1
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.