0
$\begingroup$

I work with this interaction Lagrangian density

$$\mathcal{L}_{int} = \mathcal{L}_{int}^{(1)} + \mathcal{L}_{int}^{(2)} + {\mathcal{L}_{int}^{(2)}}^\dagger = ia\bar{\Psi}\gamma^\mu\Psi Z_\mu +ib(\phi^\dagger\partial_\mu \phi - \partial_\mu\phi^\dagger \phi)Z^\mu,$$ with

$$ \mathcal{L}_{int}^{(2)} = +ib(\phi^\dagger\partial_\mu \phi)Z^\mu,$$ where $Z^\mu$ is an hermitian vector field, $\phi$ is a complex scalar field and $\Psi$ is the field of the electron.

I must work with Hamiltonian formalism.

I know that in this case I can work with $\mathcal{H}_{int} = - \mathcal{L}_{int}$ because the non-convariant term of the propagator of the vector field is cancelled by the non-covariant term of the interaction Hamiltonian (this is the case of quantizing a vector field with an interaction $J^\mu Z_\mu$ ).

My teacher told me the easiest way to derive feynman rules for the vertex $+ib(\phi^\dagger\partial_\mu \phi - \partial_\mu\phi^\dagger \phi)Z^\mu$ is to "attach" the incoming/outgoing particle momenta to the vertex. So, I get the following Feynman rules:

$\mathcal{H}_{int}^{(1)} = -ia\bar{\Psi}\gamma^\mu\Psi Z_\mu$ vertex: $-ia\gamma^\mu$ (the others feynman rules are obvious in this case)

$ \mathcal{H}_{int}^{(2)} + {\mathcal{H}_{int}^{(2)}}^\dagger = -ib(\phi^\dagger\partial_\mu \phi - \partial_\mu\phi^\dagger \phi)Z^\mu$ vertex: $\pm bp^\mu$ where $\pm$ if the sign of momenta $\vec{p}$ and the arrow of the charge are concordant.

Ok, the Feynman rules for the vertices are very simple. Let's go to the computation of feynman rules for the propagator, like that

$$ \int d^4x d^4y <0|T(\phi^\dagger(x)\partial_\mu\phi(y))|0> $$

if $\mu = 1,2,3$, then it's true that I have just to "attach" the momenta of incoming/outgoing particle to the vertex; but there is an extra term if $\mu= 0$ that yields a non-covariant term after have substituted the integral representation of theta-function $$ \int d^4x d^4y <0|T(\phi^\dagger(x)\partial_\mu\phi(y))|0> = \delta_\mu^j\int d^4x d^4y \frac{d^4p ip_j}{(2\pi)^4}e^{ipx}\Delta_F(p) + \delta_\mu^0\int d^4x d^4y d^3p d^3q \left[ ip_0\theta(x^0-y^0)[\phi^+_c(x,p),\phi^-_c(y,q)]- iq_0\theta(y^0-x^0)[\phi^+(y,p),\phi^-(x,q)]\right]. $$

How to handle it? How to find the feynman rule for the propagator??

p.s. I write the scalar field as $\phi(x) = \int d^3p \phi^+(x,p)+\phi^-_c(x,p)$

$\endgroup$
  • $\begingroup$ Your Feynman rule for ${\cal H}^{(2)}$ is wrong. It should be I believe something like $b(p_1^\mu - p_2^\mu)$ where $p_1$ and $p_2$ are the two outgoing scalar momenta in the vertex. The overall sign of the factor, I'm not sure about, but can be worked out in the way your professor told you. $\endgroup$ – Prahar Dec 26 '15 at 20:31
  • $\begingroup$ Thank you, is the same I thought. So, there is no way to get the expression of the propagator? $\endgroup$ – apt45 Dec 26 '15 at 23:53

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.