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## Homework Statement

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

[itex]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}[/itex]

to Eq (5).

[itex]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 +}[/itex]

## The Attempt at a Solution

[itex]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}[/itex]

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 [itex]k^0 \to 0[/itex], and even if it does, this surely gives [itex]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}[/itex]

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

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

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

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