# Probability density in double slit experiment

I don't understand why the probability density in the double slit experiment in the case of both slits opened, has a minimum corresponding to the maximum of intensity. Shouldn't $$P_{12}$$ have the same trend as intensity?

Intensity patterns in the double-slit experiment (a)Photon intensity $$I_1$$ on the screen with slit-1 only opened (b) Photon intensity I_2 with slit2 only opened (c) Interference pattern when both slits were opened d) probability distribution $$P_{12}$$ of photons when both the slits were opened and when the film was replaced by an array of photon detectors.


The manual says Probability distribution P_12 of photons when both the slits were opened and when the film was replaced by an array of photon detectors

• How do you define "intensity"? How is it different from the probability density? (I would say that they are the same, expect that one term is used for electrons and the other for electromagnetic waves.) – Roger Vadim Sep 30 '20 at 8:29
• In the case of em waves is defined like this $I = |\mathbf{E}(\mathbf{x},t|^2 = I_1 + I_2 + 2Re(\mathbf{E_1}^{*} \cdot \mathbf{E_2})$ – Giuliano Malatesta Sep 30 '20 at 8:32
• Precisely. And in the case of electrons we have probability density: $\rho(\mathbf{x},t) = |\psi(\mathbf{x},t)|^2 = |\psi_1|^2 + |\psi_2|^2 + 2\Re(\psi_1^*\psi_2)$ – Roger Vadim Sep 30 '20 at 8:34
• It is unclear what these curves represent. a) and b) do not show any diffraction. c) Looks asymmetric and the zero order light is quite weak. I have no idea what d) can be. Perhaps a relative 180 degrees phase was applied between the two slits. In the present form I propose to close the question. – my2cts Oct 29 '20 at 9:45

The hands on way to understand this is by assuming that a wave parting from each of the slits is a plane wave, i.e. $$\psi_1(\mathbf{x}) = e^{i\mathbf{k}_1(\mathbf{x}-\mathbf{x}_1)}, \psi_2(\mathbf{x}) = e^{i\mathbf{k}_2(\mathbf{x}-\mathbf{x}_2)}$$ Then one can look at the probability density in the plane $$x=x_0$$ and see that it will exhibit oscillatory behavior: $$|\psi_1(\mathbf{x}) + \psi_2(\mathbf{x})|^2 = 1 + 1 + 2\Re\left[e^{i\mathbf{k}_1(\mathbf{x}-\mathbf{x}_1)}e^{i\mathbf{k}_2(\mathbf{x}-\mathbf{x}_2)}\right]$$