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 $(2)$?


1 Answer 1


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

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

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

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

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

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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.