Consider the quantum double slit experiment, where a plane wave, with frequency $\omega$, travels in a direction orthogonal to the slit screen and is incident on that screen. Then, from slit 1, a radial wave emerges of the form $r^{-1}_1 e^{-i(kr_1- \omega t)}$, where $r_1$ is measured from slit 1, and, from slit 2, a radial wave emerges of the form $r^{-1}_2 e^{-i(kr_2-\omega t)}$, where $r_2$ is measured from the slit 2. Note, $r^{-1}_1 e^{-i(kr_1-\omega t)}$ satisfies Schrodinger’s equation when expressed in polar coordinates $(r_1, \theta_1)$ and is an eigenfunction of the Hamiltonian of that equation. Similarly, $r^{-1}_2 e^{-i(kr_2-\omega t)}$ satisfies Schrodinger’s equation when expressed in polar coordinates $(r_2, \theta_2)$ and is an eigenfunction of the Hamiltonian of that equation.
The total wave function is the sum of the two waves: $(r^{-1}_1 e^{-ikr_1} + r^{-1}_2 e^{-ikr_2})e^{i \omega t}$. This produces the famous interference pattern at the detector.
If both radial waves are represented in terms of the same polar coordinate system $(r, \theta)$, instead of in terms of $(r_1, \theta_1)$ and $(r_2, \theta_2)$, as above, then will the total wave function satisfy Schrodinger’s equation, in the $(r,\theta)$ coordinate system? Clearly, we need to have the total wave function be an eigenfunction of the Hamiltonian in that equation.
So I am doubting that the sum of the above two radial waves is a solution to Schrodinger’s equation. However, it must be, since we know that the experiment involves interference between the two radial waves of the wave function. So, how do we show that the sum of the above two radial waves is a solution to Schrodinger's equation?
Perhaps, to gain some insight, another way to think about this is consider two radial waves that have the same $\omega$, $k$, phase, and center point. Assume both are represented in the same polar coordinate system $(r, \theta)$ where the origin of coordinates is at the waves’ center point. We then move each wave so that they have different center points, such that neither is at the origin of coordinates. However, I do not think that the cartesian translation operators, which one might use to do this, when represented in $(r, \theta)$ polar coordinates, commutes with the Hamiltonian in $(r, \theta)$ polar form.
