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Scalar field theory - Feynman diagrams and anti-particles

  1. Aug 24, 2010 #1
    I'm working on a "draw all possible Feynman diagrams up to order 2" problem for a scalar field that obeys the Klein-Gordon equation, and I'm wondering about a few things. When I did a course on particle physics and was first introduced to Feynman diagrams in the context of QED (but not QED itself in any detail), I was allowed to flip a line of propagation around, i.e. from propagating towards a vertex to propagating away from a vertex or vice versa, the particle would be an anti-particle and I'd still have a valid diagram.

    Can I do the same thing for my scalar particles? What is the anti-particle of such a particle? It seems the propagator doesn't distinguish between directions of propagation at all, so are they their own antiparticles? What kind of particles do I identify with my scalar field in the first place?
  2. jcsd
  3. Aug 25, 2010 #2
    It depends on whether you're talking about a real or a complex scalar field. (In the operator formalism, a "real" quantum field is hermitian, but its complex counterpart isn't.) A real scalar field gives rise to particles that are their own antiparticles, wheras for a complex field you've got to make a distinction between the two.

    This is one point that is easier to think about from the point of view of the operator formalism. When you expand a real field in terms of its fourier modes, the coefficients a(k), b(k) of the positive and negative frequency modes must be complex conjugates of each other. When you promote the field to an operator, complex conjugation becomes hermitian conjugation, and you have creation and annihilation operators corresponding to a single particle. Obviously, in the complex case this restriction no longer applies.

    Finally, the klein-gordon field is mostly a kind of toy model. I think I'm right in saying that the higgs field is a a scalar field, but the whole point of the higgs is that it's introduced into a lagrangian with other fields to which it couples... I've never seen anyone actually apply these considerations to it. Apart from that, I don't know of any scalar fields that exist in nature.
  4. Aug 25, 2010 #3
    Alright, I got that. I actually got to reading a little about the complex scalar field after I had asked this question.

    So is the following a valid Feynman diagram? It seems to fit the rules, but I never saw anything like this in my particle physics course. Is this "extracting zero-point energy"? (The vertical direction is time.)

    [PLAIN]http://img217.imageshack.us/img217/3896/15530464.png [Broken]
    Last edited by a moderator: May 4, 2017
  5. Aug 25, 2010 #4
    I think the important question is - what kind of theory exactly are you writing the Feynman diagrams for? You can have many Lagrangians with scalar fields that obey the Klein-Gordon equation (actually even fermions obey the KG equation - it's simply a statement about the relativistic energy-momentum relation). This kind of diagram will be possible (among others) in a theory where the Lagrangian contains a term coupling a complex scalar field to a real one (unless you have arrow heads on the loop as well - then its simply a [tex]|\phi|^4[/tex] theory).

    This kind of diagram will give a contribution to the 2-point function (or propagator) of the scalar field. Its value will be infinite and would need to be renormalized along with other diagrams contributing to the full propagator of the scalar field.

    This means that the propagator you are used to of the scalar field gets quantum modifications from higher order diagrams, and this is one of them.
  6. Aug 25, 2010 #5
    Last edited by a moderator: May 4, 2017
  7. Aug 28, 2010 #6
    Doesn't the fact that one is an anti-particle take care of your problem with momentum conservation?
  8. Aug 28, 2010 #7
    No, because the total of energy-momentum in the initial state is not zero, while the energy-momentum of the final state (which doesn't really exist here) is zero.

    When you have particles in the initial state, you must have particles in the final state to carry the energy-momentum of the initial state.
    For example, if you consider electron-positron annihilation you have two photons (for example) in the final sate (you can't even have one photon because in the center of mass frame of the electron-positron pair they have zero spatial momentum and a massless particle can't have zero spatial momentum).

    Hope this helps a bit..
  9. Aug 29, 2010 #8
    That sounds reasonable. I guess this is encompassed in the maths behind the diagram in that it yields a delta function

    \delta^4 (k1 - k2),

    which I wouldn't really know how to satisfy. Thanks for the replies.
  10. Aug 29, 2010 #9
    That delta function is easy to satisfy. It just says that the two physical particles have the same 4-momentum. The diagram you presented should have [itex]\delta^4(k_1+k_2)[/itex]. Given the constraint of non-negative energy, this cannot be satisfied for massive particles and can only be satisfied when all components are 0 for massless particles, which makes the diagram physically irrelevant.
  11. Aug 29, 2010 #10
    You're right, thanks for correcting that.

    I have another question about these diagrams. What is the general approach towards finding all of the possible diagrams? I hope it's not writing out the Wick contraction term by term. For example, the problem I was working on requested all diagrams with up to two vertices that represented two mesons becoming four. I drew a bunch and I don't think I can draw any others, but how do I know for sure?
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