# Scintillation from wave function

Suppose we have a system with a (non-relativistic) electron whose state is described by a time-dependent wave function $\psi(x,t)$. Then I think it's correct to say that if we introduce a phosphor screen into the system, the expected brightness of the scintillation we expect to see is proportional to $|\psi(x,t)|^2$.

What if we have two electrons whose joint wave function is $\psi(x_0,x_1,t)=-\psi(x_1,x_0,t)$. What brightness of scintillation do we expect to see?

The wavefunction $\psi(x_0, x_1, t)$ gives the probability of finding one electron between $x_0$ and $x_0+dx_0$ and the other between $x_1+dx_1$: $$P(x_0, x_0+dx_0; x_1, x_1+dx_1; t) = \vert\psi(x_0, x_1, t)\vert^2dx_0dx_1$$
We expect the brightness of the scintillation at $y$ to be given by the probability (density) that either electron is found at $y$: $$p(\text{either at }y) = \sum_{x_1}P(y, y + dy; x_1, x_1+dx_1; t) + \sum_{x_0}P(x_0, x_0 + dx_0; y, y+dy; t)$$ (since $P(\text{both at } y) \sim P(y, ..., y, ..., t) = 0$ anyway). This can be written as follows $$p(\text{either at }y) = \int\left(\vert\psi(y, x, t)\vert^2 + \vert\psi(x, y, t)\vert^2\right) dx$$
Therefore, since $\psi(x_0, x_1, t) = -\psi(x_1, x_0, t)$, the intensity of scintillation will be proportional to $$p(\text{either at } y) = 2\int\vert\psi(y, x, t)\vert^2dx$$ The $2$ is indicative of the fact that the total scintillation intensity (i.e. the intensity accumulated for a large number of trials) $\int p(\text{either at } y)dy$, will be twice the total intensity one would expect from a single electron (as in both cases, the wavefunction is normalized to one).