Auxiliary field in Quantum Field Theory I'm reading this answer which explains 'What is auxiliary field in Quantum Field Theory?'

As a simple example, we can imagine that the real field must follow some constraints. In the path integral we can add this as a Dirac delta functional. We can write this dirac delta functional as a Fourier transform where there would be an extra field (over which the FT is performed). This is an auxiliary field. We have to integrate over this field but there is nothing physical about this. This is because if we were to compute the equations of motion for this field, it is easy to see that we just retrieve the constraint on the real field.

I was wondering if someone could give an explanation of this example with equations. From my QFT notes, I saw this functional form of dirac delta equation:
$$
\int[d\phi]\int[d\lambda]e^{iS[\phi]+i\lambda(x)F[\phi]} = \int[d\phi]\delta(F[\phi])e^{iS[\phi]}
$$
If we interpret this equation using that example, is $\lambda$ the extra field being integrated over? How does the dirac functional impose constraints on the path integral?
Also on the Wikipedia page, auxiliary field has no kinetic terms. I wonder if this statement is relevant to this example of auxiliary field.
Thanks for the help!
 A: Auxiliary fields usually mean non-propagating$^1$ fields, such as e.g.,

*

*Faddeev-Popov (FP) ghost and antighost fields,


*Lagrange multipliers,


*antifields in the Batalin-Vilkovisky (BV) formalism,
etc.
Concerning Lagrange multipliers, see also e.g. this related Phys.SE post.
--
$^1$ Note that they may have kinetic terms, such as e.g.,

*

*the Faddeev-Popov (FP) term often has a kinetic term for the ghost and antighost,


*Lagrange multipliers in the Batalin-Fradkin-Vilkovisky (BFV) formalism have a kinetic term.
If they have kinetic terms, they should decouple in the unitary gauge.
A: 
If we interpret this equation using that example, is  the extra field being integrated over?

Yes.

How does the dirac functional impose constraints on the path integral?

By guaranteeing $F[\Phi]=0$ in every field configuration that contributes to the path integral.

Also on the Wikipedia page, auxiliary field has no kinetic terms. I wonder if this statement is relevant to this example of auxiliary field.

Since there is no term $\sim (\partial \lambda)^2$ in the action when $\lambda$ is included, indeed the auxiliary field $\lambda$ has no kinetic term.
