Consider the following one-loop diagram:

enter image description here

The corresponding expression in momentum space is:

\begin{equation} (-i\lambda)\int \dfrac{i}{k^2-m^2+i\epsilon}\frac{d^4k}{(2\pi)^{2}} \end{equation}

(Ignoring symmetry factors).

My attempt:

I'm using the notations (metric signature, Fourier transform..) according to the textbook by P&S. Starting from Position space:

\begin{align} (-i\lambda)\langle 0 | \int d^{4}z D_{F}(x-z)D_{F}(z-z)D_{F}(z-y)| 0 \rangle, \end{align} where $x$ is first point, $y$ the second and $z$ the position of the vertix.

\begin{align} (-i\lambda)\langle 0 | \int d^{4}z \int \frac{d^{4}p_{1}}{(2\pi)^4} \dfrac{ie^{-i p_{1}\cdot(x-z)} }{p_{1}^2-m^2+i\epsilon}\int \frac{d^{4}k}{(2\pi)^4}\dfrac{i}{k^2-m^2+i\epsilon} \int \frac{d^{4}p_{2}}{(2\pi)^4} \dfrac{ie^{-i p_{2}\cdot(z-y)} }{p_{2}^2-m^2+i\epsilon} |0 \rangle \end{align}

Making use of the Fourier transform:

\begin{align} \int dz^{4}e^{-iz \cdot (p_{1}-p_{2})}=(2\pi)^4 \delta^{4}(p_{1}-p_{2}), \end{align} we find \begin{align} (-i\lambda)i^{2}\int \frac{d^{4}p_{1}}{(2\pi)^4} \dfrac{e^{-i p_{1}\cdot(x-y)} }{(p_{1}^2-m^2+i\epsilon)^{2}}\int \frac{d^{4}k}{(2\pi)^4}\dfrac{i}{k^2-m^2+i\epsilon}. \end{align} How to proceed to find the above expression?

Note, at the vertix 4-momentum is conserved. We can define $p_{1}+k=p_{2}+k$ or even naming $k_{1}, k_{2}$ for incoming\outgoing momenta for the propagator but that's al the same as $\delta^{4}(p_{1}-p_{2})$

  • 3
    $\begingroup$ What exactly are you trying to accomplish? It seems to me the 'expression in momentum space' you have written is amputated (i.e. the propagators hanging off either end of the diagram are removed), whereas the expression you have in position space is not amputated, which is why you get the additional propagators on either end $\endgroup$ Jun 22, 2022 at 15:37
  • $\begingroup$ Amputated? Isn't it the correct expression according to the Feynmann rules in momentum space?(1. Internal lines get propagators 2. Vertices get factors of iλ 3. Lines connected to external points get nothing4. Momentum is conserved at each vertex. 5. Integrate over undetermined 4-momenta) I'm trying to get an expression in position space with an exponential to recognize the expression in moment space. $\endgroup$
    – M91
    Jun 22, 2022 at 15:45
  • $\begingroup$ You can't integrate the same variable two times (p1), it might be a typo, check the third integral in the long expression, should it be dp2 and exp^p2? $\endgroup$ Jun 22, 2022 at 15:50
  • $\begingroup$ @guiablo, you're right. It is now edited. I integrate once over $p_{2}$ when evaluating the delta function $\endgroup$
    – M91
    Jun 22, 2022 at 16:00
  • 1
    $\begingroup$ "3. Lines connected to external points get nothing" You're following the Feynman rules for amputated diagrams with this statement. In contrast, you have included propagators connected to external points in your position-space calculation. That's why they're different. $\endgroup$ Jun 23, 2022 at 4:42


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