In Peskin and Schroeder, we are trying to evaluate the path integral as given by equation 9.15

$$ \int \mathcal{D} \phi(x) \phi(x_1)\phi(x_2) \exp \Bigg[ i \int_{-T}^T d^4x \mathcal{L}(\phi) \Bigg]. $$

They then state that we can break up this functional integral by imposing two constraints that $\phi(x_1^0,\mathbf{x}) = \phi_1(\mathbf{x})$ and $\phi(x_2^0,\mathbf{x}) = \phi_2(\mathbf{x})$ and then integrate over all constraints to give

$$ \int \mathcal{D} \phi(x) = \int \mathcal{D} \phi_1(\mathbf{x}) \int \mathcal{D} \phi_2(\mathbf{x}) \int_{\mathrm{constrained}} \mathcal{D} \phi(x). $$

This makes sense. So applying this to the original integral we have

$$ \int \mathcal{D} \phi_1(\mathbf{x}) \int \mathcal{D} \phi_2(\mathbf{x}) \int_{\mathrm{constrained}} \mathcal{D} \phi(x)\phi(x_1)\phi(x_2) \exp \Bigg[ i \int_{-T}^T d^4x \mathcal{L}(\phi) \Bigg] $$ However they then say this is equal to

$$\int \mathcal{D} \phi_1(\mathbf{x}) \int \mathcal{D} \phi_2(\mathbf{x}) \phi_1(\mathbf{x_1})\phi_2(\mathbf{x_2})\int_{\mathrm{constrained}} \mathcal{D} \phi(x) \exp \Bigg[ i \int_{-T}^T d^4x \mathcal{L}(\phi) \Bigg] $$

Where they have replaced $\phi(x_1)$ with $\phi_1(\mathrm{x_1)}$ and $\phi(x_2)$ with $\phi_2(\mathrm{x_2)}$. I do not understand this. Could anyone help me understand what they did there?

I have been using bold for three vectors.


1 Answer 1


It appears trivial to me.

Before, there was an integration variable for all $x$ in the closed spacetime region.

After the split, we acquire an integration over the open region plus two boundary integrations for all $\bf{x}_{1,2}$.

It is just a relabeling, there is nothing physically meaningful or nontrivial here.


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