5
$\begingroup$

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?

$\endgroup$
7
  • 1
    $\begingroup$ 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. $\endgroup$ – John Rennie Jan 28 '20 at 17:57
  • 3
    $\begingroup$ 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? $\endgroup$ – Codename 47 Jan 28 '20 at 18:00
  • 2
    $\begingroup$ 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. $\endgroup$ – John Rennie Jan 28 '20 at 18:03
  • 1
    $\begingroup$ 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? $\endgroup$ – Codename 47 Jan 28 '20 at 18:05
  • 2
    $\begingroup$ @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. $\endgroup$ – Codename 47 Sep 1 '20 at 17:30
5
+100
$\begingroup$

1D system is quite a degenerate case: it's the only case where the set of final velocities is completely defined by initial velocities. It's due to the number of degrees of freedom being equal to the number of conserved quantities: total energy and total momentum.

Let's consider a less trivial problem: 2D system of two particles. First, classical case. For simplicity we'll use the center-of-mass frame. What happens when two interacting particles come close to each other? They scatter off each other, and their velocities are generally not exchanged—they both change direction. How much—depends on impact parameter, as well as the interaction potential. If our particles have an uncertain positions or momenta (e.g. varying in different experiments), the impact parameter is uncertain, and thus the directions of outgoing particles are also uncertain.

This makes it easy to see what happens with a pair of interacting quantum particles on impact: they'll scatter around, with the only qualitative difference being the wobbles of the probability density due to self-interference of the wave.

Now what if the particles don't interact? Of course, then their wave packets (or probability bumps in classical regime) will simply pass through each other unscattered. Even if you do proper antisymmetrization of the wavefunction to yield the correct fermionic wavefunction, the resulting outgoing wave packets will still retain their overall shape (up to spreading due to dispersion).

Notice how different the outcomes are now. Scattering—even perfectly elastic—yields a cylindrical wave, while passing through retains the wave packets. So indeed, if there's a force, it's definitely not a central force.


Another, maybe even easier, way to see it is to note that the only difference between the scattering experiments with non-interacting fermions vs. bosons is the initial condition: the wavefunction that you use as the initial condition for the Schrödinger's equation is either antisymmetric in particle exchange, or it is symmetric. In principle, nothing prevents you from preparing such a state for two distinct non-interacting particles. You will get exactly the same result: the antisymmetric initial state will result in zero probability density of the particles occupying the same point, while the symmetric initial state will give you nonzero value for this. Does this mean that there's some force keeping the antisymmetrized particles away from each other?

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.