hard wall system: V(x) = In the $x\le0$ region, the two $\Ppm(x)$ components are exactly the same as in the free particle case, except that the $\Psi(0)=0$ boundary condition imposes the additional constraints, s\_-(0) = s\_+(0) + (2 ) ; r\_-(0) = r\_+(0). \[hwconst\] This also results in only [*one*]{} linearly independent solution instead of two, i.e. $\Psi(x) \propto \sin(k x)$, with $k =
\sqrt{2 m E}/\hbar$. Regarding the LMs and trajectories, in the $x<0$ region, these are identical to those of Sec. \[free\], e.g. the $\Psi_+(x)$ LM trajectories move uniformly to the right, towards the hard wall at $x=0$.

It is natural to ask what happens when the $\Psi_+(x)$ LM trajectories actually reach $x=0$. There are two reasonable interpretations. The first is that the trajectories keep moving uniformly into the $x>0$ region of configuration space. This approach treats the hard wall system as if it were the free particle system, but with the $x>0$ region effectively ignored.[@poirier00qcI] This underscores the fact that unlike $\Psi(x)$ itself, the individual $\Ppm(x)$ components [*per se*]{} are unconstrained at the origin—though the constraint implies a unique correspondence between the two. This interpretation also makes it clear that for the hard wall system, the paper 