A few questions about QFT

  • Thread starter jdstokes
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  • #1
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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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Answers and Replies

  • #2
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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 [tex]i\varepsilon[/tex] 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.
 
  • #3
From

[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]

first you do the [tex]y^0[/tex] integration as

[itex]\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*}[/itex]

from

[tex] \int\frac{d\,k^0}{2\,\pi}\,\delta(k^0-0)\,f(k^0)=f(0)[/tex]
 
  • #4
523
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Perfect, Rainbow Child. Thanks.
 

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