# Equal-time 2-point correlator and divergent (?) integral

Assume a massless scalar field in 3+1 dimensions which can be written as $$\phi(t,\vec{x})=\int\frac{d^3k}{\sqrt{(2\pi)^32k}}\left(a(\vec{k})e^{-ikx}+a^\dagger(\vec{k})e^{ikx}\right)\, ,$$ where $$\vec{k}$$ is the momentum (natrual units), $$k=\vert\vec{k}\vert$$, $$kx=k_\mu x^\mu$$ is the inner product of the four-vectors and $$a(\vec{k})$$ and $$a^\dagger(\vec{k})$$ obey the familiar commutator relations for the creation and annihilation operators.

Now I am interested in the equal-time 2-point correlator function $$\langle 0\vert\phi(t,\vec{x}_1)\phi(t,\vec{x}_2)\vert 0\rangle$$. If we pull out a piece of paper we can find $$\langle 0\vert\phi(t,\vec{x}_1)\phi(t,\vec{x}_2)\vert 0\rangle =\int\frac{d^3k}{(2\pi)^32k}e^{i\vec{k}\cdot(\vec{x}_1-\vec{x}_2)}\, .$$ My approch to solve this integral explicitly is to transform it into spherical coordinates. Let $$r:=\vert\vec{x}_1-\vec{x}_2\vert$$. Thus, one can find $$\langle 0\vert\phi(t,\vec{x}_1)\phi(t,\vec{x}_2)\vert 0\rangle =\int\limits_0^\infty dk\, k^2\int\limits_{-1}^1d\cos(\theta)\int\limits_0^{2\pi}d\varphi\,\frac{1}{(2\pi)^32k}e^{ikr\cos(\theta)}\\ =\frac{2\pi}{2(2\pi)^3ir}\int\limits_0^\infty dk\, (e^{ikr}-e^{-ikr})\, .\,\,\,\,\,\,\,\,\,(1)$$ Note, that the exponentials can be decomposed into a $$2i\sin(kr)$$. This integral looks divergent to me. But someone told me that I should define a new integral with a regulator $$\epsilon>0$$: $$I(\epsilon)=\int\limits_0^\infty dk\, (e^{ikr}-e^{-ikr})e^{-\epsilon k}\, .$$ This integral can be calculated rather easily. Now I can take the limit $$\epsilon\rightarrow 0$$ and see that $$I(0)$$ converges and hence the two-point correlator from above converges.

My questions are:

• Why is the interchange of limits possible such that I can even use $$I(\epsilon\rightarrow 0)$$ to express the two-point correlator?
• Is this integral still a Riemann-integral? Since we made, in at least my eyes, the divergent integral in eq. (1) converge somehow.

Your original integral diverges, and hence the equation defining $$\phi(x,t)$$ is not well defined as written. As you've shown, there's simply no way around that. Since your result when considering $$I(\epsilon)$$ is finite, it simply cant be equal to what you get if you use your original expression for $$\phi$$.

I believe that the correct way to think about it is to simply go back and re-define what you mean by the field $$\phi$$. That is, the real definition of the $$\phi$$ is

$$\phi(t,\vec{x})= \lim_{\epsilon\to 0} \int\frac{d^3k}{\sqrt{(2\pi)^32k}}\left(a(\vec{k})e^{-ikx}+a^\dagger(\vec{k})e^{ikx}\right)e^{-\epsilon k/2}.$$

This works since, if the integral were actually convergent and sufficiently well behaved, then you could exchange the integral and the limit and do away with the $$\epsilon$$ from the start—leaving you with an answer that would agree with the original definition. In the case that it diverges, then you've worked your exponential regulator into the theory.

• Sorry but this is wrong. OP has not specified the decay rate of $a(\vec k)$ so there is no way to check whether the original integral diverges or not. The correct answer is that $\phi$ is a distribution and therefore the integrals are just a schematic notation. The two-point function is only defined when evaluated on suitable test functions, in which case the regularization is justified by the fact that the integral is absolutely convergent. Nov 16, 2021 at 21:45
• I agree on that @AccidentalFourierTransform - the scalar field is indeed a distribution. Just to clear thing up: Equation (1) is a distribution and thus the integral is not a Riemann-integral. So we have to work with test functions, which is the regularization $e^{-\epsilon k}$ in this case. Then I can calculate $I(\epsilon\rightarrow 0)$. Formally, this can be used to express the original "integral" in eq. (1). Right? But one could have chosen any other suitable test function and have to obtain the same result? Nov 17, 2021 at 8:09

You have actually obtained the right answer in $$(1)$$ which is $$\langle0|\phi(x_1)\phi(x_2)|0\rangle=\frac{1}{8\pi |x_1-x_2|}\big[\delta(|x_1-x_2|)-\delta(|x_2-x_1|)\big]$$ Schwartz $$(12.75)$$. Looking at above expression you can see the correlator function is not analytical so special care has to be taken while using a regularization scheme. The way you have deformed UV region of your theory you obtain something like $$\frac{1}{ir\pm\epsilon}$$, since $$\epsilon\to0$$ any multiple of $$r$$ with it can be redfined to $$\epsilon$$, which have delta function hidden inside them.

You could have gone to another regularization scheme $$m\to0$$ where you obtain the same issue of $$\frac{1}{r}$$ and no $$\delta(r)$$. Sadly, I couldn't find the missing piece of $$\delta(r)$$ in this calculation.

The crux lies in non-analytical behavior of the above correlator, Schwartz sec $$12.6$$.

• I can see your point. In the solution to the problem I gave above it is written that (1) ist just equal to a real number: $(1)=1/(4\pi^2r^2)$. This now confuses me since (1) is a distribution. With your "hidden delta function" (thanks for that, I didn't know that!) I obtain a different solution. So what is going on here? Nov 17, 2021 at 8:47