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

Amplitude for scalar-proca couplings

  1. Oct 4, 2014 #1
    I'm trying to calculate the amplitude for an interaction between a scalar field $$\phi$$ and two identical spin 1 fields $$A_{\mu} \quad and \quad A^{\mu}$$ for the interaction $$\phi \longrightarrow A^{\mu} A_{\mu}$$
    with the Lagrangian density $$L_{int} = -ik\phi A^{\mu} A_{\mu}$$ where k is a constant. My first thought is that the amplitude should evaluate to $$M = k\epsilon_{\mu}^{*}\epsilon^{\mu *}$$ where each of the epsilons has a particular spin. Thus, this would evaluate to -k if the two spins are equal and 0 if they are not equal. However, I would expect this to come out in terms of energy squared, but this is clearly not the case. Where am I erring?
     
  2. jcsd
  3. Oct 4, 2014 #2

    Orodruin

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    The dimension of the matrix element squared should be energy squared so the dimension of the matrix element should be energy. This is consistent with what you have.
     
  4. Oct 4, 2014 #3
    But -k is unit less. Am I wrong to thing that the two epsilons go to -1 if the spins are the same and 0 otherwise? Or is the -1 in units of energy?
     
  5. Oct 4, 2014 #4

    Orodruin

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    No, ##k## is not unit-less. You can derive the unit of any constant in the Lagrangian on the basis that it has to have dimension energy^4 for the action to be dimensionless. Based on the kinetic terms, the scalar ##\phi## and the spin one fields ##A^\mu## both have dimension energy, which means that ##k## must have unit energy for the dimension of ##k\phi A^2## to be energy^4.

    You should be able to check your assumption of the polarisations based on a particular representation of the polarisation vectors. For example, what happens if both epsilons are ##\epsilon^\mu = (0, 1, i , 0)^\mu/\sqrt{2}##.
     
  6. Oct 4, 2014 #5
    Then $$\epsilon_{\mu}^{*}\epsilon^{\mu *}$$ evaluates to 0? The makes sense, the probability should be zero, since the spins must be opposite. So I guess it evaluates to -1 if the spins are different? If I make the other $$\epsilon_{\mu}^{*} = (0,1,-i,0)^{\mu}/\sqrt{2} $$ Then $$\epsilon_{\mu}^{*}\epsilon^{\mu *}$$ evaluates to $$-1 $$ indeed
     
  7. Oct 4, 2014 #6

    Orodruin

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    Yes, notice that it will be different if you instead of the definite spin states chose to go with linear polarised spin vectors.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook