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Can massless states be used to build massive ones?

  1. Feb 21, 2015 #1
    I think I have read that massless fermionic states can be used (as a basis?) to build up massive fermionic states. Is that true? If so, could you please give me a very short outline of how this is done, or maybe better, a mathematical statement that "says" the same thing?

    Thanks for any help!
     
  2. jcsd
  3. Feb 21, 2015 #2
    I don't know what do you mean with "build up massive fermionic states". Build up in what sense?
    We know that in QCD we have gluons which are the massless bosons that mediate the strong interactions. If you create a bound system of just gluons (the so-called glueballs) then one can show with lattice QCD computations that they have a non-zero mass even its components are all massless. The mass of the bound state is all due to binding energy.
     
  4. Feb 21, 2015 #3
    As in,

    "This tells us that a general field can be described by two Weyl fields: one left-chiral and one right-chiral. This is the advantage of talking in terms of Weyl fields: they can be seen as the building blocks for any fermion field."

    Part of a larger section,

    "5 Weyl fermions

    It has been noted that the problem with assigning a frame-independent helicity to a fermion disappears if the fermion is massless. The problem with a conserved value of γ5 also disappears in this limit, since γ5 does indeed commute with the mass-independent term in the Dirac Hamiltonian. This shows that, without any ambiguity, one can talk about a positive or negative helicity fermion or of a left or right chiral fermion when one talks about massless fermions.

    5.1 Irreducible fermion fields

    Indeed, it is very convenient to use such objects in any discussion regarding fermions. A general solution of the Dirac equation is not an irreducible representation of the Lorentz group. This is best seen by the existence of the matrix γ5 that commutes 12with all generators of the representation, a fact that was summarized in Eq. (4.11). By Schur’s lemma, no matrix other than the unit matrix should have this property if the generators pertain to an irreducible representation. We have already seen that a left-chiral fermion field retains its chirality under Lorentz transformations, implying that such fields are irreducible.4 So are right-chiral fields, of course. It is known that the proper Lorentz algebra is isomorphic to SU(2) × SU(2), so that any representation of the Lorentz algebra can be identified by its transformation properties under each of the SU(2) factors. In this language, a left-chiral fermion would be a doublet under one of the SU(2)’s and singlet under the other, a fact that is summarized by denoting the representation as ( 1 2 , 0). A right-chiral fermion is a (0, 1 2 ) representation. Either of them is called a Weyl fermion. A general fermion field transforms like a reducible representation ( 1 2 , 0) + (0, 1 2 ). This tells us that a general field can be described by two Weyl fields: one left-chiral and one right-chiral. This is the advantage of talking in terms of Weyl fields: they can be seen as the building blocks for any fermion field."

    I take the following to mean massive fermions (?),

    " ... they can be seen as the building blocks for any fermion field. "

    From,

    http://arxiv.org/pdf/1006.1718v2.pdf

    From,

    http://www.quora.com/Quantum-Field-Theory/What-is-a-Weyl-fermion [Broken]
     
    Last edited by a moderator: May 7, 2017
  5. Feb 21, 2015 #4
    How familiar are you with group theory? The mathematical explation is given in the text. The left and right Weyl spinors transform as irreducible representation of the Lorentz group SO(3,1). This essentially means that if you take, say, a left spinor and you apply a Lorentz trasnfromation to it then you will again obtain a left spinor. Same thing for a right spinor.
    Now, if a field is massless it can always be classified according to its chirality. This means that an m=0 fermion can always be either a left or a right spinor. If the field is massive, instead, you can always write:
    $$
    \psi=\left(\frac{1+\gamma_5}{2}+\frac{1-\gamma_5}{2}\right)\psi=\psi_L+\psi_R.
    $$
    Therefore, a generic massive field can always be written as a sum of a left and a right spinor. Of course, since ##\psi## itself does not belong to an irreducible representation of the Lorentz group, when applying a Lorentz trasformation its chirality can change.
     
  6. Feb 21, 2015 #5
    In simple language, does that mean massless fields can combine to form massive fields?

    Thanks for your help!
     
  7. Feb 21, 2015 #6
    It really depends on what you mean by "combine". In group theory language you can combine two states either with a tensor product or with a direct sum. In this case is a direct sum.
    In plain language, every massive field can be written as a sum of a left and a right fields.
     
  8. Feb 21, 2015 #7
    Then am I misinterpreting,

    " ... they can be seen as the building blocks for any fermion field. ",

    where I take "they" to be massless fermions and I take "any fermion field" to be massive fermion fields?

    Thanks for your time!
     
  9. Feb 22, 2015 #8
    No, when it talks about "they" it refers to the Weyl left/right spinors: every massive spinor can be seen as the sum a left and a right spinor. IF the spinor is massless, then it has a definite chirality and hence you can write it just as a left spinor or as a right one.
     
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