2

In Greiner Field Quantization book chapter 6, He derived the Feynman propagator for massive spin 1 vector field to be (page 169)

$$ i\Delta^{\mu\nu}_F(x-y)=-\bigg(g^{\mu\nu}+\frac{1}{m^2}\partial^\mu\partial^\nu\bigg)i\Delta_F(x-y)-\frac{i}{m^2}g^{\mu 0}g^{\nu 0}\delta^4(x-y).\tag{14}$$

Greiner then claims the delta function term should be omitted because something called "normal dependent terms" cancel the delta functions. I cannot find anything about this "normal dependent terms", can anyone explain?

Weinberg QFT page 278 says we can subtract a Hamiltonian $\frac{1}{2m}J^0(x)^2$ from it, but I do not understand the discussion in Weinberg either. I don't see how adding to the Lagrangian $$\tag{6.2.22} H_{NC}(x)=-H_{eff}(x)=\frac{1}{2m^2}[J^0(x)]^2$$ can cancel the delta function term. Especially I found Weinberg explanation of what $J^0(x)$ even is to be quite confusing:

"specifically, if $V_\mu(x)$ interacts with other fields through a term $V_\mu(x)J^\mu(x)$ in $H(x)$."

Qmechanic
  • 201,751
Simplyorange
  • 768

1 Answers1

1

Comments:

  1. The non-covariant contact terms in eq. (14) is related to the non-covariant time-ordering prescription in the operator formalism. For a similar issue, see e.g. my Phys.SE answer here.

  2. Concerning eq. (6.2.22) here is one line of reasoning: Recall that the free propagator can be found from the quadratic action by completing the square, cf. e.g. my related Phys.SE answer here. Hence changing the Hamiltonian with a term (6.2.22) will affect the free propagator.

Qmechanic
  • 201,751