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Photon helicity: Wigner's unitary rep. of Poincare group and gauge symmetry

  1. Oct 15, 2012 #1

    tom.stoer

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    1) Since Wigner it is well known that for massless particles of spin s the physical states are labelled by helicity h = ±s; other states are absent. So e.g. for photons the physical states are labelled by |kμ, h> with kμkμ = 0 and h = ±1 and we have two d.o.f.

    2) For gauge theories with massless gauge bosons like QED and QCD it is well known that the 4-vector Aμ carries two unphysical d.o.f. which can be eliminated by gauge fixing (a la Dirac, Gupta-Bleuler, BRST, ...). An obvious way to see this is to
    i) use the temporal gauge A° = 0 to eliminate one unphysical d.o.f. A° (∏° = 0 b/c there's no ∂°A° in the Lagrangian ~ F²)
    ii) keep the corresponding Euler-Lagrange equation (Gauss law G) as constraint to define the physical Hilbert space as its kernel G|phys> = 0 which fixes the residual gauge symmetry of time-indep. gauge transformations ∂°θ = 0
    b/c we have 4 components in Aμ and 2 gauge fixing conditions A° = 0 and G ~ 0 we arrive at 4-2 = 2 d.o.f.

    The method 2) gives us exactly the two helicity states described in 1) But 1) is using Poincare invariance whereas 2) is using gauge invariance w/o ever looking at Poincare invariance. So it seems that it's sheer coincidence that 1) and 2) arrive at the same results.

    Where's the relation???
     
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  3. Oct 15, 2012 #2

    DrDu

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    Hm, you are looking at the gauge transformation properties of a 4-vector for a massless particle.
    "4-vector" and "massless" nicely specify the representation of the Poincare group. So I don't see that 2) is never looking at Poincare invariance.
     
  4. Oct 15, 2012 #3

    DrDu

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    On the other hand, there are Weyl fermions which are also massless two component helicity eigenstates but transform differently under gauge transformations.
     
  5. Oct 15, 2012 #4

    tom.stoer

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    The simple question is: why does gauge symmetry reduce exactly from 4 to 2 d.o.f. as required by the Poincare representation. Doing the math there is no obvious relation (of couzrse everything is Poincare invaraint, but Poincare invariance does not know anything about gauge symmetries)
     
  6. Oct 15, 2012 #5

    George Jones

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    Isn't there a relationship between 1) and 2)? See VI of Weinberg's quantum field theory text, and also papers by Kim.

    Caveat: I wrote the above without giving any real thought to the matter.
     
  7. Oct 15, 2012 #6

    tom.stoer

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    I'll have a look at Weinberg's book
     
  8. Oct 15, 2012 #7
    Tom, I'm not completely sure what you mean by "Poincare invariance does not know anything about gauge symmetries", can you elaborate?
     
  9. Oct 15, 2012 #8

    tom.stoer

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    I mean that when you go through the math of 1) and 2) there is absolutely now relation between both approaches; however they both arrive at the same result.
     
  10. Oct 15, 2012 #9
    The Poincare group is the group of isometries of Minkowski space, the 4-vector Aμ of QED is brought from the Minkowskian formulation of the Maxwell equations and these equations have that gauge symmetry also in the QM context (since we are in QFT), so I guess this is how the gauge symmetry knows about the Poincare invariance (rather than the other way around), does this make any sense?
     
  11. Oct 15, 2012 #10

    tom.stoer

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    Yes, of course this guess makes sense. I think there is some deep connection, but I can't see it. That's why I am asking.

    btw.: the same applies to linearized gravity as well; there are two graviton helicity states; and there is a gauge symmetry which reduces 10 components of the metric to two d.o.f.
     
  12. Oct 15, 2012 #11
    I also think the connection between gauge symmetries and the Poincare group is worth invstigating.
    Linearized gravity has as background Minkowski spacetime so it also seems the gauge symmetries d.o.f. reduction is related to that fact.
     
  13. Oct 15, 2012 #12

    tom.stoer

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    No, it isn't. The same reduction i.e. the same number of d.o.f. holds for arbitrary curved spacetimes in GR
     
  14. Oct 15, 2012 #13

    strangerep

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    The Wigner method for massless particles requires an additional input to be put in by hand: that there are no particle types in existence with (so-called) "continuous spin". (Personally, I find that term a bit misleading, but it's in wide usage even though Weinberg doesn't seem to use that phrase.)

    In method 2, you presumably have a Lagrangian that respects Poincare invariance, and then you find unwanted degrees of freedom which must be handled/banished somehow, e.g., by gauge-fixing or a constraint approach.

    So both methods can be thought of as "Poincare + extra arbitrary input".

    IMHO, it is both puzzling and intriguing that the Poincare group does not give exactly the right set of answers for elementary particle classification without some extra empirical input.
     
  15. Oct 16, 2012 #14

    DrDu

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    I don't think that gauge invariance is necessary in that context. If you take a massive A_mu and let m tend to 0, the longitudinal and time-like photons also decouple from the transversal ones. That's how e.g. Zee calculates the photon propagator.
     
  16. Oct 16, 2012 #15
    You are right, quite disturbing, isn't it?
    Look at strangerep answer about the extra arbitrary info for both approaches, I think that is the key to the connection.
     
  17. Oct 16, 2012 #16

    tom.stoer

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    This approach fails in non-abelian gauge theories.
     
  18. Oct 16, 2012 #17
    Right, it is the non-abelian case that needs by-hand additions, not justified by the Poincare group. After all, the Poincare translations are abelian so why should it inform non-abelian gauge symmetries?
     
  19. Oct 16, 2012 #18

    tom.stoer

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    What do you mean by "by-hand additions"?

    The Poincare group always commutes with local gauge symmetries, even for the non-abelian case; they have nothing to do with each other. And it's not only about translations but about the full non-abelian Poincare group
     
  20. Oct 16, 2012 #19
    I guess the same thing as strangerep in #13.
    How does that contradict the fact that the translation subgroup is abelian?
    I must have misunderstood you, so what "fails in non-abelian gauge theories" in your opinion then?
     
  21. Oct 16, 2012 #20

    DrDu

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    How?
     
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