# Does the (non-)collision of two fermion wave packets demonstrate that there is no exchange "force"?

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?

• The Pauli force is not a fundamental force. See for example my answer to Is the electromagnetic force responsible for contact forces?. If in your experiment you took non-interacting fermions, i.e. possessing no charge of any kind, then they would not scatter off each other. They would simply pass straight though each other. Jan 28, 2020 at 17:57
• How exactly would you distinguish the case of two identical, non-interacting fermions passing through each other, and the case of the same two particles colliding elastically? Jan 28, 2020 at 18:00
• Ah I see what you mean. You're suggesting that since the particles are identical a collision and recoil is physically the same as no collision (in 1D). Or at least the end result is the same. Jan 28, 2020 at 18:03
• Exactly. What is unclear to me is how they so confidently declare that there is no collision happening, when the case of collision and the case of no collision are indistinguishable. In my understanding not only the end result, but the entire physical process would be the same. Do you have a recommendation on how to make the question clearer? Jan 28, 2020 at 18:05
• @ChiralAnomaly In principle such a theory could be correct, but would have to be indistinguishable from the clearly simpler case where the particles do not exchange places, and thus there would be no reason to consider it. My point is exactly that in the referenced paper, it is claimed with confidence that there is an unambiguous interpretation of the situation. At best, I would say that this is a physically ambiguous situation with no clear simplest interpretation. For example, in 1D quantum liquid physics, non-interacting fermions are modelled as unable to pass each other. Sep 1, 2020 at 17:30