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In order to compute the scattering probability that two particles of type 1 (associated to $\phi_1(x)$) which come from the far past with the momenta $p_1$ and $p_2$, to scatter and evolve into two particles of type 2 (associated to $\phi_2(x)$) with the momenta $p_3$ and $p_4$, I am going to apply the momentum space scattering rules, I'm just first of all trying to establish the number of distinct Feynman diagrams of $$\langle0|T( \phi_1(x_1)\phi_1(x_2)\phi_2(x_3)\phi_2(x_4))|0\rangle,$$ where the interacting Hamiltonian is $$H_\mathrm{int}= \frac g4 \phi_1^{2}(x)\phi_2^{2}(x)$$

QUESTION:

The solution lists these two Feynman diagrams only, see attachment 'dia 1'

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But I am also getting this diagram, see attachment 'dia 2'

enter image description here

Where solid lines are associated to $\phi_1(x)$, dotted to $\phi_2(x)$, and $z$ and $w$ are the internal variables I am integrating over.

Is there a reason this should be excluded? I know that 'bubble/vacuum' diagrams are excluded - disconnected diagrams with no connection to external points, but I'm unsure here since it's connected...


On another note, my lecture notes describe the contribution from the $g^{0}$ term as 'trivial' since it describes non-interacting particles. so yes in this case it's $x_1, x_2$ and $x_3, x_4$ contracted since you can not contract different fields, however, what is meant by 'trivial' in this sense? Because I know that diagrams that have no connections to external points are not included since they are vacuum contributions and will just cancel out anyway, however here it is a diagram solely between external points, so whilst it doesn't describe any interaction, it still needs to be included right?

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  • $\begingroup$ You might have seen that the S-matrix can be decomposed as $S=1+iT$. In the trivial diagram, the momenta of the particles are unchanged. Of course that diagram does exist, but it doesn't represent scattering, and is captured by the "1" in the S-matrix. The other diagrams are part of the $T$ and are what you are primarily interested in, for example, for computing the cross section. $\endgroup$
    – kaylimekay
    Dec 29 '20 at 4:18
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Without knowing the text you are referencing, it's possible that they are only considering amputated diagrams, since the third diagram is a self-energy correction to the 4-point contact diagram. If you consider amputated diagrams, you only look at the diagrams close to the interaction vertex, and you effectively discard the self energy contribution, since it could happen arbitrarily far in the future of particle $x_2$ and not around the interaction point.

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  • $\begingroup$ Okay thank you. We have just been introduced to these diagrams and are asked to draw the diagrams from the momentum space scatter feynman diagram rules, one of which is to amputate external legs- move propagators to external points. So I think this rule is saying to only consider amputated diagrams is this is consistent with what you are saying. I don't really understand what an amputated diagram is and why this is the same as 'move propagators to external points' , have you got any further information or good links on this? Many thanks. $\endgroup$ Jan 6 '17 at 20:09
  • $\begingroup$ An amputated diagram is just one where you only care about particle interactive at a particular point in space, and not the long term future/past of each particle. In your example, the correction to the propagator of x2 could happen immediately after the initial interaction, or it could happen in 100 years, so we restrict to cases of definite interaction. This type of diagram is known as 1 particle reducible, the others being 1 particle irreducible - you ought to look them up. See here bit.ly/2i1ZbbP $\endgroup$
    – Akoben
    Jan 8 '17 at 5:59
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Un-amputated diagrams contribute to the correlation function, however once you feed the correlation functions into the LSZ theorem the external leg corrections become (mostly) irrelevant. Only the location of their one-particle poles, and the residues at these poles matters and is taken into account in the LSZ theorem by using the physical pole mass and subtracting through the residues.

To be more precise, since we are taking the limit where external particles go on-shell, that is $p^2=m^2$, it is only important to know how the external leg corrections behave in this limit. If you sum up all the corrections (formally), you get the expression $$\frac{i}{p^2 - m^2 + \Sigma(p^2)}.$$ On-shell renormalization will demand that $\Sigma(m^2)=0$, and thus the resummed external leg correction has a simple pole at $p^2=m^2$ (we're treating $p^2$ as its own complex variable here). When you feed this into the LSZ, you get a calculation that looks like $$\lim_{p^2->m^2}(p^2-m^2)\frac{i}{p^2 - m^2 + \Sigma(p^2)}=i$$ And so you see that the detailed momentum dependency of the eternal leg corrections does not matter to the scattering amplitude. Note however, that when the self-energy diagram is placed on an internal line, its momentum depedency DOES affect the scattering amplitude. This is what leads to vacuum polarization and other quantum corrections to the semi-classical scattering potential.

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