Solving time dependent schrodinger equation using RK4

Question

Assuming $\hbar = 1$, $m = 1$ and $V$ only depends on $x$, the Schrodinger equations becomes: \begin{equation} \frac{\partial \Psi}{\partial t}(x,t) = \frac{i} {2}(\nabla^2 + V(x))\Psi(x,t) \end{equation} Given $x_0$, define $f(t) = \Psi(x_0,t)$: \begin{equation} \frac{\partial f}{\partial t}(t) = F(t,f(t)) \end{equation} using aproximation for $\nabla^2$: \begin{equation} F(t,f)=\frac{i}{2} \left( \frac{\Psi(x_0+h_x,t)-2\Psi(x_0,t)+\Psi(x_0-h_x,t)}{h_x^2}- V(x_0)f \right) \end{equation} I want to find $f_n \approx \psi(x_0,t_0+nh_t) = \psi(x_0,t_n)$, given $f_0 = \psi(x_0,t_0)$, using $RK4$ define: \begin{align*} k_1 &= F(t_n,f_n) \\ k_2 &= F(t_n+\frac{h_t}{2},f_n+\frac{h_t}{2}k_1) \\ k_3 &= F(t_n+\frac{h_t}{2},f_n+\frac{h_t}{2}k_2) \\ k_4 &= F(t_n+h_t,f_n + h_tk_3) \end{align*} then: \begin{equation} f_{n+1} = f_n + \frac{h_t}{6}(k_1 + 2k_2 + 2k_3 + k_4) \end{equation} The problem is, to find let's say $k_4$ y need to know $\Psi(x_0,t_n+h_t) = f_{n+1}$, how can I apply $RK4$?

Answer 1

You are mixing up notations. The RK4 method is explicit. You use $f_n+h_tk_3$ as approximation of the array $Ψ(x_j,t_n+h_t)$, $j=0,...,M$, where $x_j=x_0+jh_x$.

Also, your use of $x_0$ is obfuscating the complexity of the problem. You compute the approximations of $Ψ(x_j,t_n+h_t)$ where $x_j=x_0+jh_x$ all simultaneously, there is no way around this (esp. if $V$ is not constant, as then even the Fourier coefficients are coupled). Thus $$ F_j(t_n,f_n)=\frac{i}{2} \left( \frac{f_{n,j+1}-2f_{n,j}+f_{n,j-1}}{h_x^2}- V(x_j)f_j \right). $$ You might have to replace $V(x_j)f_j$ with some local average to fit the approximation model, for instance when starting from finite elements or some other Galerkin scheme.