How can we reconcile Zee's claims about QFT integration on page 26?

[jdstokes]

Homework Statement

In Zee's book on QFT, I'm confused on page 26 by how we gets from Eq (4)

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

to Eq (5).

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

The Attempt at a Solution

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

Zee claims that ``Integrating over y^0 we get a delta function setting k^0 to zero''.

Firstly, I don't see why we may assume $k^0 \to 0$, and even if it does, this surely gives $W(J) \to \iint dx^0dy^0 \int \frac{dk^0}{2\pi}\int \frac{d^3 k}{(2\pi)^3}\frac{e^{i \vec{k}(\vec{x}_1 - \vec{x}_2)}}{\vec{k}^2 + m^2}$

Is he trying to suggest that the term $(k^0)^2$ in the denominator of the k-integrand is somehow negligible compared to $e^{-ik^0(x^0-y^0)}$? In this case

$\int \frac{dk^0}{2\pi} e^{ik^0(x-y)^0} = \delta(y^0 - x^0)$. Doing the y^0 integration then simply gives the area under the delta function, which is $\sqrt{2\pi}$ if I recall correctly... No good.


On the same page, I also don't see what enables us to write $\langle 0 | e^{-i Ht}| 0 \rangle$ in the form $e^{-iEt}$.

 

[Sojourner01]

I almost agree with your derivation, except for a stray minus sign in the denominator of the last integral, and the fact that the factor $$i\varepsilon$$ vanishes.

As for the bra-ket - some derivations allow you to replace an operator with its generalised eigenvalue; that of the hamiltonian H obviously being E. I'm not quite sure what permits this.

 

[Rainbow Child]

From

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

first you do the $$y^0$$ integration as

$\begin{multline*}W(J) = - \int dx^0 dk^0 e^{ik^0x^0}\int \frac{d^3k}{(2\pi)^3}\frac{e^{i\vec{k} \cdot(\vec{x}_1 - \vec{x}_2)}}{(k^0)^2 - \vec{k}^2 - m^2 + i\varepsilon}\int \frac{dy^0}{2\pi}e^{-ik^0y^0} = - \int dx^0 dk^0 e^{ik^0x^0}\int \frac{d^3k}{(2\pi)^3}\frac{e^{i\vec{k} \cdot(\vec{x}_1 - \vec{x}_2)}}{(k^0)^2 - \vec{k}^2 - m^2 + i\varepsilon}\delta(k^0-0)\\=- \int dx^0 \int \frac{d^3k}{(2\pi)^3}\frac{e^{i\vec{k} \cdot(\vec{x}_1 - \vec{x}_2)}}{- \vec{k}^2 - m^2 + i\varepsilon}= \int dx^0 \int \frac{d^3k}{(2\pi)^3}\frac{e^{i\vec{k} \cdot(\vec{x}_1 - \vec{x}_2)}}{\vec{k}^2 + m^2 - i\varepsilon}\end{multline*}$

from

$$\int\frac{d\,k^0}{2\,\pi}\,\delta(k^0-0)\,f(k^0)=f(0)$$

 

[jdstokes]

Perfect, Rainbow Child. Thanks.