@SillyGoose definitely

@qwerty the "gets brighter" has nothing quantum to it. He is absolutely correct. There is nothing weird about it from a classical waves perspective. All the weirdness comes from when you realise that you must have an explanation for how photons go through them one by one.

@Feynmate miao miao rarely deals merely with memes...

Is there a reason why sometimes in order to solve the K.G equation we consider $\Phi(x^\mu)=e^{-ix_\mup^\mu}$ and sometimes a fourier transform of it $\phi(x)=\int \frac{1}{(2\pi)^3}e^{i\vec p \vec x}\tilde{\Phi(t,\vec p)}d^3p$ ?

@User198 You are so close that Tobias did not even want to stress the small difference. You have a weighted sum with classical probabilities, so the equations must be $$\begin{align}\tag1\rho(t)&=\sum_jp_j\left|\psi_j(t)\right>\!\left<\psi_j(t)\right|\\\tag2&=\sum_jp_jU(t-t_0)\left|\psi_j(t_0)\right>\!\left<\psi_j(t_0)\right|U^\dagger(t,t_0)\\\tag3&=U(t,t_0)\left(\sum_jp_j\left|\psi_j(t_0)\right>\!\left<\psi_j(t_0)\right|\right)U^\dagger(t-t_0)\end {align}$$

$$\tag4\therefore\qquad\rho(t)=U(t,t_0)\rho(t_0)U^\dagger(t,t_0)$$ as wanted

Hence, if you postulated quantum dynamics for wavefunctions, then you will automatically obtain the correct dynamics for density operators by this derivation.

another misleading statement: "the partition function for a system of non-interacting particles factors"

someone needs to take you to fock space...