Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Fundamental particles in physics

  1. Nov 6, 2013 #1
    I was reading this recent Scientific american article wich I found interesting, and was spurred on by it to ask a couple of questions.

    Is the mathematical concept of fundamental point particle currently a basic postulate in physics(let's take as current physics the QM based-QFT modern models)?
    I'm thinking about mathematical concepts like the Dirac delta and modeling point sources with Green's functions as something necessary to keep linear superposition and QM as a linear theory, (not sure we can talk about the different QFT's being linear or not).

    Is the field concept in QFT really more fundamental or not?
    Is the term "particle physics" really a misnomer?
    How should one picture the high number and diversity of quantum fields,(one for every possible fundamental particle... scalar, vectorial and tensorial... interacting and ad free...)?
  2. jcsd
  3. Nov 6, 2013 #2
    Let's answer the second part first...

    I think for sure the field concept in QFT is more fundamental. Particles are just excitations of the fields, so I think it is pretty reasonable to consider the field more fundamental.

    As for whether the name "particle physics" is a misnomer, well, I don't really think so, since one only ever deals with the particles. The existence of the underlying fields is only indirectly inferred.

    "How should one picture the high number and diversity of quantum fields" - I don't exactly know what you are looking for. They are just there, all "sitting on top of each other" at every point of spacetime. They might all be different degrees of freedom of some more fundamental underlying field though, or of some compactified dimensions of spacetime etc, string theory style.

    So back to the first question: the quantum fields are pretty much a postulate of quantum field theory, yes. It isn't exactly formalised that way, there are games one plays reconciling standard quantum mechanics with special relativity and it works out that these quantum fields emerge as the solution for doing that, but you can more or less take them as a postulate I think.
  4. Nov 7, 2013 #3
    Kurros answer pretty much covers it. I just want to point out that there really isn't a one to one correspondence between the number of fields and the number of particles. For instance, the electron is an excitation of two different fields, one of which interacts through the weak isospin interaction while the other one doesn't.
  5. Nov 7, 2013 #4


    User Avatar
    2017 Award

    Staff: Mentor

    I deal with some of those fields every day, especially with gravity in the morning (okay, not part of the SM) and electromagnetic fields every time I use electricity.
  6. Nov 8, 2013 #5


    User Avatar
    Science Advisor

    This is absolutely false. For starters, leptons are fundamental Dirac fields in SM. So there is one-to-one between particles and fields. You might be thinking about the photons and the place they take in the electroweak interactions, but even that can be fixed with a basis change. Which is, basically, what's done in SM. Even if you want to look at a photon as an excitation in two fields, the total number of fields is still the same. The U(1)xSU(2) has 4 generators, 4 gauge fields, and 4 particles associated with it, even if we consider particles to be linear combinations of the fields.

    It's a Yang-Mills gauge field. So it's not quantizable. I say, close enough.
  7. Nov 8, 2013 #6


    User Avatar
    Science Advisor

    I believe it depends on viewpoint. One could equally well regard the electron arising from two fields - left- and right-handed Weyl spinors, which the Higgs interaction couples together.

    Conversely, one could say the single Dirac field leads to two related particles - electron and positron. Just depends on how you count things.
  8. Nov 8, 2013 #7
    I'm referring to the fact that the standard model is a chiral theory so what we think as one particle - the electron for instance - must be built of two fields with opposite chirality. Those two fields don't even interact the same way with other fields. Notably, the left handed electron field (and the right handed positron field) do not interact through weak isospin at all. The usual counting of particles tell us that there are four different fermions per family in the standard model. For the lightest family those are the electron, the electron-neutrino, the up quark and the down quark. There many different sensible ways you could chose to count different fields but none of them gives four as the total number of fields per family. So it is absolutely TRUE that there is no sensible one to one correspondence between particles and fields.
    Last edited: Nov 8, 2013
  9. Nov 8, 2013 #8
    The left- and right-handed Weyl spinors belong to two different representations of the electro-weak symmetry group. There is no way to sensibly buid a Dirac field before electroweak symmetry breaking happens. That's why the electroweak theory is built out of 5 different spinor fields per family, each one belonging to its own representation under the standard model gauge symmetry but we get only four particles out of those fields - the electron, the electron neutrino, the up quark, and the down quark.
  10. Nov 8, 2013 #9


    User Avatar
    Science Advisor

    Please explain this in more detail. How do you get five?
  11. Nov 8, 2013 #10
    Here is the list of fields followed by the multiplet they belong to under the SU(2)Weak isospin x SU(3)Color Symmetry Group

    E_R (1,1) for a total of 1 field
    L_L (2,1) for a total of 2 fields
    U_R (1,3) for a total of 3 fields
    D_R (1,3) for a total of 3 fields
    Q_L (2,3) for a total of 6 fields,

    So if you count a multiplet as a single field, you have 5 different fields and if you count each individual member of a multiplet as a separate field, you have 15 fields. There is no way to get only four fields. Those fields lead to only four standard model particles as I mentioned earlier.

    All these fields are chiral fields and are accordingly labeled either with a "_L" or with a "_R". The first two fields are lepton fields while the last 3 are quark fields.

    The electron mass is generated by the interaction E_R-L_L-Higgs,
    the up quark mass is generated by the interaction U_R-Q_L-Higgs, and
    the down quark mass is generated by the interaction D_R-Q_L-Higgs.

    One of the components of L_L is left un-paired leading to a massless neutral particle - the neutrino (which is not a Dirac spinor in the Standard Model).
    Last edited: Nov 8, 2013
  12. Nov 8, 2013 #11


    User Avatar
    Science Advisor

    That's supersymmetry, which is beyond SM. None of this is true in SM.

    Also, electron current mass is generated via interaction with Higgs. Same for the quarks. Most of the mass is actually dynamically generated. But that's aside from the discussion.
  13. Nov 8, 2013 #12
  14. Nov 8, 2013 #13


    User Avatar
    Science Advisor

    dauto is correct. Furthermore, if you take the hermitian conjugate of a L field, you get a R field (in the complex-conjugate rep of the gauge group), and vice-versa. So it's just a matter of convention that some fields are listed as L and some as R; we could list them all as L. This is all explained in great detail in Srednicki's QFT text.
  15. Nov 9, 2013 #14
    What? No, I'm not talking about Supersymmetry. Supersymmetry would require additional particles and additional fields. The fields I described are the Standard Model chiral fermion fields.
    Last edited: Nov 9, 2013
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook