Amplitude for scalar-proca couplings

[c++guru]

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?

 

[Orodruin]

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.

 

[c++guru]

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.

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?

 

[Orodruin]

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 $\epsilon^\mu = (0, 1, i , 0)^\mu/\sqrt{2}$.

 

[c++guru]

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 $\epsilon^\mu = (0, 1, i , 0)^\mu/\sqrt{2}$.

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

 

[Orodruin]

Yes, notice that it will be different if you instead of the definite spin states chose to go with linear polarised spin vectors.