Young's double slit Am I right to think the (general) probability distribution of photon in a double slit experiment at the screen has the form $|\psi|^2 = c e^{\alpha x^2}\cos^2(\beta x)$? (Due to the superposition of the 2 Gaussian shaped wavefunctions of the independent slits)  Thank you.
 A: In one word, yes.
Now here are some tips for the proof. The exact proof depends directly on the given exercise.


*

*Suppose two cohered rays $R_1$ and $R_2$, which were emitted from the same source S. After traveling through two splits, these rays interfere together in $M(X,Y,Z)^T$.

*Consider the screen on which the rays interferes is far enough that the spherical waveform due to the diffraction of the rays on the edges of the splits (which produce the interference) may be considered as their tangent plane.

*We know that the intensity at any given interference point of two waves of the same amplitude is the expressed as follows, where $\Delta\Phi$ is the travel difference between both rays: $$I(M)=2I_0(1+\cos{\Delta\phi})=4 I_0 \cos^2{\frac{\Delta\Phi}2}$$

*Keep in mind that in order to create the two coherent rays which ought to interfere, $R_1$ and $R_2$, one of the rays goes through a mirror. This creates a travel difference between both rays: $\Delta\phi=\pi$

*By calculating the length difference $S_2M - S_1M$. Supposing that $W_y$ is the width of the Young splits, we then realize that $$S_1M \simeq S_2M \gg W_y$$

*We can now use Taylor's expansion at the first order for the sum of squares. This yields to the exact travel distance, from which we can determine the phase difference between both rays. 

*We now sum all the identical contributions of all the points of interests, and determine the photon intensity at an interference point, such as M: $$I(M)=4I_0\cos^2(\frac{\pi{ax}}{z\lambda})$$
I hope this helps!
