# QFTs which are pure constraint

I am interested in (typically topological) field theories arising from Lagrangians of the form.

$f(\Phi) \lambda$,

where $\lambda$ is a Lagrange multiplier field not appearing in $f(\Phi)$. Perturbatively, the partition function just counts the solutions to $f(\Phi)=0$. What about nonperturbatively? Are there exponentially suppressed kinks that affect the partition function?

Has anyone considered these?

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I am confused--- the result that this counts solutions is only nonperturbative. What perturbation theory reproduces it? The field $\lambda$ doesn't have a quadratic term, and it's integrated over, so it gives a delta-function. – Ron Maimon Aug 21 '12 at 6:10
I meant perturbative in the sense that the saddle points are all the solutions $f(\phi)=0$ and each contributes 1 in an expansion around that saddle point (assuming solutions are isolated). – Ryan Thorngren Aug 21 '12 at 7:33
I see--- in this case, this is an exact result of path integration. I'll make it an answer. – Ron Maimon Aug 21 '12 at 7:36

$$\int e^{i\int \lambda(x) F(x)} d\lambda = \prod_x (2\pi) \delta(F(x))$$
+1 Thanks, Ron. I agree that (modulo a factor of $1/|det F'(x)|$, what happened to that?) this is morally the case, and I think I see what you mean about there being no regulator ambiguity (we used no symmetry arguments). I thought that there might be a nice example where this argument fails, but I haven't been able to come up with one yet. – Ryan Thorngren Aug 21 '12 at 17:28
@user404153: The factor is $\mathrm{det}({d\over d\lambda} (\lambda F))$, any remaining factors come from popping the delta function when you do another path integral. – Ron Maimon Aug 21 '12 at 20:28