Constraints in path integral and the Lagrange multiplier

Question

I was reading some references on the slave-particle approach to the Kondo problem and Anderson model. 
It is known that the slave-particle is introduced in the large Hubbard $U$ limit of the system so that the Hilbert space for the localized orbital (e.g., $f$ electron) on a single site only include: ${ | 0 rangle, |uparrow rangle, |downarrowrangle }$.

The constraint $n_f le 1$ then becomes $$b^dagger b + sum_{sigma} s^dagger_sigma s_sigma =1$$ (here $b$ is boson and $s_sigma$ is fermion). In the path integral, people say the measure of the integration becomes:

$$int D(s^dagger,s)D(b^dagger,b)delta(b^dagger b + sum_{sigma} s^dagger_sigma s_sigma-1)$$

and a Lagrange multiplier is then introduced which replace the Dirac delta function above by: $$int Dlambda  e^{ilambda(b^dagger b + sum_{sigma} s^dagger_sigma s_sigma-1)}.$$

So here I want to ask how does the constraint transformed into the Dirac delta function of the field (at each space-time) above?

Alexander Nikolaenko · Answer

Delta function can be defined in the following way:
$$2 pi delta(x)=int e^{i lambda x }dlambda$$
Now, we can represent path intefral as multidimensional integral and integrate it using the expression above.

mike stone · Answer

Yes. You need a field $lambda(x)$ in @Alexandro Nikolaenko's answer above. One lambda for each site or point $x$.
$$
int d[lambda(x)]e^{int dx lambda(x)(b^dagger(x) b(x)+...)} 
$$

Constraints in path integral and the Lagrange multiplier

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