I've always thought that the integration notation in physics is weird, but I understood it nevertheless for a single variable, until I started reading Zee's QFT in a nutshell, where 4 dimensions are used. So I was wondering how the integration notation works.

For example, for a free theory with sources at $x_1$ and $x_2$ we have: $$ W(J)= -\frac{1}{2}\int\frac{d^4k}{(2\pi)^4}J_2(k)^*\frac{1}{k^2-m^2+i\epsilon}J_1(k) $$ Now we let: $$ J_a(x)=\delta^{(3)}(\vec{x}-\vec{x_a}) $$ where $a=1,2$. Using the Fourier transform into momentum space: $$ J(k)=\int d^4xe^{-ik\cdot x}J(x) $$ What I don't understand is that W(J) becomes: $$ W(J)=-\int\int dx^0dy^0\int \frac{dk^0}{2\pi}e^{ik^0(x^0-y^0)}\int\frac{d^3k}{(2\pi)^3}\frac{e^{ik\cdot(x_1-x_2)}}{k^2-m^2+i\epsilon} =\int dx^0\int\frac{d^3k}{(2\pi)^3}\frac{e^{ik\cdot(x_1-x_2)}}{k^2+m^2} $$ I know that the $2$ comes from the two terms being equal so don't mind that.

It says integrating over $y^0$ will get a delta function setting $k^0$ to zero, and that I don't understand.

The only thing I know (and I'm guessing but I'm not sure yet) is that: $$ J_a(k)=\int d^4xe^{-ik\cdot x}\delta^{(3)}(x-x_a)=\int dx^0e^{-ik^0\cdot x^0}\int d^3xe^{-ik\cdot x}\delta^{(3)}(x-x_a)= \int dx^0e^{-ik^0\cdot x^0}e^{-ik\cdot x_a} $$ and the integral over $x^0$ (or $y^0$) is independent of the spatial $k$ so it's outside the integral of spatial $k$, but inside the integral of temporal $k$ in the $W(J)$. And the one with the $x_a$ part is inside the spatial integral $k$. Am I correct about this?

But yeah, my questions are:

  • What's with the integral over $y^0$?

  • Why is there even a $y$? Is it equal to $x_2$?

  • How did the sign changed for the $m^2$ part?

  • How do you actually know what's inside the integral using this notation? Or actually how to read integrals like this?

  • $\begingroup$ 1, 2) You need to Fourier transform both factors of $J$ in $W$. For the other one, use integration variable $y$. 3) Note that $k^2 = (k^0)^2 - \vec{k}^2$, causing a sign flip. 4) I'm not sure what you're asking here. What's inside an integral is pretty clear, it's everything to the right of the integral sign. $\endgroup$
    – knzhou
    Commented Mar 30, 2018 at 22:19

1 Answer 1


$$ J_a(k) = \int dx^0 e^{-ik^0x^0} \int d^3x e^{i\vec{k}\cdot\vec{x}}\delta^3(\vec{x}-\vec{x}_a) = e^{i\vec{k}\cdot\vec{x}_a}\int dx^0 e^{-ik^0x^0} $$

$$ J^{*}_a(k) = \int dy^0 e^{ik^0y^0} \int d^3y e^{-i\vec{k}\cdot\vec{y}}\delta^3(\vec{y}-\vec{x}_a) = e^{-i\vec{k}\cdot\vec{x}_a}\int dy^0 e^{ik^0y^0} $$

So you have

$$ J^*_2(k)J_1(k) = e^{i\vec{k}\cdot(\vec{x}_1-\vec{x_2})} \int dx^0 \int dy^0 e^{ik^0(y^0-x^0)} $$

Now let me work with the term $k^2-m^2$:

$$k^2-m^2 = (k^0)^2-\vec{k}^2-m^2 = (k^0)^2-(\vec{k}^2+m^2) $$

Now let's talk about $\int \frac{dy^0}{2\pi} e^{ik^0(x^0-y^0)}$: by changing the integration variable $y^0-x^0 = \tilde{y}^0$ (so that $dy^0 = d\tilde{y}^0$), you get $\int \frac{d\tilde{y}^0}{2\pi} e^{i\tilde{y}^0k^0} $ which is the one dimensional Dirac Delta (you can quickly Google it): $\delta(k^0-0)$. Now taking the integral over $dk^0$ simply means setting $k^0=0$ since you have this Dirac Delta function. The sign change comes as a consequence since you have $-(\vec{k}^2+m^2)$ at the denominator and you finally get:

$$ W[J] = \int dx^0 \int \frac{d^3k}{(2\pi)^3} \frac{e^{i\vec{k}\cdot(\vec{x}_1-\vec{x}_2)}}{\vec{k}^2+m^2}$$

I hope this helps!


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