# Constraints in path integral and the Lagrange multiplier

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: $$\{ | 0 \rangle, |\uparrow \rangle, |\downarrow\rangle \}$$.

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: $$\int D\lambda \ e^{i\lambda(b^\dagger b + \sum_{\sigma} s^\dagger_\sigma s_\sigma-1)}.$$

Here I would like to ask how does the imaginary time dependence entered into the constraint? Because originally we just have a constraint of the particle number operators, which can be transformed into a delta function at a specific time, say, at $$\tau = 0$$.

• Which references? Which pages? Nov 27, 2020 at 19:10

The constraint $$\delta\left(z^{2}-1\right)$$ is actually added for any time and space, rather than just $$\tau=0$$, thus, it need to be written as: $$\prod_{x, \tau} \delta\left(|z(x, \tau)|^{2}-1\right)$$ since $$\delta\left(|z(x, \tau)|^{2}-1\right)=\int_{-\infty+i c}^{+\infty+i c} d \lambda(x, \tau) e^{-i \lambda(\tau, x)\left(|z(x, \tau)|^{2}-1\right)}$$ thus, $$\prod_{x, \tau} \delta\left(|z(x, \tau)|^{2}-1\right)=\prod_{x, \tau}\left(\int_{-\infty+i c}^{+\infty+i c} d \lambda(x, \tau) e^{-i \lambda\left(|z(x, \tau)|^{2}-1\right)}\right)=\int \mathcal{D}[\lambda] e^{-i \int d^{D} x \lambda(x, \tau)\left(|z(x, \tau)|^{2}-1\right)}$$ where $$\int \mathcal{D}[\lambda]=\prod_{x, \tau} \int_{-\infty+i c}^{+\infty+i c} d \lambda(x, \tau)$$ Now, imaginary time dependence entered into the constraint $$\lambda$$. CH.8.1 of Polyakov also gives this detail.
Yes. You need a field $$\lambda(x)$$ in @Alexandro Nikolaenko's answer above. One lambda for each site or point $$x$$. $$\int d[\lambda(x)]e^{\int dx \lambda(x)(b^\dagger(x) b(x)+...)}$$
• thanks for the ansewer. But what makes me a bit confused is how the (imaginary) time dependence of $\lambda$ is introduced? Nov 28, 2020 at 20:38
• I should have said one $\lambda$ for each space-time point. Nov 28, 2020 at 20:40
The delta function can be defined in the following way: $$2 \pi \delta(x)=\int e^{i \lambda x }d\lambda$$ Now, we can represent the path integral as a multidimensional integral and integrate it using the expression above.