In the paper "Quantum statistics: Is there an effective fermion repulsion or boson attraction?" by Mulling and Blaylock (2003), it is claimed that
We can demonstrate there is no real force due to Fermi/ Bose symmetries by examining a time-dependent wave packet for two noninteracting spinless fermions.
They then proceed to demonstrate that, in a 1D system of two identical non-interacting fermions, letting two Gaussian wave packets of unequal widths $\alpha$ and $\beta$ propagate towards each other leaves both wave packets completely unaffected after the collision.
In other words, at $t = 0$ the $\alpha$ packet is located at $x=-a$ with velocity $v > 0$, while the $\beta$ package is located at $x = a$ with velocity $-v$. At the time $t = 2 a/v$ the packets have switched places and are moving in each their direction with unchanged speed.
In the words of the authors:
Now the $\alpha$-packet is peaked at $a$, but still moving to the right and the $\beta$-packet is peaked at $-a$ and still moving to the left. The packets have moved through one another unimpeded because, after all, they represent free-particle wave functions. Describing this process in terms of effective forces would imply the presence of scattering and acceleration, which do not occur here, and would be highly misleading.
But since the fermions are identical, let us try to consider this as a perfectly elastic collision. In such a case the two particles would exchange momentum, corresponding to the two packages exchanging width. Is this not an equally valid description? And if so, does that not invalidate their claim that this demonstrates that there is no force, since this other description includes both scattering and acceleration?