In the case of the double slit experiment, in the space between the slits and the detector, the free particle’s wave function $\psi$consists of two interfering radial waves. How is that wave function mathematically represented in that space, in a manner that satisfies the free particle Schrodinger’s equation? It must be something like $$\psi = f(r, \theta, r_a, \theta_a)e^{-i(k \lvert \vec(r)-\vec(r_a) \rvert -\omega t)} + g(r, \theta, r_b, \theta_b)e^{-i(k \lvert \vec(r)-\vec(r_b) \rvert -\omega t)} = (f(r, \theta, r_a, \theta_a)e^{-ik \lvert \vec(r)-\vec(r_a) \rvert} + g(r, \theta, r_b, \theta_b)e^{-ik \lvert \vec(r)-\vec(r_b) \rvert})e^{i \omega t}=\psi(r, \theta) e^{i \omega t} $$ where a and b are the two slits and $(r, \theta)$ are polar coordinates. Clearly, $\psi(r, \theta)$ must be an eigenfunction of the free particle Hamiltonian. But I wonder if this can be more fleshed out.
I think the key to the above is: What is the general solution to Schrodinger’s equation, for a free particle, in $(r,\theta)$ polar coordinates? I know a particular solution is the polar free-particle-Hamiltonian eigenfunction $r^{-1}e^{-i(kr-\omega t)}$, which is a radial wave centered at the origin of coordinates. That certainly is not sufficient for the double slit experiment, where the two radial waves have different centers. But we do know that when $r_a$ and $g$, above, equal $0$, then $f$, above, equals $r^{-1}$, as a clue to determining $f$. The analogous clue applies for $g$.
Best Answer
The solutions to the free Schrodinger's equation in polar coordinates are the same as the solutions in Cartesian coordinates -- arbitrary superpositions of plane waves. The radial wave centered at zero is one such superposition, which just happens have a nice (i.e. separable) formula in polar coordinates. So is a radial wave centered anywhere else -- just with a much nastier formula that I'm going to guess isn't separable.