Calc Commutator in Infinitely Potential Well: Is it Possible?

[vanhees71]

My reasoning was based on the premiss that the particle IS a FREE particle.



This linear combination DOES NOT represent a FREE PARTICLE:
1)free particles do not just flip the sign of their momenta.
2) it can be made proportional to $\sin ( p x )$ which is not a free particle wavefunction. So to restrict the solutions of $P^{ 2 } \psi = p^{ 2 } \psi$ to a free particle, you should set $b = 0$ or $a = 0$.

The main issue regarding the particle in the box is the following: Inside the box, the particle is never a free particle.

I do not understand what you mean. Also for a free particle (a special case for a particle in a finite potential, namely $V=0$ by the way) the two generalized momentum eigenfunctions are $u_{p}$ and $u_{-p}$ both are eigenfunctions of $\hat{p}^2=-\partial_x^2$ and so is any linear combination. Any non-trivial linear combination is, however, obviously not a generalized eigenfunction of $\hat{p}$, which disproves your claim that any eigenfunction of $\hat{p}^2$ must be also one of $\hat{p}$. The other way is right, i.e., any eigenfunction of $\hat{p}$ is also an eigenfunction of any operator that is a function of $\hat{p}$.