1
$\begingroup$

Consider $4d$ $\mathcal{N}=2$ susy gauge theory. Why is the propagator of the non-dynamical auxiliary field $D$ (the field needed such that the susy algebra is closed off-shell) proportional to a delta function? I.e. why is it

$$ \langle D(x)D(y) \rangle \backsim \delta(x-y)? $$

This suggests that it is given by some sort of Gaussian integration (in the path integral). What is the exact form of this 2-pt function then?

$\endgroup$
1
$\begingroup$

One way of looking at it will be to read off the Feynman rule from the Lagrangian directly. Auxiliary field does not have any kinetic term in the Lagrangian, so if you replace $D$ by a free field and collect the coefficient then it is just some constant term. So the momentum space Feynman rule is just inverse of some constant term. To go to coordinate space you have to take the four dimensional Fourier transform of a constant, which is a delta function.

$\endgroup$
  • $\begingroup$ So what does this look like in the path integral? $\endgroup$ – Marion Nov 22 '16 at 10:39
  • $\begingroup$ I mean, it should be something like $\langle D(x) D(y) \rangle = \int [dD \, ] D(x)D(y) e^{-\int dx D(x)^2}$ right? $\endgroup$ – Marion Nov 22 '16 at 10:52
  • $\begingroup$ Two point function is always calculated as the Green function and is an inhomogenous differential equation in all cases. Here since no derivative appear for auxiliary field, the eqn. will look like $(constant)G_F(x,y)= \delta(x-y)$ and hence the propagator is simply proportional to $\delta(x-y)$. Successive differentiation are done in path-integral formalism to obtain higher order correlation function. Noone is going to do a messy calculation like dividing it into intervals and then.... $\endgroup$ – ved Nov 22 '16 at 12:19
  • $\begingroup$ Yes, sure, thanks for the answer. I just want to make sure that what I have written is schematically correct. $\endgroup$ – Marion Nov 22 '16 at 14:22

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.