# Correlation Functions in Path Integral Formulation of QFT

1. Sep 6, 2009

### maverick280857

Hi,

I was going through section 9.2 of Peskin and Schroeder, and came across equation 9.16 which reads

$$\int\mathcal{D}\phi(x) = \int \mathcal{D}\phi_{1}({{\bf{x}}}) = \int \mathcal{D}\phi_{2}({{\bf{x}}}\)int_{\phi(x_{1}^{0},{\bf{x}})\\\phi(x_{1}^{0},{\bf{x}})}\mathcal{D}\phi(x)$$

What does the right hand side mean, and how does this follow from

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

? What is the relationship between $x_{1}^{0}$ and $T$?

Last edited: Sep 6, 2009
2. Sep 7, 2009

### xepma

You gotta look at it from a slighty different perspective. The functional integral:

$$\int\mathcal{D}\phi(x)$$

is a functional integral over "all possible functions" $$\phi(x)$$. We now set certain boundary conditions at times $$x_1^0$$ and $$x_2^0$$. To be precise, we demand "restrict" the function $$\phi(x)$$ to obey the boundary conditions:

$$\phi(x_1^0,\mathbf{x}) = \phi_1(\mathbf{x})$$
$$\phi(x_2^0,\mathbf{x}) = \phi_2(\mathbf{x})$$

You should interpret that as $$\phi_1(\mathbf{x})$$ and $$\phi_2(\mathbf{x})$$ being certain fixed configurations of the field $$\phi(x)$$ at the corresponding timeslices. Plugging this into the functional integral would give:

$$\int_{\left(\phi(x_1^0,\mathbf{x}) = \phi_1(\mathbf{x})\right)}^{\left(\phi(x_2^0,\mathbf{x}) = \phi_2(\mathbf{x})\right)}\mathcal{D}\phi(x)$$

This is written in the suggestive notation that it's just like an ordinary integral with an upper and lower limit - only they are not really limits, but rather constraints on the function we integrate over.

This functional integral is obviously not the same as the one we started out with, since we are not integrating over all possible functions - but rather only over the class which respects the boundary conditions, i.e. the constraints. However, if we now integrate over all possible boundary conditions, i.e. over the configurations $$\phi_1(\mathbf{x})$$ and $$\phi_2(\mathbf{x})$$, we integrate over all possible functions again! Hence we end up with the identity:

$$\int\mathcal{D}\phi(x) = \int\mathcal{D}\phi_1(\mathbf{x})\int\mathcal{D}\phi_2(\mathbf{x})\left(\int_{\left(\phi(x_1^0,\mathbf{x}) = \phi_1(\mathbf{x})\right)}^{\left(\phi(x_2^0,\mathbf{x}) = \phi_2(\mathbf{x})\right)}\mathcal{D}\phi(x) \right)$$

So all in all it's just a different way of writing your functional integral. Why is this trick useful? Well, we use it to compute the two-point correlator:

$$\langle\Omega|T\phi(x_1)\phi(x_2)|\Omega\rangle = \int\mathcal{D}\phi(x) \phi(x_1)\phi_(x_2) \exp\left[i\int_{-T}^T d^4x\mathcal{L}\right]$$

which is what the remainder of the chapter explains. Basically, you fix the field $$\phi(x)$$ at two times, which allows you to pull out the two factors $$\phi(x_1)\phi_(x_2)$$. What remains are three seperate Gaussian integrals, over three different time intervals, which are "glued together" at the two time slices. These Gaussian integrals can be performed, and what remains is some expression which only integrates over the boundary conditions $$\phi_1$$ and $$\phi_2$$.

Now, the story continues, but the moral of it all is that this expression in terms of the path integral:
$$\langle\Omega|T\phi(x_1)\phi(x_2)|\Omega\rangle = \int\mathcal{D}\phi(x) \phi(x_1)\phi_(x_2) \exp\left[i\int_{-T}^T d^4x\mathcal{L}\right]$$

gives you precisely the time-ordered two point correlator. This is actually quite a suprise, since the two-point correlator can come in many varieties, if you start out with the operator formalism. The path integral makes no reference to operators, yet its two-point correlators correspond to a specific ordering of the operators - namely the time ordering.

In conclusion: the proof shows how the path integral is related to the time orderded correlator.

3. Sep 8, 2009

### maverick280857

Thanks for the detailed reply xepma.