Consider the following Feynman diagram: enter image description here

I've read that it will have associated with it one integral over the loop. The issue is, in Schwartz book the Feynman rules for momentum space are:

  1. Internal lines get propagators;
  2. Vertices get factors of $i\lambda$;
  3. Lines connected to external points get nothing;
  4. Momentum is conserved at each vertex;
  5. Integrate over undetermined 4-momenta;

Now in this diagram I let $p_1$ be the incoming momentum on the left and $p_2$ the outgoing momentum on the right.

If in the loop I assign a counterclockwise momentum $k$, we must have at the vertex $p_1 - p_2 - k = 0$ so that $k = p_1-p_2$. In that case, $k$ is not undetermined.

I would guess by these rules that the amplitude is

$$i\mathcal{M}= (i\lambda)\dfrac{i}{k^2-m^2+i\epsilon}=(i\lambda) \dfrac{i}{(p_1-p_2)^2-m^2+i\epsilon}$$

but searching the internet it seems that the correct form would be

$$i\mathcal{M}=(i\lambda)\int \dfrac{i}{k^2-m^2+i\epsilon}d^4k,$$

but $k$ is not undetermined momentum, so why integrate over it following this set of rules?

What is actually the right viewpoint on this diagram?


1 Answer 1


The momentum conservation at the vertex is NOT $p_1-p_2-k=0$ but rather $p_1-p_2-k+k=0$ which leaves $k$ completely free. The vertex has two half-edges carrying the momentum $k$ but in opposite directions.

  • $\begingroup$ So whenever I see a line that starts and end at the same point, I should consider it really as two lines, one starting at the point and other ending at the point? This confuses me because the line gets just a single propagator, so I thought it was just a single line. $\endgroup$
    – Gold
    May 17, 2017 at 14:30
  • 1
    $\begingroup$ Yes you have to think like a vertex. The latter is where the momentum conservation lives and the vertex doesn't know about the rest of the graph. $\endgroup$ May 17, 2017 at 14:37
  • $\begingroup$ I get your point now. The issue isn't with the line, that gets a single propagator really, but the fact that in the vertex it "sees" the one line going with momentum $k$ and another coming with momentum $-k$, without any sensitivity to whether they are the same line or not, right? $\endgroup$
    – Gold
    May 17, 2017 at 14:55
  • 1
    $\begingroup$ yes that's right. $\endgroup$ May 17, 2017 at 15:12

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