# Fourier transform for $W(J)$ in a free QFT

In chapter I.3 and I.4 of A. Zee's QFT in a Nutshell, he starts with the theory

$$\mathcal{L}(\varphi) = \frac 12[(\partial\phi)^2-m^2\varphi^2]+J\varphi$$

Using the path integral approach, he obtains the amplitude

$$\tag{1}W(J) = -\frac 12\int\int d^4xd^4yJ(x)D(x-y)J(y)$$

Then he applies the Fourier transform $J(k) = \int d^4xe^{-ikx}J(x)$ to transform $(1)$ into

$$\tag{2}W(J) = -\frac 12\int\frac{d^4k}{(2\pi)^4}J^*(k)\frac{1}{k^2-m^2+i\epsilon}J(k)$$

The only hint he gives is $J(k)* = J(-k)$ because $J(x)$ is real.

I don't understand how he "gets rid" of one of the integrals. There are two fourier transforms involved, which should give us two additional integrals on top of the two integrals of $(1)$. I can't see how to reduce them to one integral.

So, how can I get from $(1)$ to $(2)$?

In the following calculation, I ignore some coefficients.

According to $$J(x)=\int d^4 k_1 e^{ik_1 x}$$ , $$J(y)=\int d^4 k_2 e^{ik_2 y}$$ and $$D(x-y)=\int d^4k \frac{e^{ik(x-y)}}{k^2-m^2+i\epsilon}$$

We have

$$W(J) = \int d^4x d^4y d^4 k d^4 k_1 d^4 k_2 J(k_1)e^{ik_1 x} J(k_2)e^{ik_2 x} \frac{e^{ik(x-y)}}{k^2-m^2+i\epsilon}$$

$$W(J)=\int d^4x d^4y d^4 k d^4 k_1 d^4 k_2 J(k_1) J(k_2) \frac{1}{k^2-m^2+i\epsilon} e^{i(k_1 +k)x} e^{i(k_2 -k)y}$$

$$W(J)=\int d^4 k d^4 k_1 d^4 k_2 J(k_1) J(k_2) \delta(k_1 +k) \delta(k_2 -k)\frac{1}{k^2-m^2+i\epsilon}$$

$$W(J)=\int d^4 k J(-k) \frac{1}{k^2-m^2+i\epsilon} J(k)$$

$$W(J)=\int d^4 k J^{\ast}(k) \frac{1}{k^2-m^2+i\epsilon} J(k)$$