# 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

For a path integral, the integral

$$\int e^{i\int \lambda(x) F(x)} d\lambda = \prod_x (2\pi) \delta(F(x))$$

Where the product is over all points x (imagine a lattice). So that the result is not just a saddle point identity (saddle points and perturbative approximations are not the same, perturbative is by expanding the exponential in a series), it is an exact path integration identity.

While the result is not fully rigorous, that is only because of the issue of path integration definition, the limit of continuous space as the lattice gets finer. Ignoring this mathematical point, there is no way this identity is avoided in any proper definition of path integration, since it holds without any regulator ambiguity.

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+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
Oh you're right. Thanks. –  Ryan Thorngren Aug 21 '12 at 20:33