Question: I know that when AC is passed through the primary coil of a step-up transformer a higher emf is induced in the secondary coil (with more turns) of the transformer. Since energy is conserved, and P = VI, this would mean a drop in current. But I'm not able to visualize what is really happening to the electrons in the secondary coil. Does a higher voltage in the secondary coil mean that all electrons in the secondary coil are taken to a higher potential? And why is the current (flow of electrons) low when that happens?

What is really happening to the charges in the secondary coil?

Answer: 
I'll assume your wavefunction, say $\Psi$, is a function of the positions $\mathbf{x},\,\mathbf{y}$ of two particles. Then $|\Psi|^2$ is the PDF of their positions' joint position. So if $S\subseteq(\Bbb R^3)^2$,$$P((\mathbf{x},\,\mathbf{y})\in S)=\int\_S|\Psi(\mathbf{x},\,\mathbf{y})|^2\rm{d}^3\mathbf{x}\rm{d}^3\mathbf{y}.$$
The six-dimensional space is the [configuration space](https://en.wikipedia.org/wiki/Configuration_space_(physics)) of the particles. If the particles have positions $\mathbf x$ and $\mathbf y$ in physical space, this configuration is represented by a single point $(\mathbf x , \mathbf y)$ in configuration space. The square of the amplitude of the wave function $|\Psi (\mathbf x , \mathbf y)|^2$ at a point $(\mathbf x , \mathbf y)$ in this configuration space is, roughly speaking, the probability that the particles will have positions $\mathbf x$ and $\mathbf y$ respectively in physical (more precisely, $|\Psi|^2$ is a [probability density function](https://en.wikipedia.org/wiki/Probability_density_function)).