An update from hardmath's comment: here are the equationsequation I'm attempting to solve is the time-dependent Schroedinger equation, with no potential. $$i \hbar \frac{ \partial }{ \partial t} u(x, t) = - \frac{ \hbar^{2} }{2m} \frac{\partial^2}{\partial x^2} u(x, t)$$ Applying the Crank-Nicolson method results in (for me): $$u_{j}^{n+1}+\frac{1}{2}\frac{i\Delta t}{\hbar}\frac{-\hbar^{2}}{2m} \frac{u_{j+1}^{n+1}-2u_{j}^{n+1}+u_{j-1}^{n+1}}{\Delta x^2}= u_{j}^{n}-\frac{1}{2}\frac{i\Delta t}{\hbar}\frac{-\hbar^{2}}{2m} \frac{u_{j+1}^{n}-2u_{j}^{n}+u_{j-1}^{n}}{\Delta x^2}$$ Superscript is for time stepping, subscript for space stepping. $$A_{l} u^{n+1} = A_{r} u^{n}$$ where: $$A_{l} = \begin{pmatrix} 1-\alpha & \alpha & \cdots & \cdots & 0 \\ \alpha & 1-2\alpha & \alpha & \cdots & 0 \\ \vdots & \ddots & \ddots & \ddots & \vdots \\ 0 & \cdots & \alpha & 1-2\alpha & \alpha \\ 0 & \cdots & \cdots & \alpha & 1-\alpha \\ \end{pmatrix}$$ $$A_{r} = \begin{pmatrix} 1-\beta & \beta & \cdots & \cdots & 0 \\ \beta & 1-2\beta & \beta & \cdots & 0 \\ \vdots & \ddots & \ddots & \ddots & \vdots \\ 0 & \cdots & \beta & 1-2\beta & \beta \\ 0 & \cdots & \cdots & \beta & 1-\beta \\ \end{pmatrix}$$ The coefficients $\alpha$ and $\beta$ collect terms arising from applying the Crank-Nicolson method to the form of the Schroedinger equation I'm attempting to solve here (and working with natural units $\hbar=m=1$): $$\alpha = \frac{1}{4} \frac{i \Delta t}{ \Delta x^{2} }$$ $$\beta = -\frac{1}{4} \frac{i \Delta t}{ \Delta x^{2} }$$ The matrices are diagonally symmetric, except the corners, where the boundary conditions imposed (Dirichlet: $u(x_{min},t)=u(x_{max},t)=0$) result in elements 'outside' the matrices being known as $0$ and changing the corner elements appropriately.
