Let's consider two particles, moving, for simplicity, in a 2D plane, represented by their wavefunctions in the form of (when being modulus squared) very very very narrow Gaussians (let's say close to the infinitely narrow ones). So, it means that the positions of the particles are quite well defined. And according to the Heisenberg's uncertainty principle, the momenta of two particles are not well defined. The last means that wavefunctions being represented in momentum space have very very very large width of the corresponding Gaussians. And in momentum space they are definitely overlapped. Two wavefunctions, being overlapped, interact with each other. So scattering event in momentum space should take place, even if in the real space they are not overlapped. If there is scattering event in the momentum space, then it means that momenta of two particles were affected, which implies in turn that something should change in the motion of the particles in the real space. Am I right? It means in turn that particles can scatter from each other even they "don't touch" each other... Where am I, if I am, wrong in my thinking flow?

  • $\begingroup$ A nearly parallel laser beam is a perfect example of both momentum and position being defined well without any violation of the uncertainty principle. You are over-thinking this. "Collisions" don't depend on these things, at all. They only depend on the density and the scattering cross sections. For visible light, for instance, the collision cross section between photons is essentially zero. There are no photon-photon interactions. Between light and a solid object the cross section is large and there will be scattering etc.. $\endgroup$
    – CuriousOne
    Commented Jul 28, 2016 at 7:52
  • $\begingroup$ Wikipedia's explanation of QED. en.m.wikipedia.org/wiki/Quantum_electrodynamics. Apologies if you have already read it, but it's a good intuitive summary, imo. $\endgroup$
    – user108787
    Commented Jul 28, 2016 at 9:08
  • $\begingroup$ OK, John Forkosh, I can rephrase my question by adding the measurement procedure in between lines, but it doesn't change the point of the question. $\endgroup$ Commented Jul 28, 2016 at 11:03

2 Answers 2


Overlapping of wavefunctions and scattering are two different aspects of quantum mechanical solutions to elementary particle set ups.

As stressed in the comments also, photons are a good example: photons do not scatter off each other for visible wavelengths , the two photon crossection below gamma ray energies is essentially zero. Nevertheless, two laser beams will show interference effects. What is happening at the photon level is described in this answer .

Two wavefunctions, being overlapped, interact with each other.

The wavefunctions overlap, so their summed complex conjugate squared carries information about the presence of each in the probability distribution which gives an interference pattern. There is no interaction, just overlap. Interference can exist without interaction in quantum mechanics.

Scattering presupposes the existence of interactions, a potential and wavefunctions which are the solutions for the problem, and that a scattering cross section can be calculated.


This seems to be a basic mixup; the argument would apply just as well to any two spin up particles, which overlap in 'spin space', and any two protons in the universe, which overlap in 'isospin space'.

Locality guarantees that only overlaps in position space directly lead to interactions. That is, the interaction term for a generic QFT in position space will look like $$H_{\text{int}} \sim \int d\mathbf{x} \, \phi(\mathbf{x}) \psi(\mathbf{x}) \ldots.$$ This does not mean interactions are local in momentum space. Upon taking a Fourier transform, we would instead have $$H_{\text{int}} \sim \int d\mathbf{p}_1 d\mathbf{p}_2 \ldots \, \delta(\mathbf{p}_1 + \mathbf{p}_2 + \ldots) \, \tilde{\phi}(\mathbf{p}_1) \tilde{\psi}(\mathbf{p}_2) \ldots$$ which is nonlocal.

You might say that two fermions can't have the exact same momentum by the Pauli exclusion principle, but that doesn't contradict locality: particles with definite momentum are completely delocalized, so they do overlap in position space.


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