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B Collision of identical fermions

  1. Oct 5, 2016 #1
    What happens if I shoot a fermion at another identical fermion at rest? For example, do the fermions stick together, or do they bounce?

    Let's ignore gravity, electro-magnetism, weak force and strong force.

    Edit: We are considering the Pauli exclusion principle.
     
    Last edited: Oct 5, 2016
  2. jcsd
  3. Oct 5, 2016 #2
    If you ignore ALL the forces? I would hazard a guess that they will totally ignore each other!
     
  4. Oct 5, 2016 #3

    DrChinese

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    Normally they scatter if of the same charge or have internal structure.

    But if you ignore EM and other forces, that's not a factor and they would act as point particles. I don't believe you could get them to interact. What is your question trying to elicit?
     
  5. Oct 5, 2016 #4
    Pauli exclusion principle prevents the particles getting too close, right?
     
  6. Oct 5, 2016 #5

    DrClaude

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    It depends on their spin state.
     
  7. Oct 5, 2016 #6
    Does it matter if they are headed in opposite directions?
     
  8. Oct 5, 2016 #7
    Let's say the spin states are identical.
     
  9. Oct 5, 2016 #8
    If it mattered, wouldn't it collapse the neutron stars? Or are the neutrons in a neutron star completely still?
     
  10. Oct 5, 2016 #9
    Good point. I would suggest their direction of motion is underdetermined though. Unlike a beam.
     
  11. Oct 5, 2016 #10

    DrChinese

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    If they are electrons, that's not an issue unless they are bound (to the same nucleus). Right?

    At any rate, what is the specific thing you are looking to understand? The question is so hypothetical as to have no meaningful answer.
     
  12. Oct 5, 2016 #11
    They scatter just like any other particles but with an angular distribution determined by even L+S (composite orbital and spin states) in their CM frame.
     
  13. Oct 5, 2016 #12

    Vanadium 50

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    No. It prevents them from being in the same state. Two fermions with vastly different momenta are obviously not in the same state.
     
  14. Oct 6, 2016 #13
    I see.

    But let's say two identical fermions are moving in a line formation, they have the same momentum, then the first one collides on a wall, now as the momentums are different there is no Pauli exclusion effect anymore.

    It is said that Pauli exclusion effect causes two identical fermions to behave like a rigid object, but the identical fermions did not behave like rigid object when colliding on a wall.

    So I don't quite understand what this Pauli exclusion thing does.
     
  15. Oct 6, 2016 #14

    DrClaude

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    If two spin-1/2 particles have the same spin, or more precisely are part of a spin triplet, then the proper spatial wave function for the two particles is anti-symmetry with respect to the exchange of the two particles, which means that indeed the probability of finding the two fermions at the same place is 0. But in the absence of any interaction potential between the two particles, they will not scatter off each other.

    If there is an interaction, for instance if they have a charge, then the Pauli principle can come into play in the scattering amplitude of the particles at different angles. It can be found in the Feynman lectures: http://www.feynmanlectures.caltech.edu/III_04.html and http://www.feynmanlectures.caltech.edu/III_03.html#Ch3-S4
     
  16. Oct 6, 2016 #15

    Vanadium 50

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    • If there are no interactions, why did one fermion bounce off the wall? You need a consistent model to get sensible answers.
    • If you have two identical fermions with the same position, momentum, etc. as they approach a wall, you have set up an impossible situation. You won't get sensible answers from an impossible scenario.
    • "It is said that Pauli exclusion effect causes two identical fermions to behave like a rigid object" By whom? You've been around PF long enough to know that "I heard a guy say once" is an unacceptable reference. There's no way to tell if you misunderstood or if he said something wrong.
    • "So I don't quite understand what this Pauli exclusion thing does." No you don't. If you want to, you need to learn quantum mechanics. Asking what happens in impossible scenarios won't improve your understanding.
     
  17. Oct 6, 2016 #16

    A. Neumaier

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    In this case there is no interaction and the moving particle simply passes through the other. Note that the two particles are in different states since thy have different momentum. Hence the Pauli exclusion principle excludes nothing. If they are moving in a line formation, they differ in state, because the position is different. In a neutron star they are interacting, hence your assumptions don't hold.
     
  18. Oct 6, 2016 #17
    The Pauli Exclusion Principle is not always the best way to approach this problem. A better (and more general) way is to look at the symmetry properties of the state vector (or wavefunction) under permutation.
     
  19. Oct 6, 2016 #18
  20. Oct 6, 2016 #19
    So there must be enough of uncertainty of position and enough of uncertainty of momentum. Then some kind of destructive and constructive interference can occur, which affects the chances of finding the fermions at some positions. Right?

    Is there perhaps also such interference that affects the changes of finding the fermions with some momentums?
     
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