quantum mechanics - Relation between Wave equation of light and photon wave function?

Suppose in our double slit experimental setup with the usual notations $d,D$, we have a beam of light of known frequency $(\nu)$ and wavelength $(\lambda)$ - so we can describe it as

$$ξ_0 = A\sin(kx-\omega t). \tag{1}$$ It passes through the two holes and moves ahead doing the usual interference stuff, so the final form of the wave will be $$ξ = ξ_1 + ξ_2 = 2A\cos(u/2)\sin(kx-\omega t+0.5*u) \tag{2}$$ where $u$ is the phase difference.

We can convert phase difference $u$ to path difference $q$. Now we choose the point of interest on the screen $(s)$ ,(which depends on path difference q and hence phase difference u). The amplitude at $s$ will be

$$ξ = 2A\cos(as)\sin(kx-wt+as), \tag{3}$$ where $a$ is constant.

Now this amplitude is a set of waves which interfere with different phases, and is function of the variables $s,x,t$. Since I placed the screen at some fix distance $x=D$ from the wall with slits, $ξ$ reduces to a function of two variables $s,t$. Rewriting

$$ξ_D=2A\cos(as)\sin(as -wt +kD), \tag{4}$$ this is also a wave description (but with different meaning).

The screen is along our $x$-axis (or to be precise $s$-axis). The intensity obtained on the screen is proportional to to absolute square of the wave amplitude written above, which in turn depends on $s$ (and t as well).

But the intensity is also proportional to number of photons. So we postulate that the probability that a photon hit a certain $s$ is proportional to the

$$\text {intensity} = |\text {amplitude}|^2. \tag{5}$$

Now, the function $ξ$ I have written above is the wave function ($\psi$) from the quantum mechanics with $s$ acting as $x$ (in $\psi$)? If not, then what is the relation between them? (I will have some additional things to ask depending upon your response.) Thank you!