Computing a Feynman integral

Question

Reading the lectures notes by A. Connes and M. Marcolli I have some difficulty undestanding how they compute the Feynman integrals using integration by parts. Consider the following integral of formulas (1.32), (1.34): $$ I = \mathcal N_0∫ φ(x)φ(y) \, \exp(iS_0(φ)) \, \mathcal D[φ], $$ where $\mathcal N_0$ is a normalization factor $$ \mathcal N_0^{-1} = ∫ \exp(iS_0(φ)) \, \mathcal D[φ], $$ and $$ S_O(φ) = (2\pi)^{-D} ∫ \frac 1 2 (p^2 - m^2) \hat φ(p) \hat φ(-p) \, d^D p, \\ \hat φ(p) = ∫ \exp(-i p x) φ(x) \, d^D x. $$ The authors say that using formal integration by parts (note that the the integrand is a product of two linear forms in the fields times the exponent of a quadratic form in the fields) one gets something like: $$ I = i (2\pi)^D ∫ \frac{\delta(p_1+p_2)}{p_1^2-m^2} e^{\pm ip(x-y)} d^D p_1 d^D p_2 $$ I do not understand this computation. Could you please explain it?

Answer 1

Usually integration by parts under a Gaussian measure refers to the following formula. Let $X$ and $X_i$ be Gaussians. Then $$ \mathbb{E}[Xf(X_1,...,X_n)] = \sum_{i=1}^n\mathbb{E}[X X_i] \mathbb{E}[\frac{\partial}{\partial X_i} f(X_1,...,X_n)]\,. $$ For continuum fields this becomes $$ \int \varphi(x) f(\varphi) e^{i S(\varphi)} D \varphi =\int dz \int \varphi(x) \varphi(z) e^{i S(\varphi)} D \varphi \int \frac{\delta f(\varphi)}{\delta \varphi(z)} e^{iS(\varphi)} D \varphi\,, $$ where $e^{iS(\varphi)} D \varphi$ denotes an infinite dimensional Gaussian measure. Your example corresponds to $f(\varphi) = \varphi(y)$, and the implications of the above formula are rather trivial in this example, thus I find the choice of words in the lecture notes a bit confusing.

In the end the answer to your question depends on how you define your "Gaussian" measure $\mathcal{N}_0 \exp{iS_0(\varphi)} \mathcal{D}[\varphi]$. Formally, this is a Gaussian with the covariance operator $i(\Delta-m^2)^{-1}$ (up to some multiplicative constant, I guess). What is the two point function (the integral $I$) of this?

As an analogy, consider the discrete case, that is, let $X$ be a Gaussian vector in $\mathbb{R}^n$ with covariance matrix $A$. Then \begin{align} \mathbb{E}[X_i X_j] &= \int x_i x_j e^{-\frac{1}{2} (x,A^{-1} x)} d^n x \\ &= \frac{\partial^2}{\partial J_i \partial J_j} \Big|_{J=0} \int e^{-\frac{1}{2} (x,A^{-1} x) + J \cdot x} d^nx\,. \end{align} Thus we can express the two-point function in terms of derivatives of the moment generating function, which can be explicitly computed for Gaussian measures. In the above example, the end result should be $A_{ij}$.

Similarly, in the case of fields, we have \begin{align} \mathcal{N}_0 \int \varphi(x) \varphi(y)\exp{iS_0(\varphi)} \mathcal{D}[\varphi] &= \frac{\delta^2}{\delta J(x) J(y)}\Big|_{J=0} \mathcal{N}_0 \int e^{\int J(z) \varphi(z) dz} \exp{iS_0(\varphi)} \mathcal{D}[\varphi]\,, \end{align} and the rest of the computation are completely analogous to the discrete case. This time the end result is (formally) $$ (\delta_y, i(\Delta-m^2)^{-1} \delta_x)_{L^2} = (\mathcal{F}{\delta_y}, i \mathcal{F}{(\Delta-m^2)^{-1} \delta_x})_{L^2} $$ where $\mathcal{F}$ is the Fourier transform. This simplifies to the form $$ i\int e^{-ik \cdot y} \frac{e^{ik\cdot x}}{|k|^2-m^2} dk $$ This can be rewritten as $$ i \int \frac{\delta(k+k')}{|k|^2-m^2}e^{i(k\cdot x + k'\cdot y)} dkdk' \,, $$ if you prefer.