# Tag Info

Test your code. "I don't know if my linear algebra routines work or not" is a problem you can solve easily. It is easy to check if you have computed the correct eigenvalues or not; just check that $VDV^{-1}=H$, or if you don't fancy the inversion $HV=VD$. If you are not sure that your individual pieces work, you are walking in the dark. Possibly a hot take: ...
As David said, absorbing boundary conditions won't be completely reflectionless. That said, we can reduce relfections quite a bit, which helps to avoid influence from the boundaries while the particle still travelling inside. Since this is a time dependent problem, one simple choice of boundary conditions will look like this. At the left boundary: $$\frac{\... 2 I don't have enough reputation to comment, but I wanted to give a basic complement of gIS's answer. Simply put, if we have the decomposition over V=V_1\otimes V_2, which we represent with, say, the reshaping [2,4,2,4] why do we einsum over the first and third indices for trace over V_1 and second and fourth for trace over V_2? Maybe this is obvious ... 2 Your problem was the lower limit of integration. It should have been -x_e instead of 0, since x_e is the equilibrium point for the potential and not the minimum distance. After correcting that, you get the following #%% Solution xe, lam = 1.0, 6.0 # parameters for potential xmax = 10 # Bval, Bval2 = wavefunction values at x = bound1, bound2 bound1, ... 1 Problem solved: I just have to include the solution of the homogeneous equation into the solution of the differential equation. The solution to the differential equation is$$V_H(r) = -\frac{K}{r} - \frac{r+1}{r} \exp(-2r).$$By requiring that \lim_{r \to 0} V_H(r) is real we get K=-1 and$$V_H(r) = \frac{1}{r} - \frac{r+1}{r} \exp(-2r).$$1 Your whole code is not understandable to me. Try this alternative components: int grid(int i, double a, double h) { return a+i*h; } int index(int i, int j, int N) { return i*N+j; } //potential function. returns the integral over the grid square indexed by i double V(int i, double x0, double hx, int j, double y0, double hy) { double xi=grid(i,... 1 I think your approach is very good. Several studies show that doing something stays deeper in memory than reading about it and testing things for yourself will give you a good intuitive feeling. Now, to your questions: Is MATLAB capable of all this? Yes, it has strong numerical and GUI capabilities If not, what is? As mentioned by Abdullah Ali Sivas, you ... 1 I can describe the simplified version of this algorithm which might help you to understand what's going on here. Let's you have this Time-dependent Schrodinger equation:$$i \hbar \partial_{t} \Psi = -\frac{\hbar^{2}}{2m} \nabla^{2} \Psi For a moment forget about potential cause that complicate things here, which might be necessarily helpful for ...