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The Pauli-Lubanski vector for massless particles

  1. May 12, 2009 #1

    nrqed

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    Does anyone know how to prove that [itex] W^2 = 0 [/itex] for massless particles, where W is the Pauli-Lubanski vector

    [tex] W^\mu = \frac{1}{2} \epsilon^{\mu \nu \rho \sigma} M_{\rho \sigma} P_\nu [/tex]
    ??

    The books I have either give no proof, or they rely on the expression obtained in the massive case,

    [tex] W^2 = - m^2 \vec{J}^2 [/tex]

    However, the derivation of this formula relies entirely on working in the rest frame of the particle, so it makes no sense to infer anything about the massless limit from this result.

    I have found that, in general,

    [tex] W^2 = - \frac{1}{2} P^2 M_{\mu \nu} M^{\mu \nu} + M_{\mu \nu} M^{\sigma \nu} P^\mu P_{\sigma} [/tex]

    In the massless limit, the first term is zero but I can't see why the second term is also zero. I have considered working in the frame

    [tex] p_\mu = (E,0,0,E) [/tex]

    but I still don't get zero. The antisymmetry of the Lorentz generators M does not help.


    Thanks in advance
     
  2. jcsd
  3. May 12, 2009 #2

    Hans de Vries

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    There is only a trivial reason: (edit: the reason you gave I see)
    The definition of [itex]P_\mu P^\mu=m^2[/itex] as the first Casimir invariant of the
    Poincaré group in the massive case.

    The Pauli Lubanski vector transforms as a spin vector and the square
    of a spin vector is in principle never zero. In the rest frame we have
    per definition [itex]W_0=0[/itex] and [itex]W^2=W_x^2+W_y^2+W_z^2[/itex].


    Regards, Hans
     
    Last edited: May 12, 2009
  4. May 12, 2009 #3

    strangerep

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    Which books? It sounds wrong to me, for the reasons you gave, (but don't
    take that as gospel :-)

    Analysis of massless Poincare irreps is very different from the massive case,
    involving the non-compact E(2) little group instead of SO(3), among other things.

    Anyway, in this case one should probably be studying the Casimirs and irreps of
    the conformal group instead of Poincare.
     
  5. May 12, 2009 #4

    nrqed

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    Thanks Hans for the feedback.

    I am trying to work out the massless case, so I can't go to the rest frame.
    I am trying to prove that W^2 =0 in that case.
     
  6. May 12, 2009 #5
    Are you sure that [tex] W^2 = 0[/tex] in the massless case? Massless particles can have spins (actually, helicities). This seems to suggest that [tex] W^2[/tex] could be non-zero in this case.

    I know how to prove [tex] W^2= m^2 j^2[/tex] for arbitrary mass m, where [tex] \mathbf{j} [/tex] is the spin operator. However [tex] \mathbf{j} [/tex] diverges as [tex] 1/m [/tex], when mass goes to zero. So, this does not prove that [tex] W^2 = 0[/tex].
     
  7. May 12, 2009 #6

    Hans de Vries

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    Ok, but mass-less particles can have spin, photons, gluons, mass-less neutrinos,
    In general the individual chiral currents.

    [tex]j_R ~=~\psi^\dag_R\sigma^\mu\psi_R, ~~~~~~
    j_L ~=~ \psi^\dag_L\tilde{\sigma}^\mu\psi_L[/tex]

    transform also light-like but they do have spin.

    True but W^2 is reference frame independent. Otherwise, the "official" response
    would be strangerep's: look at the reduced group. (E.Wigner, page 197)

    http://www.pages.drexel.edu/~gln22/On%20Unitary%20Representations%20of%20the%20Inhomogeneous%20Lorentz%20Group.pdf [Broken]


    Regards, Hans
     
    Last edited by a moderator: May 4, 2017
  8. May 12, 2009 #7

    nrqed

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    All the references I have read say that W^2 is zero for massless particle. Do you agree?
    I am not saying that [itex] W^\mu [/itex] is zero, only that W^2 = 0. For massless particles, one then have that

    [tex] W^\mu = \lambda P^\mu [/tex]

    where [itex] \lambda [/itex] is the helicity. So I am not saying that massless particles have no spin, simply that W^2 =0.

    Once I accept that W^2, I can get all the other results, it's just that step which is unclear to me.

    I guess there are two possible answers: either there is a trick that I am missing to show that [itex]W^2=0[/itex] or [itex] W^\mu[/itex] has to be defined differently when the particles are massless.


    Regards,

    Patrick

    EDIT: Thanks for the reference by the way, it's neat to see the original paper!
     
    Last edited by a moderator: May 4, 2017
  9. May 12, 2009 #8

    strangerep

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    To classify elementary particles under the Poincare group, one must start from the Casimirs,
    i.e., P^2 and W^2, which we call mass squared and total angular momentum
    squared (or spin squared), respectively. You can't "define [itex] W^\mu[/itex] differently",
    because what matters here is the Casimir W^2. If you insist that W^2=0, all you're doing
    is restricting your considerations to spinless particles. (Could that be perhaps what those
    unnamed books of yours are doing?)
     
  10. May 12, 2009 #9

    nrqed

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    Bailin and Love in Supersymmetric Gauge Theory and String Theory say (page 16):

    For massless particles [itex] W^2 = P^2 = 0 [/itex]


    Ryder (Quantum Field Theory) says W^2 |k> = 0 .

    Baer and Tata (Weak Scale Supersymmetry), page 45, says W^2 =0


    Actually, I just found out that there is a detailed discussion in Maggiore (A Modern Introduction to Quantum Field Theory) who proves W^2 =0 in the massless case in a careful manner, so I will check it out.
     
  11. May 12, 2009 #10

    nrqed

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    No, because the conclusion is that we can write
    [tex] M^\mu = \lambda P^\mu [/tex] where lambda is the helicity. Again, I never implied that [itex] W^\mu [/itex] is zero, just that [itex] W^2 [/itex] =0 .
     
  12. May 13, 2009 #11

    Hans de Vries

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    You're right, I'm looking at Maggiore right now. So the spin is
    not zero but W^2 is.


    Regards, Hans
     
    Last edited: May 13, 2009
  13. May 13, 2009 #12

    Hans de Vries

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    OK. Trying to derive things simpler I think this makes sense:

    Associating W and P with spin and momentum we have for
    mass less particles:

    [tex]P^0=|\vec{P}|, ~~~~~W^0=|\vec{W}|[/tex]

    The general rule [itex]W_\mu P^\mu=0[/itex] now forces W and P to be either parallel
    or anti-parallel, which gives you the two helicity states.


    Regards, Hans
     
    Last edited: May 13, 2009
  14. May 13, 2009 #13

    strangerep

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    Hi Hans,

    I don't have a copy of Maggiore.
    Where did your "[itex]W^0=|\vec{W}|[/itex]" come from?

    Also, Patrick, when you mentioned Ryder, were you referring to his
    eq(2.214) on p64? (If so, that seems to rely on eq(2.210) which is
    for the massive case, and therefore subject to the same objections
    you raised in your first post.)

    EDIT: I was able to read enough of Maggiore on Amazon...
    What's happening is this: the little group for massless particles
    is E(2), which preserves the momentum vector (k,0,0,k).
    The two E(2) "translation" generators [itex]N_1:=K_1- J_2[/itex]
    and [itex]N_2:=K_2- J_1[/itex] are called A,B by Maggiore.
    For the massless case, one can show that

    [tex]W^2 ~=~ -k^2 (A^2 + B^2)[/tex]

    Then Magiorre gives the standard argument (also given in
    Weinberg vol1) that if we had particle eigenstates of A,B,
    (A,B commute, btw), then acting on them with [itex]J_3[/itex]
    produces a whole continuum of eigenvalues -- unless the
    eigenvalues of A,B are both 0. Then, "because we don't observe
    such particles
    " we restrict the Hilbert space just to those
    particle types. This of course also implies that the eigenvalue
    of [itex]W^2[/itex] for such states is also 0, but this is a rather
    different statement than saying that [itex]W^2=0[/itex] as a more
    general algebraic result (which does not hold).

    I.e., the restriction to states whose eigenvalues for A,B are 0
    is imposed by fiat. It is not "derived" from the general representation
    theory of the Poincare group. It is an arbitrary restriction.

    Other authors (e.g., Kim, Han, et al) see things differently, and show that
    the A,B generators can instead be interpreted as generators of gauge
    transformations. Compounding this puzzle, when one tries to construct
    a 4-vector massless spin-1 field, one finds it to be impossible. Weinberg
    covers this in more detail.
     
    Last edited: May 13, 2009
  15. May 13, 2009 #14

    Hans de Vries

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    Just from how it (spin) transforms in the limit case of beta=1.

    [tex]
    \begin{aligned}
    &\{~P^o~,~ \vec{P}~\} ~~~~&\mbox{transforms like}~~~~&\{~\gamma~,~\vec{\beta}\gamma}~\} \\
    &\{~W^o,\vec{W}~\} ~~~~&\mbox{transforms like}~~~~&\{~\beta\gamma~,~\gamma}~\}
    \end{aligned}
    [/tex]

    Regards, Hans
     
  16. May 16, 2009 #15

    Hans de Vries

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    One remark, it's the component of W along the velocity which transforms like this.
    In general we have of course.

    [tex]
    \vec{W} ~\implies~ \vec{W}_\bot+\gamma \vec{W}_\|[/tex]

    Regards, Hans
     
  17. May 16, 2009 #16

    Hans de Vries

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    The fields corresponding to the Poincaré generators:

    [tex]
    \begin{array}{|c|c|c|c|c|c|}
    \hline &&&&& \\
    &
    \mbox{transl.} &
    \mbox{Lorentz} &
    \mbox{rotation} &
    \mbox{boost} &
    \mbox{Pauli Lubanski vect.} \\
    \hline &&&&& \\
    \mbox{Poincar\'{e}:} &
    P^\mu & J^{\mu\nu} & J^i & K^i & W^\mu=\tfrac12\varepsilon^{\mu\nu\alpha\beta}\,J_{\alpha\beta}\,P_\nu \\
    &&&&& \\ \hline &&&&& \\
    \mbox{EM field:} &
    A^\mu & F^{\mu\nu} & B^i & E^i & \,C^\mu\,=\tfrac12\varepsilon^{\mu\nu\alpha\beta}\,F_{\alpha\beta}\,A_\nu \\
    &&&&& \\ \hline &&&&& \\
    \mbox{Spin $\tfrac12$ field} &
    j_V^\mu & M^{\mu\nu} & M^i & P^i & j_A^\mu \\
    \mbox{bilinear} &
    = & = & = & = & = \\
    \mbox{observables:} &
    \bar{\varphi}\, \gamma^\mu \varphi &
    \bar{\varphi}\, \sigma^{\mu\nu} \varphi &
    \bar{\varphi}\, \Sigma^i \varphi &
    \bar{\varphi}\, \Pi^i \varphi &
    \bar{\varphi}\, \gamma^\mu\gamma^5 \varphi \\
    &&&&& \\
    \hline
    \end{array}
    [/tex]

    The field [itex]C^\mu[/itex] is the important Chern Simons current, the spin of the electro-
    magnetic field. I wrote a chapter about it in a classical EM context here:

    http://physics-quest.org/Book_Chapter_EM2_ChernSimonsSpin.pdf

    [itex]M^{\mu\nu}[/itex] is the magnetization [itex](M^i)[/itex] and polarization [itex](P^i)[/itex] tensor of the electron field [itex]\varphi[/itex]

    [itex]\Sigma^i[/itex] and [itex]\Pi^i[/itex] are the spinor rotation and boost generators.


    Regards, Hans
     
    Last edited: May 16, 2009
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