# Two-point correlation function of a scalar field $\langle 0 | \phi(x) \phi(0)| 0 \rangle$

I'm trying to find the two point correlation function for a massless scalar field obeying $$\square \phi = 0$$. I can write

$$\langle 0 | \phi(x) \phi(0)| 0 \rangle = \int \frac{d^dk}{(2\pi)^d} \delta(k^2)\theta(k_0)\langle 0| \phi(x) | k\rangle \langle k | \phi(0)|0\rangle= \int \frac{d^dk}{(2\pi)^d} e^{ikx}\delta(k^2)\theta(k_0)$$

where I've inserted the identity operator for 1-particle momentum eigenstates. But elsewhere I see the expression

$$\langle 0| \phi(x) \phi(0)|0 \rangle = \int \frac{d^dk}{(2\pi)^d} \frac{e^{ikx}}{k^2}$$

for instance, the first line of this page. Are these forms equivalent? Which is correct and why?

• For free fields (i.e. with quadratic action), the computation of the correlation functions amounts to a Gaussian integral. It's the same as computing the variance of a Gaussian -but in infinite dimensions-.
– lcv
Mar 15, 2019 at 17:13
• – SRS
May 6, 2020 at 7:44

After correcting this, to see that the two approaches give the same result, perform the integral with respect to $$k^0$$.
For the first case you can use the delta function relation $$\delta(f(x)) = \sum_i \frac{\delta(x - x_i)}{\vert f'(x_i)\vert}$$, where $$x_i$$ are the zeros of $$f(x)$$.
For the second case you can use a contour integral and the residue theorem. You'll need to use the Feynman $$i\epsilon$$ prescription (see https://en.wikipedia.org/wiki/Propagator#Feynman_propagator).
I don't understand the very first formula and you discussion of identity insertion. I would like you to start from the mode decomposition of scalar field: $$\phi(x)=\int_{p}\frac{1}{\sqrt{2\omega_p}}\left(a_pe^{-ipx}+a^{\dagger}_pe^{ipx}\right)$$ and similar decomposition of $$\phi(0)$$ (keep in mind that the momenta should be another, fo r instance $$k$$). Then, you can see that from four terms only one gives contribution into the average and this term can be rewritten with help of $$[a_p,\,a^{\dagger}_k]=(2\pi)^d\delta^{(d)}(p-k)$$. Finally, you will obtain the desired result (only one integral survive after integration with delta-function). I hope this help.