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Why should force carriers be only boson?

  1. Jul 29, 2012 #1
    I am wondering why all force carriers we know are boson.

    Is there any special reason that fermion cannot be force carrier? or All known force carriers happen to be boson?
     
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  3. Jul 29, 2012 #2

    tom.stoer

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    Looking at the formalism in quantum field theory the notion "force" is somehow meaningless. There are simply interactions of different fields. One can do perturbation theory using a Feynman diagram language and can interpret an interaction as an exchange of 'virtual particles' (I don't like this term, but for the context of this question it may be helpful). Doing it that way one observes that there are both bosons exchanging fermions and fermions exchanging bosons, so both possibilities are realized in nature.

    Now one could reformulate your questions as follows: I am wondering why all interactions which can be ascribed a kind of 'classical potential' are represented by exchange of bosons.
     
  4. Jul 31, 2012 #3
    Thank you for your reply.

    Your reply reminds me of the comment that only boson mediators could have the classical limit.

    I think this comment is somehow related to your reply. Is it right?
    Could you explain more detail why all interactions which can be ascribed a kind of 'classical potential' are represented by exchange of bosons?
     
  5. Jul 31, 2012 #4

    tom.stoer

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    good question; mathematically one reason is the structure of the differential equations (like Maxwell's equations); only the equations for massless boson fields allow for macroscopic long-range solutions which can form a "potential"
     
  6. Jul 31, 2012 #5

    arivero

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    Mathematically too, you can put the units in formulae for the fields. You will see that hbar is in the fermion field, so it disappears in the classical limit.
     
  7. Jul 31, 2012 #6

    tom.stoer

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    Very good point!
     
  8. Jul 31, 2012 #7
    Isn't the other reason why only bosonic fields lead to classical potentials the fact that only Bose-Einstein statistics allow sufficient superpositions of particles to achieve macroscopically measurable effects?
     
  9. Jul 31, 2012 #8

    tom.stoer

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    I don't think so. If you derive the Coulomb potential in QED via gauge fixing the potential is essentially a "classical potential between charges" i.e. it's not a "superposition of identical particles"
     
  10. Jul 31, 2012 #9

    fzero

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    Another important point is that we seem to be talking about gauge theories here. If you gauge a bosonic symmetry, you have to introduce a spin 1 boson that has a natural interpretation as a "force carrier." If you gauge a fermionic symmetry, you will have to introduce a fermionic field.

    There's no good reason not to interpret exchange of this fermion as a "force." In fact, in (extended) supersymmetric theories, there are a class of objects known as BPS states, for which the exchange of fermions between static states precisely cancels the gauge interaction, so that the force between two BPS states at rest vanishes.
     
  11. Jul 31, 2012 #10

    arivero

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    Let me mention than Lucretius explains, somehow empirically and just by lucky guessing, that atoms must mainly scatter when crossing, to make solid bodies, but eidola must cross without interact, carrying to our eyes the shapes of such bodies. So if you want to translate modern QFT to a classics friend, you can tell him to think of bosons as eidola and fermions as individua. (sorry the pig greek/latin).
     
    Last edited: Jul 31, 2012
  12. Aug 18, 2012 #11
    Hi.

    Let us suppose a test particle is in initial state [itex]|i\rangle[/itex] with spin projection eigenvalue [itex]S_i[/itex]. Let it emit a mediator of spin eigenvalue [itex]S_M[/itex]. Let test particle's final state be with spin eigenvalue [itex]S_f[/itex].

    Spins are conserved and therefore one has

    [itex]S_i=S_f+S_M[/itex]

    All spin eigenvalues are unique up to a signum. Vectors can only have spin eigenvalues [itex]\pm 1[/itex], spinors [itex]\pm 1/2[/itex] and scalars [itex]\pm 0[/itex]. So one can write

    [itex]S_f=\pm S_i[/itex]

    Therefore

    [itex]S_M=S_i (1 \mp 1) \in \{0,2S_i\}[/itex]

    Hence:
    a) If test particle is scalar with spin eigenvalue [itex]S_i=0[/itex], mediator is scalar.
    b) If test particle is spinor with spin eigenvalue [itex]S_i=\pm 1/2[/itex], mediator is either scalar or vector.
    c) If test particle is vector with spin eigenvalue [itex]S_i=\pm 1[/itex], mediator is scalar.

    From a), b) and c) we conclude that mediators are always bosons.
    From c) we conclude that photons cannot split nor absorb other photon.

    Cheers.
     
  13. Aug 18, 2012 #12

    tom.stoer

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    The conclusion c) is correct for photons but wrong for gluons; something is missing ...
     
    Last edited: Aug 19, 2012
  14. Aug 18, 2012 #13

    bcrowell

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    I think Kraflyn is on the right track, but is wrong in some details, e.g., a spin-1 particle can have z component -1, 0, or -1, not just [itex]\pm1[/itex].

    The simplest way to put this argument is that when a particle changes its spin along a given axis, it does so by an integer amount. The difference of two integers is an integer, and so is the difference of two half-integers.

    But I get confused when trying to apply this to the magnitude of the spin rather than its z-component. It seems to imply that scalar particles can't interact electromagnetically, which I guess is fine since the Higgs is neutral. But it also seems to imply that particles with spin 0 or 1/2 can't interact gravitationally, since the graviton has spin 2. Is this resolved only by higher-order Feynman diagrams, in which the low-spin particle makes virtual pairs of higher-spin particles, which can then interact gravitationally???

    Could you give an example of the former? I don't see how the spin coupling can work out at the vertices, since you can't couple an integer spin to a half-integer spin to get an integer spin.
     
    Last edited: Aug 18, 2012
  15. Aug 18, 2012 #14
    Hi.

    If test vector particle's final spin z-projection is 0 while initial state's spin being [itex]\pm1[/itex], then the z-projection conservation of spin reads [itex]S_M = S_i=\pm1[/itex]. Therefore mediator is boson again.

    Spinors cannot have spin eigenvalue 0, so this does it: mediators are all bosons. Yes, this detail was missing, thank You Bcrowell for pointing this out. I apologize for the inconvenience. This allows photon splitting theoretically, apparently.

    Photon splitting has been experimentally detected recently, though. The process is extremely rare, because it is believed to involve vacuum loops. Photon decays into electron-positron virtual pair for a brief moment. Virtual pair emits some photons. Then virtual pair re-combines back to physical photon. If vacuum interactions are taken into account, funky predictions occur. Funky measurable predictions.

    Bcrowell, could You expand Your train of thought on particles not interacting gravitationally into a bit more mathematically elaborated theory here? I'd like to see how it may work out. Or do You have it published already? I'm a bit confused with the magnitude of the spin of fermions, because one cannot measure fermion spin in any two directions simultaneously, because spin matrices do not commute. This observation does not affect the bosonic sector of theory, of course. Looking forward to read more about it, if allowed and expected to, of course.

    Cheers.
     
  16. Aug 18, 2012 #15

    bcrowell

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    It's just standard angular momentum algebra. The operator [itex]\mathbf{J}^2[/itex] has eigenvalues j(j+1), where, e.g., j=1/2 for an electron. The operators [itex]\mathbf{J}^2[/itex] and Jz commute, even though Jz doesn't commute with Jx or Jy. You can't have [itex]\mathbf{J}_1=\mathbf{J}_2+\mathbf{J}_3[/itex] unless j1, j2, and j3 satisfy the triangle inequality, just as you'd expect for ordinary vector addition.

    This issue about gravitons is really a different one than the one this thread was originally about, so I've started a new thread on this topic in BTSM: https://www.physicsforums.com/showthread.php?t=629616
     
    Last edited: Aug 18, 2012
  17. Aug 18, 2012 #16
    One can perfectly well have charged scalar particles, however. For instance in the MSSM we have charged scalar Higgses. I think there is still no problem though, since virtual photons are allowed to have unphysical polarisations, i.e. they are allowed to be in the spin-0 longitudinal polarisation state.

    I think the unphysical polarisation argument must solve the problem for the spin-2 particles also. Although in either case I am not sure what the term in the Lagrangian would actually look like. For the charged scalar case there must be a term [itex] h^+h^+A_\mu [/itex] somewhere, except that leaves an uncontracted Lorentz index floating around, so I guess there must be something to fix this up (like the gamma matrix does for fermions). Similarly the graviton case is presumably something like [itex] hhg_{\mu\nu}[/itex], and so now there are two indices to deal with somehow...
     
  18. Aug 18, 2012 #17
  19. Aug 19, 2012 #18

    tom.stoer

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    @bcrowell: regarding "bosons exchanging fermions" and your question
    In QED you have light-light scattering at one loop level where two photons exchange an electron-positron pair in a 4-fermion box. The reason why we do not call this a "force" is simply b/c it's too weak to be observed directly.
     
  20. Aug 19, 2012 #19
    Hi.

    Yes, the vacuum loop... Interesting observation.

    Hm, photon here is scattered off electron-positron pair that briefly emerged from vacuum sea. In such description, photon might be considered a mediator again. I guess so...

    Looking forward to see where this takes us. Cheers.
     
  21. Aug 19, 2012 #20

    tom.stoer

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    I wouldn't call that a vacuum loop
     
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