So I managed to visualise the 1d schrodinger equation using the following algorithm: First solving the time independent schrodinger equation (1d) for the particle in a box potential,
$$\psi_n(x) = \sqrt{\frac{2}{L}}\sin\left(\frac{n\pi x}{L}\right)$$
where $L$ is the size of the box. Then, using the following infinite sum and the time evolution factor to get the time-dependent wave function,
$$\Psi(x,t)=\displaystyle\sum_{n=1}^\infty c_n\psi_n(x)e^{\frac{-in^2\pi^2\hbar t}{2mL^2}}$$
where
$$c_n=\displaystyle\int_0^L\psi_n(x)^*\Psi(x,0)dx$$ and $\Psi(x,0)$ could be the wave packet.
This is all fairly basic stuff but I want to try to extend the concepts into the 2d world. My attempt is as follows for the particle in a box using this as a guide.
$$\psi_{n_xn_y}(x,y)=\frac{2}{L}\sin\left(\frac{n_x\pi x}{L}\right)\sin\left(\frac{n_y\pi y}{L}\right)$$ From here I thought it would make sense to have a double sum for both $n_x$ and $n_y$ with something like $$\Psi(x,y,t) = \displaystyle\sum_{n_x=1}^\infty\displaystyle\sum_{n_y=1}^\infty c_{n_xn_y}\psi_{n_xn_y}(x,y)\exp\bigg(\frac{-i\hbar^2\pi^2}{2m}\bigg(\frac{n_x^2+n_y^2}{L^2}\bigg)\bigg)$$ I thought a double integral would also make sense for the coefficients $$c_{n_xn_y} = \displaystyle\int_0^L\int_0^L\psi_{n_xn_y}(x,y)^*\Psi(x,y,0)dxdy$$ and have the wave packet equal to $$\exp(\frac{-(x-c)^2}{2\sigma^2})\exp(\frac{-(y-c)^2}{2\sigma^2})$$ I'm still fairly new to QM and I'm not sure if I'm going about this correctly. Can someone tell me if I did something wrong or point me in the right direction? I have found almost no articles on time evolving the 2d equation which is why I'm asking here
