QFT in a nutshell: From field to particle

[jdstokes]

Homework Statement

I don't understand how Zee gets Eq. (2) on p. 24:

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

Homework Equations

$W(J) := - \frac{1}{2}\int d^4x\int d^4y J(x)D(x-y)J(y)$

The Attempt at a Solution

I don't see where the $d^4k$ and factors of 2pi are coming from. Manipulating the definition of W(J) I was able to show that

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

but this obviously doesn't agree.

Thanks.

 

[CompuChip]

I see you have a power E^{ikx} but I don't see a k integral... the idea is to substitute J by its Fourier transform:
$$J(x) = \int \frac{d^4 k}{(2\pi)^4} e^{i k x} J(k)$$
and
$$J(y)^* = \int \frac{d^4 k'}{(2\pi)^4} e^{-i k' x} J(k')^*$$
Now combine the exponentials and perform the x-integration, this will cancel out one of the k integrals
$$\int dx e^{i (k - k') x} = \delta(k - k')$$
and then do the manipulations you did to get the $$\left(k^2 - m^2 + i\epsilon\right)^{-1}$$ and the other stuff right.

 

[jdstokes]

Hi CompuChip,

Thanks for replying.

Why are you taking the complex conjugate of J(x)? The definition of W(J) is

$W(J) := - \frac{1}{2}\int d^4x\int d^4y J(x)D(x-y)J(y)$

No complex conjugates there. My approach was to separate the $e^{ik(x-y)}$ term in the expression for the propagation D(x - y) giving

$W(J) := - \frac{1}{2}\int d^4x\int d^4y J(x)D(x-y)J(y)$
$W(J) := - \frac{1}{2}\int d^4x\int d^4y J(x)\frac{e^{ik(x-y)}}{k^2-m^2+i\epsilon}J(y)$
$W(J) := - \frac{1}{2}\int d^4x\int d^4y e^{ikx}J(x)\frac{1}{k^2-m^2+i\epsilon}e^{-iky}J(y)$
$W(J) := - \frac{1}{2}\left(\int d^4x e^{ikx}J(x)\right)\frac{1}{k^2-m^2+i\epsilon}\left(\int d^4ye^{-iky}J(y)\right)$
$W(J) := - \frac{1}{2}J(-k)\frac{1}{k^2-m^2+i\epsilon}J(k)$.

Note quite right.

 

[jdstokes]

Omg I'm so dumb! I forgot to put the integral in the propagator. Problem solved now (I think). If J is complex, why is it $J^\ast(k)$ and not J(-k)?

 

[CompuChip]

Ah, my mistake. You get a $$\delta(k + k')$$ from the (real)space-integral hence J(k) and J(-k).
I might still be a little off here, but the general idea was as follows: J(x) is real. This implies that
$$J(x) = \sum_{k = -\infty}^\infty e^{ikx} J(k) <br /> \stackrel{!}{=}<br /> \sum_{k = -\infty}^\infty e^{-ikx} J(k)^* = J(x)^*$$
so comparing the coefficients of the (basis) exponentials, we conclude that $$J(k) = J(-k)^*$$, therefore we can write J(k)* instead of J(-k).