I imagine the following thought experiment: Let us prepare two identical neutral fermions with identical spin and antisymmetric spatial wave function in one spatial dimension. Let the initial state be two Gaussian wave packets which start at some large distance and have identical speed, but opposite direction. Since the particles are neutral, the wave packages should move through each other undisturbed. But at the time, where the particles are at the same place, the antisymmetrized wave function adds up to zero, so you cannot normalize the state. The exchange interaction should be zero because the particles are not charged. For the same reason there should be no magnetic moment. Where is the error in the argument?

  • $\begingroup$ Ok, the answer probably has to do with the fact, that the wave packets move in opposite directions. Therefore the phases will be different and the wave packets will not add up to zero. $\endgroup$
    – Mikael
    Commented Oct 21, 2023 at 9:57
  • $\begingroup$ The antisymmetric psi function is for at least two particles, and thus depends on coordinates of two particles, so in 1D physical space, it depends on two variables. The proper psi function to consider from start to end lives in a plane. We can imagine at the start it can be expressed as $\psi(x_1,x_2) = \frac{1}{\sqrt{2}}(L(x_1-r_{left})R(x_2-r_{right}) - L(x_2 - r_{left})R(x_1-r_{right}))$, where $L$ is a wave packet function centered at 0. $\endgroup$ Commented Oct 21, 2023 at 23:03
  • $\begingroup$ and describing particle moving from left to right, and $R$ the same but moving from right to left. When both centers meet, e.g. at 0 ($r_{left}=r_{right} = 0$), then we get $\psi(x_1,x_2) \propto L(x_1)R(x_2)-L(x_2)R(x_1)$. $\endgroup$ Commented Oct 21, 2023 at 23:14
  • $\begingroup$ This can indeed become vanishing function in the whole plane if both $L$ and $R$ are the same function. So here is maybe one of the reasons for why we stick to theories where two fermions can't be at the same state simultaneously - it breaks the Born interpretation, we get zero instead of a normalizable state. $\endgroup$ Commented Oct 21, 2023 at 23:20
  • $\begingroup$ Related: physics.stackexchange.com/questions/31565/… $\endgroup$ Commented Oct 21, 2023 at 23:36


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