been able to find anything that explicitly explains this process or how the matrix is constructed. If this is the correct procedure, how is such a matrix constructed?
Best Answer
I'm glad I can finally answer this.
Numerov's method as described on Wikipedia is not how you want to proceed here. To give you an idea of how to proceed, let's start with a simplified version of the method. What I'm going to do is to just naively discretize the differential operator, like so:
$$ \frac{d^2}{dx^2}\psi \approx \frac{\psi(x-d)-2\psi(x) +\psi(x+d)}{d^2} \equiv \frac{\psi_{n-1} - 2 \psi_n + \psi_{n+1}}{d^2}$$
The last equation is just a definition -- I'm treating space as if it's a lattice with points $d$ apart and I'm calling $\psi_n$ the value of the wavefunction on the $n$th point.. Now the Schrödinger equation reads in this notation:
$$-\frac{\hbar}{2m}\frac{\psi_{n-1} - 2 \psi_n + \psi_{n+1}}{d^2} + V_n \psi_n = E \psi_n$$
But this is a matrix equation! Let me be more explicit:
\begin{align} -\frac{\hbar}{2md^2}&\left(\begin{array}{cccccccc} -2 & 1 & & & \\ 1 & -2 & 1 & & \\ &\ddots&\ddots&\ddots& \\ & & 1 & -2 & 1 \\ & & & 1 & -2 \\ \end{array}\right) \left(\begin{array}{c} \psi_1 \\ \psi_2 \\ \vdots \\ \psi_{N-1} \\ \psi_N \end{array}\right) +\\ &\qquad\qquad\quad\left(\begin{array}{cccccccc} V_1 & & & & \\ & V_2 & & & \\ & & \ddots & & \\ & & & V_{N-1} & \\ & & & & V_N\\ \end{array}\right)\left(\begin{array}{c} \psi_1 \\ \psi_2 \\ \vdots \\ \psi_{N-1} \\ \psi_N \end{array}\right) = E \left(\begin{array}{c} \psi_1 \\ \psi_2 \\ \vdots \\ \psi_{N-1} \\ \psi_N \end{array}\right) \end{align}
Of course, I had to pick some integer $N$ that just corresponds to placing the system in a box that's "big enough" in order to have a finite system.
It is clear now that what you have is a matrix eigenvalue problem of the form
$$A\psi = E \psi$$
and you may proceed to diagonalize it in whatever way you choose. Note that we call $A$ a tridiagonal matrix, for obvious reasons. You might want to take care of the boundary conditions first -- you do that by setting $\psi_1 = \psi_n = 0$ before you construct the matrix, which corresponds to setting the first and last columns to zero. This will give you some spurious eigenfunctions with zero eigenvalue that you can just discard. If you have different boundary conditions, you're out of luck -- I don't know of a way that makes it work in general.
Now you just have to redo the above with the full Numerov's method, which will be a bit more complicated, and you're all set.
This seems to be the way you're looking for but of course it's not the only way to do this. Griffiths describes one called "wag the dog" that consists of the following: you guess an initial value for an energy and compute the wavefunction however you want (RK, for instance). Chances are it won't be an eigenvalue so the wavefunction will blow up at infinity. It might go to $+\infty$ for large $x$ or it might go to $-\infty$. You now slowly vary the energy until the behavior at infinity "flips", that is, until the tail "wags". That will allow you to constrain the value of a single energy eigenvalue and give you the form of the wavefunction out to some maximum size that increases as the chosen $E$ approaches the exact eigenvalue.