wavefunction - Applying operators to the wave function, getting the physical units

Reading the wikipedia entry about operators, in particular the table at the end listing all operators, I have several questions regarding an $N$-particle system or statements that I wonder whether they are true or not:

a) The $\Psi$ has $3N+1$ parameters for an $N$-particle system?

b) The operators listed are all applied to the same $\Psi$ to model measurements or, put another way, there are not several $\Psi$ wave functions, different for modelling the measurment of different features?

c) The value of $\Psi$ has no physical unit.

d) Physical units come about by the operators as described in the Wikipedia table and the example given below the table. For example if $[a]$ denotes the physical unit of $a$, I get $$[\hat p] = [-i\hbar \frac{\partial}{\partial x}] = [\hbar]/[x] = kg\cdot m/s$$ as momentum.

e) Given (d), how does this take care of the different masses of different particles, i.e. if I model a system of several atoms with protons, neutrons and electrons?

f) Related to (e): which mass does the $m$ in the kinetic energy operator $\frac{\hat p\cdot \hat p}{2m}$ refer to? Would this be the total (rest) mass?

Answer

a) No, it has $N$ sets of parameters. Usually, you need two parameters per particle the position vector $\vec r_n$ and the spin projection $\sigma$. $\sigma$ is a discrete variable taking $2S + 1$ values.

b) No, $\Psi$ encodes all information about the system (just as the positions and momenta of the particles encode all information in classical point mechanics, all observables there can also be formulated as functions of the positions and momenta).

c) Yes, $\Psi$ has no unit dimension.

d) Yes.

e) Momentum is the fundamental quantity, not velocity. When you ask for the velocity you have to consider the energy, due to the relation for the group velocity $\vec v = \frac 1 \hbar \nabla_\vec k E(\vec k)$. The energy eigenstates of the free particle encode the mass as they have the form $E(\vec k) = \frac{\hbar^2\vec k^2}{2m}$.

f) It is the rest mass of the particle in question. When considering the Schrödinger equation you are working non-relativistically anyway, so the qualification rest mass is unimportant. The (non-relativistic) kinetic energy operator for an $n$-particle system is given by: $$ T = \sum_{i=1}^n \frac{p_i^2}{2m_i} $$ Where $\vec p_i$ acts only on $\vec r_i$. (But you can separate the motion of the center of gravity, then you will have an operator $\vec P$ for the total momentum, and an according energy term $\vec P^2/2M$ with the total mass $M$ and a modified term for the kinetic energy of the relative motion).