# Why this loop carries an integral if there is no undetermined momenta?

Consider the following Feynman diagram: 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?

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.
• 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? – user1620696 May 17 '17 at 14:55