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
Best Answer
It turns out my intuition is correct because of the orthogonalality of the wave function and the way the kronecker delta “kills” the terms in a sum. As long as you can solve the time independent wave function analytically, you can use the sums and coefficients to get the time evolution part of it. I used a sum up to 5 for each in my simulation (25 coefficients and iterations) and it seems to work just fine