Double slit experiment: how to derive the figure of interference? We know that by performing the Double Slit Experiment we get an interference pattern on the detector screen. To explain this most sources talk about the state of the particle: calling $|\psi _l \rangle $ the state in which the particle crossed the left slit, and $|\psi _r\rangle$ the state in which the particle crossed the right slit, we have the following total state for the single unmeasured particle ($|\psi\rangle$):
$$|\psi \rangle=\frac{1}{\sqrt{2}}\Big(|\psi _l\rangle+|\psi _r\rangle \Big)$$
Then usually books on the topic talk about the particle's probability of landing left versus the probability of landing right, calculate those, and show that there are interference terms in the expression of the probability, job done for them usually.
However I am not satisfied: experimentally seeing a continuous interference pattern on the screen (with a certain shape!) implies that the single particle crossing the slits must have, at the time of impact, a wave function with the same shape.
My question, calling the time of impact $t_i$, is: what is the simplest (easiest to understand) way to derive the monodimensional wave function $\psi(x,t_i)$ of the single particle crossing the slits? (Better if the explanation starts from the postulates of QM)
I feel that theoretically finding the explicit form of this wave function is the only proper way of explaining the double slit experiment fully. In fact this is the only way to properly predict the shape of the figure of interference that will appear on the screen. Based on experimental results the wave function we find should of course look something like this:

 A: I think the most straightforward way to get the pattern is to first consider an isotropic point source of particles with momentum $k$. Here I consider the $d=2$ case. The wave function for the particles coming out of the source can be found by using the relation for probability current density:
$$\vec{J}=\frac{i\hbar}{2m}(\psi \nabla \bar{\psi}-\bar{\psi}\nabla \psi) \tag{1}$$
and the fact that to have particle number conservation we should have (again for $d=2$):
$$\vec{J}\propto \frac{\hat{r}}{r} \tag{2}$$
where the source is assumed to be at the origin. It is easy to check that $\psi$ has the following functional form:
$$\psi(r)=A\frac{e^{ikr}}{\sqrt{r}} \tag{3}$$
Where A depends on the intensity $A\propto \sqrt{I}$.
Now consider two sources of with amplitudes $A_1$ and $A_2$ at points $r_1$ and $r_2$ with initial phase difference $\Delta \varphi$, then the total wave function is the superposition of these two and we have:
$$\psi(\vec{r})=A_1 \frac{e^{ik|\vec{r}-\vec{r}_1|}}{\sqrt{|\vec{r}-\vec{r}_1|}}+A_2 \frac{e^{ik|\vec{r}-\vec{r}_2|+i\Delta \varphi}}{\sqrt{|\vec{r}-\vec{r}_2|}} \tag{4}$$
For the double slit experiment, we can put $A_1=A_2=A$ and $\Delta \varphi=0$. We also assume the slits to be at $(\pm a,0)$. Then, the wave function on a wall at $y=L$ is given by (working in units where $A=1$):
$$\psi(x,L)=\frac{e^{ik\sqrt{L^2+(x-a)^2}}}{(L^2+(x-a)^2)^{1/4}}+\frac{e^{ik\sqrt{L^2+(x+a)^2}}}{(L^2+(x+a)^2)^{1/4}} \tag{5}$$
If you plot $|\psi|^2$ you'll see the interference pattern with a decay.
The more complicated way of doing this is to solve the time independent Schrodinger equation between two walls in the presence of two holes, which would give qualitatively the same picture for $L\gg a$ while for smaller values of $L$ it would also take the effect of successive reflections of particles of the walls into account.
