The time-dependent Schrodinger Equation in an infinite square well of length $L$ is given by
$$ i \hbar \frac{\partial}{\partial t} \psi(x,t) = -\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2}\psi(x,t) + V(x) \psi(x,t), $$
where we have the boundary condition $\psi(0, t) = \psi(L,t)=0$.
Multiplying by $mL^2$ gives
$$ i mL^2 \hbar \frac{\partial}{\partial t} \psi = -\frac{1}{2}\hbar^2 L^2\frac{\partial^2}{\partial x^2}\psi + mL^2V(x) \psi. $$
Now to simplify, use natural units $\hbar=1$ so that the unit Joules is equivalent to 1/seconds (from $E=\hbar f$) and make the transformations:
$$ \begin{align*} x'&=\frac{x}{L}\\ t' &= \frac{t}{mL^2}\\ V' &= mL^2V \end{align*} $$
It is trivial to show that $x’$, $t’$ and $V’$ are dimensionless quantities. The Schrodinger Equation under this transformation becomes
$$ i \frac{\partial}{\partial t'} \psi = -\frac{1}{2}\frac{\partial^2}{\partial x'^2}\psi + V'(x) \psi $$
We can solve this equation for the primed variables and then use the transformation rules to convert them back to the original variables. We will numerically solve this equation using two methods.
Choose a discrete grid with spacing $\Delta x$ and $\Delta t$ and let $\psi_j^{m} = \psi(j\Delta x, m\Delta t)$. The Schrodinger Equation can be represented in the discrete form:
$$ i\frac{\psi_{j}^{m+1} - \psi_{j}^{m}}{\Delta t} = - \frac{1}{2}\frac{\psi_{j+1}^{m} -2\psi_j^m + \psi_{j-1}^{m}}{\Delta x^2} +V(x)\psi_j^m $$
and rearranging,
$$ \psi_{j}^{m+1} = \psi_j^m + \frac{i}{2}\frac{\Delta t}{\Delta x^2}\left(\psi_{j+1}^{m} -2\psi_j^m + \psi_{j-1}^{m}\right) - i \Delta tV(x)\psi_j^m $$
This method involves solving the time-independent Schrodinger’s Equation
$$ -\frac{1}{2}\frac{\partial^2}{\partial x^2}\psi + V(x) \psi = E\psi $$